Let $a,b,c,d\in \mathbb{Z}$. Solve $$a^3=a(b^2+c^2+d^2)+2bcd$$

I've tried everything but I haven't been able to find a general solution.

Note: We may assume $\gcd(a,b,c,d)=1$ because of homogeneity.

  • 1
    $\begingroup$ This equation has the form: $$a(a^2-b^2-c^2-d^2)=2bcd$$ This means that relatively simple solutions no. Some is necessarily a multiple of. $\endgroup$
    – individ
    Commented Jan 19, 2015 at 18:08
  • $\begingroup$ @individ: Thanks, if you manage to solve it please post your solution. (I know there is a nice solution.) $\endgroup$
    – user45220
    Commented Jan 19, 2015 at 18:13
  • $\begingroup$ Source of the problem???? $\endgroup$
    – Will Jagy
    Commented Jan 19, 2015 at 18:22
  • $\begingroup$ Source of the nice solution? $\endgroup$ Commented Jan 19, 2015 at 18:43
  • $\begingroup$ @WillJagy Sorry, I should have been more clear there, I meant "I know" as in "there has got to be". The equation is from Ptolemy's Theorem (geometry). In the pythagorean theorem the equivalent equation is $a^2=b^2+c^2$ and notice that multiplying by $a$ gives $a^3=a(b^2+c^2)$ which is SLIGHTLY similar to the one in the question. With Ptolemy's theorem the equation is the one given in the question. I thought that since the Pythagorean theorem has a nice characterization for the integer solutions, Ptolemy's theorem should have one too. $\endgroup$
    – user45220
    Commented Jan 19, 2015 at 18:58

2 Answers 2


The solution of the equation:


If you use Pythagorean triple.


Then the formula for the solution of this equation can be written.





$p,s$ - any integer asked us.

  • $\begingroup$ Excellent, so it's related to pythagorean triangles? I can show you the geometry diagram of this equation if you want $\endgroup$
    – user45220
    Commented Jan 22, 2015 at 17:24
  • $\begingroup$ @user45220 Why? I don't like the geometric approach. I prefer hard algebra. $\endgroup$
    – individ
    Commented Jan 22, 2015 at 17:29
  • $\begingroup$ Geometry is the original language of God $\endgroup$
    – user45220
    Commented Jan 22, 2015 at 18:16
  • $\begingroup$ @user45220 No. The number is the measure of all! Even logical reasoning may be flawed. Everything can be questioned - not yet decided, and not made the experiment. $\endgroup$
    – individ
    Commented Jan 22, 2015 at 18:28
  • $\begingroup$ individ, tell me about Pell's equation. $\endgroup$
    – user45220
    Commented Jan 22, 2015 at 18:48

Below mentioned equation from above has another parametrization, but without the use of a pythagorean triple,






  • $\begingroup$ Could you perhaps indicate how you obtained that parameterisation? Other people referring to this question may not find it as obvious :) $\endgroup$
    – postmortes
    Commented Jul 7, 2017 at 4:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .