Integral Solutions

Find the total number of integral solutions to the equation x 2 y 72 y + x 2 = 0 x^2y-72y+x^2=0


The answer is 9.

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1 solution

Tom Engelsman
Jan 6, 2019

Solving for y y in terms of x x gives y = x 2 x 2 72 = 1 + 72 x 2 72 y = \frac{x^2}{x^2 - 72} = 1 + \frac{72}{x^2 - 72} . We are interested in those x Z x \in \mathbb{Z} such that ( x 2 72 ) 72 (x^2 - 72)|72 , and the integer divisors of 72 include:

± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 8 , ± 9 , ± 12 , ± 18 , ± 24 , ± 36 , ± 72 \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm9, \pm12, \pm18, \pm24, \pm36, \pm72 .

Testing each of these divisors gives integral values for x , y x, y for:

72 x = ± 12 , y = 2 ; 72 \Rightarrow x = \pm12, y = 2;

9 x = ± 9 , y = 9 ; 9 \Rightarrow x = \pm9, y = 9;

8 x = ± 8 , y = 8 ; -8 \Rightarrow x = \pm8, y = 8;

36 x = ± 6 , y = 1 ; -36 \Rightarrow x = \pm6, y = -1;

72 x = 0 , y = 0. -72 \Rightarrow x = 0, y = 0.

Thus, there are a total of 9 \boxed{9} integral pairs that solve our original equation.

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