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 $3 \cdot 13^2$ . 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 $x^2 = 4$ . This leaves $y^2 - 10 = 39$ , so $y^2 = 49$ . Thus, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$ .

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=4699 ~ pi_is_3.14

https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=3704 - AMBRIGGS

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1987 AIME Problems/Problem 5

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Problem

Solution

Video Solution

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