Competition problem — Number theory

Number theory Difficulty 4.8 AIME Prove it Romania

A positive integer is square full if it is divisible by the square of each of its prime divisors. Prove that $n$ and $n + 1$ are both square full for infinitely many positive integers $n$ .

Solution

Note that $8 = 2^3$ and $8 + 1 = 9 = 3^2$ are square full. Now, if $n$ and $n + 1$ are both square full, then so are $4n(n + 1)$ and $4n(n + 1) + 1 = (2n + 1)^2$ . As $4n(n + 1) > n + 1$ , the conclusion follows.

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