Уравнение 2x^4+2Ax^3+(A^2 -96)x^2 +Bx+C=0 в целых числах

[sasa]

The roots of equation $\displaystyle 2x^4 +2Ax^3+\left(A^2 -96\right)x^2 +Bx+C=0 $ are four non-negative integers. Solve the equation.

[maxal]

sasa писал(а):

The roots of equation $\displaystyle 2x^4 +2Ax^3+\left(A^2 -96\right)x^2 +Bx+C=0 $ are four non-negative integers. Solve the equation.

It follows that the sum of squares of the roots is
$A^2 - (A^2-96)=96$
and 96 has an unique representation as the sum of four squares of non-negative integers:
$96=0^2+4^2+4^2+8^2$.