Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

Five, (17 points) Find the positive integer solutions of the equation
x2+6xy7y2=2009 x^{2}+6 x y-7 y^{2}=2009

A number or a short expression. Spacing and $ signs are ignored.

Solution

Five, factoring on the left side, we get
(xy)(x+7y)=2009 (x-y)(x+7 y)=2009 \text{. }

Since 2009=7×7×412009=7 \times 7 \times 41, we have:
When xy=1,7,41,49,287,2009x-y=1,7,41,49,287,2009, correspondingly, x+7y=2009,287,49,41,7,1x+7 y=2009,287,49,41,7,1.
Also, since x,yx, y are positive integers, then
xy<x+7y x-y<x+7 y \text{. }

Therefore, among the six relationships above, only three can hold:
(1) {xy=1,x+7y=2009;\left\{\begin{array}{l}x-y=1, \\ x+7 y=2009 ;\end{array}\right.
(2) {xy=7,x+7y=287\left\{\begin{array}{l}x-y=7, \\ x+7 y=287\end{array}\right.
(3) {xy=41,x+7y=49.\left\{\begin{array}{l}x-y=41, \\ x+7 y=49 .\end{array}\right.

Solving these, we get (x,y)=(252,251),(42,35),(42,1)(x, y)=(252,251),(42,35),(42,1).
These are all the positive integer solutions to the equation.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.