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

Question

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.

Answer 1

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.

Answer 2

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

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

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

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

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

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