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Saudi Arabian Mathematical Competitions, 2012

Number theory Difficulty 4.7 AIME Prove it Saudi Arabia

Let aa, bb, cc be positive integers. Prove that if the numbers a2a+b\frac{a^2}{a+b}, b2b+c\frac{b^2}{b+c}, c2c+a\frac{c^2}{c+a} are integers and primes, then a=b=ca = b = c.

Solution

We will use the following result:

Lemma. If xx and yy are positive integers such that x2x+y\frac{x^2}{x+y} is an integer and prime, then y≥xy \ge x.

Proof of Lemma. Assume that x2x+y=p\frac{x^2}{x+y} = p, where pp is a prime.
We have py=x(x−p)py = x(x - p), so p∣x(x−p)p \mid x(x - p), i.e. p∣xp \mid x. Let x=upx = up, for some positive integer uu. We get
y=x(up−p)=x(u−1)p≥x, y = x(up - p) = x(u - 1)p \ge x,
and we are done, since u≥2u \ge 2.

Applying Lemma in our situation, it follows b≥ab \ge a, c≥bc \ge b, a≥ca \ge c, meaning that a=b=ca = b = c.

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