Track / Stage 4 / 158 of 340 #418 of 1964

Problem 418

AMC 12 late, AIME early

Algebra Difficulty 4.8 Find the answer

1. Given that $x, y$ are integers, and satisfy $\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=-\frac{2}{3}\left(\frac{1}{x^{4}}-\frac{1}{y^{4}}\right)$ . Then the possible values of $x+y$ are ( ) .
(A) 1
(B) 2
(C) 3
(D) 4

Your answer

Multiple choice: answer with the letter of the option you want.

Official solution

-1. .
From the given equation, we have
$\frac{x+y}{x y} \cdot \frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{2}{3} \cdot \frac{x^{4}-y^{4}}{x^{4} y^{4}} \text {. }$

Clearly, $x$ and $y$ are not equal to 0.
Thus, $x+y=0$ or $3 x y=2(x-y)$ .
If $3 x y=2(x-y)$ , then
$(3 x+2)(3 y-2)=-4 \text {. }$

Since $x$ and $y$ are integers, we solve to get
$(x, y)=(-1,2) \text { or }(-2,1) \text {. }$

Therefore, $x+y=1$ or -1.
In summary, there are 3 possible values for $x+y$ .

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