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1987 AIME Problems/Problem 5

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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 (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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