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.

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.
