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.

It follows from the condition that $x,y$ are positive integers.
Since $7^4=2401>1994$, $y$ should be less than 7. 
Thus you just check 6 cases that
$y=1$,
$y=2$,
$y=3$,
$y=4$,
$y=5$,
$y=6$,
