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.
 A: 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 $x=37$.
A: Hints: 
Considered as a quadratic in $x$, you need from the discriminant that $y^4-2y^2-4y+70$ must be a perfect square. 
Now show that the quartic lies between the squares $(y^2-2)^2$ and $(y^2+1)^2$ and then you should be able to conclude $y=5$ is the only solution with $y>3$. 
A: 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.
A: 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
$y=1$,
$y=2$,
$y=3$,
$y=4$,
$y=5$,
$y=6$,
