# Diophantine Equation: $a^3=a(b^2+c^2+d^2)+2bcd$

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.

• 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. Commented Jan 19, 2015 at 18:08
• @individ: Thanks, if you manage to solve it please post your solution. (I know there is a nice solution.) Commented Jan 19, 2015 at 18:13
• Source of the problem???? Commented Jan 19, 2015 at 18:22
• Source of the nice solution? Commented Jan 19, 2015 at 18:43
• @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. Commented Jan 19, 2015 at 18:58

The solution of the equation:

$$a^3=a(b^2+c^2+d^2)+2bcd$$

If you use Pythagorean triple.

$$x^2+y^2=z^2$$

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

$$a=z(zp^2-2yps+zs^2)$$

$$b=y(zp^2-2yps+zs^2)$$

$$c=zx(s^2-p^2)$$

$$d=2x(zp-ys)s$$

$p,s$ - any integer asked us.

• Excellent, so it's related to pythagorean triangles? I can show you the geometry diagram of this equation if you want Commented Jan 22, 2015 at 17:24
• @user45220 Why? I don't like the geometric approach. I prefer hard algebra. Commented Jan 22, 2015 at 17:29
• Geometry is the original language of God Commented Jan 22, 2015 at 18:16
• @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. Commented Jan 22, 2015 at 18:28
• individ, tell me about Pell's equation. Commented Jan 22, 2015 at 18:48

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

$a^3=a(b^2+c^2+d^2)+2bcd$

$a=(k^2-4k+4)$

$b=(2k^2-4k)$

$c=-(7k^2+4k-4)$

$d=(2k^2-4k)$

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