Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 JBMO Problems/Problem 1

Revision as of 02:18, 25 December 2018 by KRIS17 (talk | contribs)

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 $y \le 15$

Also, taking $mod 3$ on both sides we see that $y$ cannot be a multiple of $3$. Also, note that $249 - y^2$ has to be even since $(498 - 2y^2) = 2(249 - y^2)$ is a perfect square. So, $y^2$ cannot be even, implying that $y$ is odd.

So we have only $\{1, 5, 7, 11, 13\}$ to consider for $y$.

Trying above 5 values for $y$ we find that $y = 7, 11$ result in perfect squares.

Thus, we have $2$ cases to check:

$Case 1: y = 7$

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

$Case 2: y = 11$

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

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


$Kris17$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_JBMO_Problems/Problem_1&oldid=99735"