2005 JBMO Problems/Problem 1

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Problem 1

Find all positive integers $x,y$ satisfying the equation \[9(x^2+y^2+1) + 2(3xy+2) = 2005 .\]


Solution

We can re-write the equation as:

$3x^2 + y^2 + 2(3x)(y) + 8y^2 + 9 + 4 = 2005$

or $(3x + y)^2 = 4(498 - 2y^2)$

The above equation tells us that $(498 - 2y^2)$ is a perfect square. Since $498 - 2y^2 \ge 0$. this implies that $y \le 15$

Trying all values of $y$ from $0$ to $15$, we find that $y = 7, 11$ result in perfect squares.

Thus, we have $2$ cases to check:

$Case 1: y = 7$

$(3x + y)^2 = 4(498 - 2y^2)$ $=> (3x + 7)^2 = 4(498 - 2(7^2))$ $=>  (3x + 7)^2 = 4(400)$ $=> x = 11$

$Case 2: y = 11$

$(3x + y)^2 = 4(498 - 2y^2)$ $=> (3x + 11)^2 = 4(498 - 2(11^2))$ $=> (3x + 11)^2 = 4(256)$ $=> x = 7$

Thus all solutions are $(7, 11)$ and $(11, 7)$.


$Kris17$

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