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.

Rearrange the equation given into
$$
  (x-12)^2 = 71 + (y^2-1)^2 - 4y.
$$
$(y^2-1)^2$ is a square.  The square before $(y^2-1)^2$ is $(y^2-2)^2$, which is $2y^2-1$ less.  For $y > 2$, $2y^2-1 > 4y > 4y-71$.  The square after $(y^2-1)^2$ is $y^4$, which is $2y^2+1$ greater.  For $y\geq 6$, $2y^2+1 > 71 > 71-4y$.
You therefore only need to consider $y=4$ and $y=5$.  It happens that $y=5$ gives you $x=37$.
