Prove $x^2+y^4=1994$ Let $x$ and $y$ positive integers with $y>3$, and $$x^2+y^4=2(x-6)^2+2(y+1)^2$$
Prove that $x^2+y^4=1994$.
I've tried finding an upper bound on the value of $x$ or $y$, but without sucess. Can anyone help me prove this problem? Note that $x^2+y^4=1994$ is the result we are trying to prove, not an assumption.

Hint replacing the $x^2,y^4$ with given condition we get $(x-6)^2+(y+1)^2=992$ so thats equal to the equation of a circle located at $h,k$ ie(6,-1) so it got only $4$ integer points which can be proved by using symmetry and at $x,y$ axis as its radius is approximately $31.5$ but out of those $4$ integer points only $37,5$ are the points which satisfy the original equation.
