Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Romania

Let $E(x, y) = \frac{x}{y} + \frac{x+1}{y+1} + \frac{x+2}{y+2}$ .

a) Solve in positive integers the equation $E(x, y) = 3$ .

b) Prove that there exist infinitely many positive integers $n$ for which the equation $E(x, y) = n$ has solutions in positive integers.

Solution

a) Rewrite the equation as
$\left(\frac{x}{y} - 1\right) + \left(\frac{x+1}{y+1} - 1\right) + \left(\frac{x+2}{y+2} - 1\right) = 0,$
which is equivalent to $(x - y) \left(\frac{1}{y} + \frac{1}{y+1} + \frac{1}{y+2}\right) = 0$ , and the solutions are all pairs $(k, k)$ , with $k \in \mathbb{N}$ .

b) Observe that $E(6k + 1, 1) = \frac{6k + 1}{1} + \frac{6k + 2}{2} + \frac{6k + 3}{3} = 6k + 1 + 3k + 1 + 2k + 1 = 11k + 3$ , $k \in \mathbb{N}$ , hence for $n = 11k + 3$ , $k \in \mathbb{N}$ , the equation has the solution $(6k + 1, 1)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.