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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?

Kenta S
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Math Tise
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1 Answers1

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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.

Parcly Taxel
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