Valid term from quadratic sequence?Fibonacci function or sequenceSylvester's sequenceThe Squaring SequenceYet Unused PairsThe lowest initial numbers in a Fibonacci-like sequenceReconstruct an arithmetic sequenceCollection from a sequence that constitute a perfect squareFind Integral Roots of A PolynomialGenerate lowest degree polynomial from sequenceThe Written Digits Sequence

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A Journey Through Space and Time



Valid term from quadratic sequence?


Fibonacci function or sequenceSylvester's sequenceThe Squaring SequenceYet Unused PairsThe lowest initial numbers in a Fibonacci-like sequenceReconstruct an arithmetic sequenceCollection from a sequence that constitute a perfect squareFind Integral Roots of A PolynomialGenerate lowest degree polynomial from sequenceThe Written Digits Sequence













10












$begingroup$


You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:



$$T_n=an^2+bn+c$$



You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.



This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.



Test cases



a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>
4 |8 |5 |2 |N









share|improve this question











$endgroup$
















    10












    $begingroup$


    You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:



    $$T_n=an^2+bn+c$$



    You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.



    This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.



    Test cases



    a |b |c |T_n |Y/N
    ------------------------
    1 |1 |1 |1 |Y #n=0
    2 |3 |5 |2 |N
    0.5 |1 |-2 |-0.5|Y #n=1
    0.5 |1 |-2 |15.5|Y #n=5
    0.5 |1 |-2 |3 |N
    -3.5|2 |-6 |-934|Y #n=-16
    0 |1 |4 |7 |Y #n=3
    0 |3 |-1 |7 |N
    0 |0 |0 |1 |N
    0 |0 |6 |6 |Y #n=<anything>
    4 |8 |5 |2 |N









    share|improve this question











    $endgroup$














      10












      10








      10


      0



      $begingroup$


      You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:



      $$T_n=an^2+bn+c$$



      You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.



      This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.



      Test cases



      a |b |c |T_n |Y/N
      ------------------------
      1 |1 |1 |1 |Y #n=0
      2 |3 |5 |2 |N
      0.5 |1 |-2 |-0.5|Y #n=1
      0.5 |1 |-2 |15.5|Y #n=5
      0.5 |1 |-2 |3 |N
      -3.5|2 |-6 |-934|Y #n=-16
      0 |1 |4 |7 |Y #n=3
      0 |3 |-1 |7 |N
      0 |0 |0 |1 |N
      0 |0 |6 |6 |Y #n=<anything>
      4 |8 |5 |2 |N









      share|improve this question











      $endgroup$




      You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:



      $$T_n=an^2+bn+c$$



      You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.



      This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.



      Test cases



      a |b |c |T_n |Y/N
      ------------------------
      1 |1 |1 |1 |Y #n=0
      2 |3 |5 |2 |N
      0.5 |1 |-2 |-0.5|Y #n=1
      0.5 |1 |-2 |15.5|Y #n=5
      0.5 |1 |-2 |3 |N
      -3.5|2 |-6 |-934|Y #n=-16
      0 |1 |4 |7 |Y #n=3
      0 |3 |-1 |7 |N
      0 |0 |0 |1 |N
      0 |0 |6 |6 |Y #n=<anything>
      4 |8 |5 |2 |N






      code-golf number decision-problem equation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 4 at 17:13







      Artemis Fowl

















      asked Apr 3 at 14:19









      Artemis FowlArtemis Fowl

      23610




      23610




















          5 Answers
          5






          active

          oldest

          votes


















          6












          $begingroup$

          JavaScript (ES7), 70 bytes



          Returns a Boolean value.





          (a,b,c,t)=>(t-=c,(a*=2)?(x=(b*b+2*a*t)**.5-b)%a&&(x+b+b)%a:b?t%b:t)==0


          Try it online!



          How?



          For the sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)



          Case $aneq0$



          The equation really is quadratic:



          $$T_n=an^2+bn+c\
          an^2+bn-d=0$$



          With $a'=2a$, the discriminant is:



          $$Delta=b^2+2a'd$$



          and the roots are:



          $$n_0=frac-b-sqrtDeltaa'\
          n_1=frac-b+sqrtDeltaa'$$



          The equation admits an integer root if $sqrtDelta$ is an integer and either:



          $$beginalign&-b-sqrtDeltaequiv 0pmoda'\ text or &-b+sqrtDeltaequiv 0pmoda'endalign$$



          Case $a=0, bneq0$



          The equation is linear:



          $$T_n=bn+c\
          bn=d\
          n=fracdb$$



          It admits an integer root if $dequiv0pmod b$.



          Case $a=0, b=0$



          The equation is not depending on $n$ anymore:



          $$T_n=c\
          d=0$$






          share|improve this answer











          $endgroup$




















            4












            $begingroup$


            Jelly,  11  10 bytes



            _/Ær1Ẹ?%1Ạ


            A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.



            * admittedly taking a little liberty with "You may take input of these four numbers in any way".



            Try it online! Or see a test-suite.



            How?



            _/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
            / - reduce by:
            _ - subtraction [c-T_n, b, a]
            ? - if...
            Ẹ - ...condition: any?
            Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
            1 - ...else: literal 1
            %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
            - note: (a+bi)%1 yields nan which is truthy
            Ạ - all? i.e. all had fractional parts?
            - note: all([]) yields 1



            If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)






            share|improve this answer











            $endgroup$








            • 2




              $begingroup$
              That input is perfectly fine.
              $endgroup$
              – Artemis Fowl
              Apr 3 at 18:21


















            1












            $begingroup$


            05AB1E, 35 bytes



            Æ©²Āi²4P³n+tÐdi(‚³-IJ·Ä%P}뮳Āi³%]_


            Port of @Arnauld's JavaScript answer, so make sure to upvote him!



            Takes the input in the format $[t,c], a, b$.



            Try it online



            Explanation:





            Æ # Reduce the (implicit) input-list by subtraction (`t-c`)
            © # Store this value in the register (without popping)
            ²Āi # If the second input `a` is not 0:
            ²4P # Calculate `(t-c)*a*4`
            ³n+ # Add the third input `b` squared to it: `(t-c)*a*4+b*b`
            t # Take the square-root of that
            # (NOTE: 05AB1E and JS behave differently for square-roots of
            # negative integers; JS produces NaN, whereas 05AB1E leaves the
            # integer unchanged, which is why we have the `di...}` here)
            Ð # Triplicate this square
            di # If the square is non-negative (>= 0):
            (‚ # Pair it with its negative
            ³- # Subtract the third input `b` from each
            Ä # Take the absolute value of both
            ²·Ä% # Modulo the absolute value of `a` doubled
            # (NOTE: 05AB1E and JS behave differently for negative modulos,
            # which is why we have the two `Ä` here)
            P # Then multiply both by taking the product
            } # And close the inner if-statement
            ë # Else (`a` is 0):
            ® # Push the `t-c` from the register
            ³Āi # If the third input `b` is not 0:
            ³% # Take modulo `b`
            ] # Close both if-else statements
            _ # And check if the result is 0
            # (which is output implicitly)





            share|improve this answer









            $endgroup$












            • $begingroup$
              Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
              $endgroup$
              – Arnauld
              Apr 4 at 10:37











            • $begingroup$
              @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
              $endgroup$
              – Kevin Cruijssen
              Apr 4 at 11:58






            • 1




              $begingroup$
              Besides, the results of Ų on negative inputs are inconsistent.
              $endgroup$
              – Arnauld
              Apr 4 at 12:30










            • $begingroup$
              @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
              $endgroup$
              – Kevin Cruijssen
              Apr 4 at 12:31



















            0












            $begingroup$


            Wolfram Language (Mathematica), 38 bytes



            Solve[n^2#+n#2+#3==#4,n,Integers]!=&


            Try it online!






            share|improve this answer









            $endgroup$




















              0












              $begingroup$


              Jelly, 15 bytes



              _3¦UÆr=Ḟ$;3ị=ɗẸ


              Try it online!



              Built-in helps here but doesn’t handle a=b=0 so this is handled specially.






              share|improve this answer











              $endgroup$













                Your Answer





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                5 Answers
                5






                active

                oldest

                votes








                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                JavaScript (ES7), 70 bytes



                Returns a Boolean value.





                (a,b,c,t)=>(t-=c,(a*=2)?(x=(b*b+2*a*t)**.5-b)%a&&(x+b+b)%a:b?t%b:t)==0


                Try it online!



                How?



                For the sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)



                Case $aneq0$



                The equation really is quadratic:



                $$T_n=an^2+bn+c\
                an^2+bn-d=0$$



                With $a'=2a$, the discriminant is:



                $$Delta=b^2+2a'd$$



                and the roots are:



                $$n_0=frac-b-sqrtDeltaa'\
                n_1=frac-b+sqrtDeltaa'$$



                The equation admits an integer root if $sqrtDelta$ is an integer and either:



                $$beginalign&-b-sqrtDeltaequiv 0pmoda'\ text or &-b+sqrtDeltaequiv 0pmoda'endalign$$



                Case $a=0, bneq0$



                The equation is linear:



                $$T_n=bn+c\
                bn=d\
                n=fracdb$$



                It admits an integer root if $dequiv0pmod b$.



                Case $a=0, b=0$



                The equation is not depending on $n$ anymore:



                $$T_n=c\
                d=0$$






                share|improve this answer











                $endgroup$

















                  6












                  $begingroup$

                  JavaScript (ES7), 70 bytes



                  Returns a Boolean value.





                  (a,b,c,t)=>(t-=c,(a*=2)?(x=(b*b+2*a*t)**.5-b)%a&&(x+b+b)%a:b?t%b:t)==0


                  Try it online!



                  How?



                  For the sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)



                  Case $aneq0$



                  The equation really is quadratic:



                  $$T_n=an^2+bn+c\
                  an^2+bn-d=0$$



                  With $a'=2a$, the discriminant is:



                  $$Delta=b^2+2a'd$$



                  and the roots are:



                  $$n_0=frac-b-sqrtDeltaa'\
                  n_1=frac-b+sqrtDeltaa'$$



                  The equation admits an integer root if $sqrtDelta$ is an integer and either:



                  $$beginalign&-b-sqrtDeltaequiv 0pmoda'\ text or &-b+sqrtDeltaequiv 0pmoda'endalign$$



                  Case $a=0, bneq0$



                  The equation is linear:



                  $$T_n=bn+c\
                  bn=d\
                  n=fracdb$$



                  It admits an integer root if $dequiv0pmod b$.



                  Case $a=0, b=0$



                  The equation is not depending on $n$ anymore:



                  $$T_n=c\
                  d=0$$






                  share|improve this answer











                  $endgroup$















                    6












                    6








                    6





                    $begingroup$

                    JavaScript (ES7), 70 bytes



                    Returns a Boolean value.





                    (a,b,c,t)=>(t-=c,(a*=2)?(x=(b*b+2*a*t)**.5-b)%a&&(x+b+b)%a:b?t%b:t)==0


                    Try it online!



                    How?



                    For the sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)



                    Case $aneq0$



                    The equation really is quadratic:



                    $$T_n=an^2+bn+c\
                    an^2+bn-d=0$$



                    With $a'=2a$, the discriminant is:



                    $$Delta=b^2+2a'd$$



                    and the roots are:



                    $$n_0=frac-b-sqrtDeltaa'\
                    n_1=frac-b+sqrtDeltaa'$$



                    The equation admits an integer root if $sqrtDelta$ is an integer and either:



                    $$beginalign&-b-sqrtDeltaequiv 0pmoda'\ text or &-b+sqrtDeltaequiv 0pmoda'endalign$$



                    Case $a=0, bneq0$



                    The equation is linear:



                    $$T_n=bn+c\
                    bn=d\
                    n=fracdb$$



                    It admits an integer root if $dequiv0pmod b$.



                    Case $a=0, b=0$



                    The equation is not depending on $n$ anymore:



                    $$T_n=c\
                    d=0$$






                    share|improve this answer











                    $endgroup$



                    JavaScript (ES7), 70 bytes



                    Returns a Boolean value.





                    (a,b,c,t)=>(t-=c,(a*=2)?(x=(b*b+2*a*t)**.5-b)%a&&(x+b+b)%a:b?t%b:t)==0


                    Try it online!



                    How?



                    For the sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)



                    Case $aneq0$



                    The equation really is quadratic:



                    $$T_n=an^2+bn+c\
                    an^2+bn-d=0$$



                    With $a'=2a$, the discriminant is:



                    $$Delta=b^2+2a'd$$



                    and the roots are:



                    $$n_0=frac-b-sqrtDeltaa'\
                    n_1=frac-b+sqrtDeltaa'$$



                    The equation admits an integer root if $sqrtDelta$ is an integer and either:



                    $$beginalign&-b-sqrtDeltaequiv 0pmoda'\ text or &-b+sqrtDeltaequiv 0pmoda'endalign$$



                    Case $a=0, bneq0$



                    The equation is linear:



                    $$T_n=bn+c\
                    bn=d\
                    n=fracdb$$



                    It admits an integer root if $dequiv0pmod b$.



                    Case $a=0, b=0$



                    The equation is not depending on $n$ anymore:



                    $$T_n=c\
                    d=0$$







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Apr 4 at 8:21

























                    answered Apr 3 at 14:37









                    ArnauldArnauld

                    80.5k797333




                    80.5k797333





















                        4












                        $begingroup$


                        Jelly,  11  10 bytes



                        _/Ær1Ẹ?%1Ạ


                        A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.



                        * admittedly taking a little liberty with "You may take input of these four numbers in any way".



                        Try it online! Or see a test-suite.



                        How?



                        _/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
                        / - reduce by:
                        _ - subtraction [c-T_n, b, a]
                        ? - if...
                        Ẹ - ...condition: any?
                        Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
                        1 - ...else: literal 1
                        %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
                        - note: (a+bi)%1 yields nan which is truthy
                        Ạ - all? i.e. all had fractional parts?
                        - note: all([]) yields 1



                        If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)






                        share|improve this answer











                        $endgroup$








                        • 2




                          $begingroup$
                          That input is perfectly fine.
                          $endgroup$
                          – Artemis Fowl
                          Apr 3 at 18:21















                        4












                        $begingroup$


                        Jelly,  11  10 bytes



                        _/Ær1Ẹ?%1Ạ


                        A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.



                        * admittedly taking a little liberty with "You may take input of these four numbers in any way".



                        Try it online! Or see a test-suite.



                        How?



                        _/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
                        / - reduce by:
                        _ - subtraction [c-T_n, b, a]
                        ? - if...
                        Ẹ - ...condition: any?
                        Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
                        1 - ...else: literal 1
                        %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
                        - note: (a+bi)%1 yields nan which is truthy
                        Ạ - all? i.e. all had fractional parts?
                        - note: all([]) yields 1



                        If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)






                        share|improve this answer











                        $endgroup$








                        • 2




                          $begingroup$
                          That input is perfectly fine.
                          $endgroup$
                          – Artemis Fowl
                          Apr 3 at 18:21













                        4












                        4








                        4





                        $begingroup$


                        Jelly,  11  10 bytes



                        _/Ær1Ẹ?%1Ạ


                        A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.



                        * admittedly taking a little liberty with "You may take input of these four numbers in any way".



                        Try it online! Or see a test-suite.



                        How?



                        _/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
                        / - reduce by:
                        _ - subtraction [c-T_n, b, a]
                        ? - if...
                        Ẹ - ...condition: any?
                        Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
                        1 - ...else: literal 1
                        %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
                        - note: (a+bi)%1 yields nan which is truthy
                        Ạ - all? i.e. all had fractional parts?
                        - note: all([]) yields 1



                        If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)






                        share|improve this answer











                        $endgroup$




                        Jelly,  11  10 bytes



                        _/Ær1Ẹ?%1Ạ


                        A monadic Link which accepts a list of lists* [[c, b, a], [T_n]] and yields 0 if T_n is a valid solution or 1 if not.



                        * admittedly taking a little liberty with "You may take input of these four numbers in any way".



                        Try it online! Or see a test-suite.



                        How?



                        _/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
                        / - reduce by:
                        _ - subtraction [c-T_n, b, a]
                        ? - if...
                        Ẹ - ...condition: any?
                        Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
                        1 - ...else: literal 1
                        %1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
                        - note: (a+bi)%1 yields nan which is truthy
                        Ạ - all? i.e. all had fractional parts?
                        - note: all([]) yields 1



                        If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ would also work for 10 (it yields 1 when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)







                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited Apr 3 at 19:00

























                        answered Apr 3 at 17:23









                        Jonathan AllanJonathan Allan

                        53.8k535173




                        53.8k535173







                        • 2




                          $begingroup$
                          That input is perfectly fine.
                          $endgroup$
                          – Artemis Fowl
                          Apr 3 at 18:21












                        • 2




                          $begingroup$
                          That input is perfectly fine.
                          $endgroup$
                          – Artemis Fowl
                          Apr 3 at 18:21







                        2




                        2




                        $begingroup$
                        That input is perfectly fine.
                        $endgroup$
                        – Artemis Fowl
                        Apr 3 at 18:21




                        $begingroup$
                        That input is perfectly fine.
                        $endgroup$
                        – Artemis Fowl
                        Apr 3 at 18:21











                        1












                        $begingroup$


                        05AB1E, 35 bytes



                        Æ©²Āi²4P³n+tÐdi(‚³-IJ·Ä%P}뮳Āi³%]_


                        Port of @Arnauld's JavaScript answer, so make sure to upvote him!



                        Takes the input in the format $[t,c], a, b$.



                        Try it online



                        Explanation:





                        Æ # Reduce the (implicit) input-list by subtraction (`t-c`)
                        © # Store this value in the register (without popping)
                        ²Āi # If the second input `a` is not 0:
                        ²4P # Calculate `(t-c)*a*4`
                        ³n+ # Add the third input `b` squared to it: `(t-c)*a*4+b*b`
                        t # Take the square-root of that
                        # (NOTE: 05AB1E and JS behave differently for square-roots of
                        # negative integers; JS produces NaN, whereas 05AB1E leaves the
                        # integer unchanged, which is why we have the `di...}` here)
                        Ð # Triplicate this square
                        di # If the square is non-negative (>= 0):
                        (‚ # Pair it with its negative
                        ³- # Subtract the third input `b` from each
                        Ä # Take the absolute value of both
                        ²·Ä% # Modulo the absolute value of `a` doubled
                        # (NOTE: 05AB1E and JS behave differently for negative modulos,
                        # which is why we have the two `Ä` here)
                        P # Then multiply both by taking the product
                        } # And close the inner if-statement
                        ë # Else (`a` is 0):
                        ® # Push the `t-c` from the register
                        ³Āi # If the third input `b` is not 0:
                        ³% # Take modulo `b`
                        ] # Close both if-else statements
                        _ # And check if the result is 0
                        # (which is output implicitly)





                        share|improve this answer









                        $endgroup$












                        • $begingroup$
                          Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                          $endgroup$
                          – Arnauld
                          Apr 4 at 10:37











                        • $begingroup$
                          @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 11:58






                        • 1




                          $begingroup$
                          Besides, the results of Ų on negative inputs are inconsistent.
                          $endgroup$
                          – Arnauld
                          Apr 4 at 12:30










                        • $begingroup$
                          @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 12:31
















                        1












                        $begingroup$


                        05AB1E, 35 bytes



                        Æ©²Āi²4P³n+tÐdi(‚³-IJ·Ä%P}뮳Āi³%]_


                        Port of @Arnauld's JavaScript answer, so make sure to upvote him!



                        Takes the input in the format $[t,c], a, b$.



                        Try it online



                        Explanation:





                        Æ # Reduce the (implicit) input-list by subtraction (`t-c`)
                        © # Store this value in the register (without popping)
                        ²Āi # If the second input `a` is not 0:
                        ²4P # Calculate `(t-c)*a*4`
                        ³n+ # Add the third input `b` squared to it: `(t-c)*a*4+b*b`
                        t # Take the square-root of that
                        # (NOTE: 05AB1E and JS behave differently for square-roots of
                        # negative integers; JS produces NaN, whereas 05AB1E leaves the
                        # integer unchanged, which is why we have the `di...}` here)
                        Ð # Triplicate this square
                        di # If the square is non-negative (>= 0):
                        (‚ # Pair it with its negative
                        ³- # Subtract the third input `b` from each
                        Ä # Take the absolute value of both
                        ²·Ä% # Modulo the absolute value of `a` doubled
                        # (NOTE: 05AB1E and JS behave differently for negative modulos,
                        # which is why we have the two `Ä` here)
                        P # Then multiply both by taking the product
                        } # And close the inner if-statement
                        ë # Else (`a` is 0):
                        ® # Push the `t-c` from the register
                        ³Āi # If the third input `b` is not 0:
                        ³% # Take modulo `b`
                        ] # Close both if-else statements
                        _ # And check if the result is 0
                        # (which is output implicitly)





                        share|improve this answer









                        $endgroup$












                        • $begingroup$
                          Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                          $endgroup$
                          – Arnauld
                          Apr 4 at 10:37











                        • $begingroup$
                          @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 11:58






                        • 1




                          $begingroup$
                          Besides, the results of Ų on negative inputs are inconsistent.
                          $endgroup$
                          – Arnauld
                          Apr 4 at 12:30










                        • $begingroup$
                          @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 12:31














                        1












                        1








                        1





                        $begingroup$


                        05AB1E, 35 bytes



                        Æ©²Āi²4P³n+tÐdi(‚³-IJ·Ä%P}뮳Āi³%]_


                        Port of @Arnauld's JavaScript answer, so make sure to upvote him!



                        Takes the input in the format $[t,c], a, b$.



                        Try it online



                        Explanation:





                        Æ # Reduce the (implicit) input-list by subtraction (`t-c`)
                        © # Store this value in the register (without popping)
                        ²Āi # If the second input `a` is not 0:
                        ²4P # Calculate `(t-c)*a*4`
                        ³n+ # Add the third input `b` squared to it: `(t-c)*a*4+b*b`
                        t # Take the square-root of that
                        # (NOTE: 05AB1E and JS behave differently for square-roots of
                        # negative integers; JS produces NaN, whereas 05AB1E leaves the
                        # integer unchanged, which is why we have the `di...}` here)
                        Ð # Triplicate this square
                        di # If the square is non-negative (>= 0):
                        (‚ # Pair it with its negative
                        ³- # Subtract the third input `b` from each
                        Ä # Take the absolute value of both
                        ²·Ä% # Modulo the absolute value of `a` doubled
                        # (NOTE: 05AB1E and JS behave differently for negative modulos,
                        # which is why we have the two `Ä` here)
                        P # Then multiply both by taking the product
                        } # And close the inner if-statement
                        ë # Else (`a` is 0):
                        ® # Push the `t-c` from the register
                        ³Āi # If the third input `b` is not 0:
                        ³% # Take modulo `b`
                        ] # Close both if-else statements
                        _ # And check if the result is 0
                        # (which is output implicitly)





                        share|improve this answer









                        $endgroup$




                        05AB1E, 35 bytes



                        Æ©²Āi²4P³n+tÐdi(‚³-IJ·Ä%P}뮳Āi³%]_


                        Port of @Arnauld's JavaScript answer, so make sure to upvote him!



                        Takes the input in the format $[t,c], a, b$.



                        Try it online



                        Explanation:





                        Æ # Reduce the (implicit) input-list by subtraction (`t-c`)
                        © # Store this value in the register (without popping)
                        ²Āi # If the second input `a` is not 0:
                        ²4P # Calculate `(t-c)*a*4`
                        ³n+ # Add the third input `b` squared to it: `(t-c)*a*4+b*b`
                        t # Take the square-root of that
                        # (NOTE: 05AB1E and JS behave differently for square-roots of
                        # negative integers; JS produces NaN, whereas 05AB1E leaves the
                        # integer unchanged, which is why we have the `di...}` here)
                        Ð # Triplicate this square
                        di # If the square is non-negative (>= 0):
                        (‚ # Pair it with its negative
                        ³- # Subtract the third input `b` from each
                        Ä # Take the absolute value of both
                        ²·Ä% # Modulo the absolute value of `a` doubled
                        # (NOTE: 05AB1E and JS behave differently for negative modulos,
                        # which is why we have the two `Ä` here)
                        P # Then multiply both by taking the product
                        } # And close the inner if-statement
                        ë # Else (`a` is 0):
                        ® # Push the `t-c` from the register
                        ³Āi # If the third input `b` is not 0:
                        ³% # Take modulo `b`
                        ] # Close both if-else statements
                        _ # And check if the result is 0
                        # (which is output implicitly)






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Apr 4 at 9:56









                        Kevin CruijssenKevin Cruijssen

                        42.3k570217




                        42.3k570217











                        • $begingroup$
                          Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                          $endgroup$
                          – Arnauld
                          Apr 4 at 10:37











                        • $begingroup$
                          @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 11:58






                        • 1




                          $begingroup$
                          Besides, the results of Ų on negative inputs are inconsistent.
                          $endgroup$
                          – Arnauld
                          Apr 4 at 12:30










                        • $begingroup$
                          @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 12:31

















                        • $begingroup$
                          Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                          $endgroup$
                          – Arnauld
                          Apr 4 at 10:37











                        • $begingroup$
                          @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 11:58






                        • 1




                          $begingroup$
                          Besides, the results of Ų on negative inputs are inconsistent.
                          $endgroup$
                          – Arnauld
                          Apr 4 at 12:30










                        • $begingroup$
                          @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                          $endgroup$
                          – Kevin Cruijssen
                          Apr 4 at 12:31
















                        $begingroup$
                        Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                        $endgroup$
                        – Arnauld
                        Apr 4 at 10:37





                        $begingroup$
                        Would Ų save some bytes? (Probably not since we later need to compute the square root anyway.)
                        $endgroup$
                        – Arnauld
                        Apr 4 at 10:37













                        $begingroup$
                        @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 4 at 11:58




                        $begingroup$
                        @Arnauld Unfortunately not for three reasons: 1. Ų with negative values somehow gives the value itself instead of 0.. 2. Ų with decimal values (even with .0) gives 0 instead of 1 whether they're a square or not (this is a bug which I will report to Adnan). 3. Even if both would have worked and -4.0 would result in 0 instead of -4.0 and 4.0 would result in 1 instead of 0, it would still be +2 bytes since we need the square-root and the triplicate would be separated duplicates: tÐdi vs DŲitD; or currently DÄïŲitD to fix the other two mentioned issues.
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 4 at 11:58




                        1




                        1




                        $begingroup$
                        Besides, the results of Ų on negative inputs are inconsistent.
                        $endgroup$
                        – Arnauld
                        Apr 4 at 12:30




                        $begingroup$
                        Besides, the results of Ų on negative inputs are inconsistent.
                        $endgroup$
                        – Arnauld
                        Apr 4 at 12:30












                        $begingroup$
                        @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 4 at 12:31





                        $begingroup$
                        @Arnauld Wth.. that's indeed pretty weird. And the legacy version even gives a different, just as weird result.. :S I've reported the bugs, including your test TIO to Adnan in the 05AB1E chat.
                        $endgroup$
                        – Kevin Cruijssen
                        Apr 4 at 12:31












                        0












                        $begingroup$


                        Wolfram Language (Mathematica), 38 bytes



                        Solve[n^2#+n#2+#3==#4,n,Integers]!=&


                        Try it online!






                        share|improve this answer









                        $endgroup$

















                          0












                          $begingroup$


                          Wolfram Language (Mathematica), 38 bytes



                          Solve[n^2#+n#2+#3==#4,n,Integers]!=&


                          Try it online!






                          share|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$


                            Wolfram Language (Mathematica), 38 bytes



                            Solve[n^2#+n#2+#3==#4,n,Integers]!=&


                            Try it online!






                            share|improve this answer









                            $endgroup$




                            Wolfram Language (Mathematica), 38 bytes



                            Solve[n^2#+n#2+#3==#4,n,Integers]!=&


                            Try it online!







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Apr 3 at 21:33









                            J42161217J42161217

                            13.8k21253




                            13.8k21253





















                                0












                                $begingroup$


                                Jelly, 15 bytes



                                _3¦UÆr=Ḟ$;3ị=ɗẸ


                                Try it online!



                                Built-in helps here but doesn’t handle a=b=0 so this is handled specially.






                                share|improve this answer











                                $endgroup$

















                                  0












                                  $begingroup$


                                  Jelly, 15 bytes



                                  _3¦UÆr=Ḟ$;3ị=ɗẸ


                                  Try it online!



                                  Built-in helps here but doesn’t handle a=b=0 so this is handled specially.






                                  share|improve this answer











                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$


                                    Jelly, 15 bytes



                                    _3¦UÆr=Ḟ$;3ị=ɗẸ


                                    Try it online!



                                    Built-in helps here but doesn’t handle a=b=0 so this is handled specially.






                                    share|improve this answer











                                    $endgroup$




                                    Jelly, 15 bytes



                                    _3¦UÆr=Ḟ$;3ị=ɗẸ


                                    Try it online!



                                    Built-in helps here but doesn’t handle a=b=0 so this is handled specially.







                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited Apr 4 at 7:39

























                                    answered Apr 3 at 16:46









                                    Nick KennedyNick Kennedy

                                    1,33649




                                    1,33649



























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