Determine the number of ordered pairs $(x, y)$ of positive integers satisfying the equation $x^2+y^2-16y=2004$.
My solution:
$x^2+y^2-16y=2004$
$\Rightarrow x^2=2004-y^2+16y$
$\Rightarrow x=\sqrt{2004-y^2+16y}$
Now, plugging in integers on the R.H.S. for $y=1$ to $53$ (since both $x$ and $y$ are positive integers, and for $y>53$, $x$ is imaginary), (on Excel);
I do not get any pair of $x$ & $y$; where both are positive integers.
Answer : $0$
Is this solution correct?, and is there a smarter way to solve this problem?
Zero is correct, but a much faster way follows. Complete the square first: $$x^2+y^2-16y+64=2004+64$$ $$x^2+(y-8)^2=2068=4×11×47$$ $11\equiv3\bmod4$ and it appears only once in the prime factorisation, so no pair $(x,y)$ satisfies the given equation by the sum of two squares theorem.