1987 AIME Problems/Problem 5

Problem

Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$.

Solution

If we move the $x^2$ term to the left side, it is factorable:

\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]

$507$ is equal to $3 * 13^2$. Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. $169$ doesn't work either, so $3x^2 + 1 = 13$, and $x = \pm 2$. This leaves $y^2 - 10 = 39$, so $y = \pm 7$. Thus, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$.

See also

1987 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions

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