AIME 1987 · 第 5 题

Solution 1

If we move the $x^2$ term to the left side, it is factorable with Simon's Favorite Factoring Trick:

$$(3x^2 + 1)(y^2 - 10) = 517 - 10$$ $507$ 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, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$.

Solution 2

First factor out the $y^2$ term on the left side.

$(y^2)(1+3x^2) = 30x^2 + 517$

Then divide both sides by $(1+3x^2)$ so we isolate the $y^2$ term.

$y^2 = (30x^2 + 517)/(1+3x^2)$

Now we long divide the right side to get

$y^2 = (30x^2 + 517)/(1+3x^2)$

Ok now we have a Diophantine equation to proceed we long divide the right-side of the equation

$y^2 = 10 + 507/(3x^2 + 1)$

Due to the fact that $x$ and $y$ are integers $507/(3x^2 + 1)$ must be a integer too.

We thus prime factorize and list the factors of $507$

$507 = 13^2 \times 3$

The factors of $507$ are $1, 3, 13, 39, 169, 507$

Now we set the denominator equal to $1, 3, 13, 39, 169, 507$

We see that $13$ works as $x = 2$ and thus $y = 7$

So $3x^2 y^2 = 3(xy)^2 = 3(14^2) = 3(200 - 4) = 600-12 = \boxed{588}$. ~blankbox

Video Solution by OmegaLearn

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

Video Solution

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