Competition problem — Number theory

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Number theory Difficulty 6.7 National olympiad Prove it Netherlands

Given a positive integer $n$ , we denote by $n!$ (‘n factorial’) the number we get if we multiply all integers from $1$ to $n$ . For example: $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$ .

a. Determine all integers $n$ with $1 \le n \le 100$ for which $n! \cdot (n+1)!$ is a perfect square. Also, prove that you have found all solutions $n$ .

b. Prove that no positive integer $n$ exists such that $n! \cdot (n+1)! \cdot (n+2)! \cdot (n+3)!$ is a perfect square.

Solution

a. We observe that $(n+1)! = (n+1) \cdot n!$ , and therefore that $n! \cdot (n+1)! = (n!)^2 \cdot (n+1)$ . That product is a perfect square if and only if $n+1$ is a perfect square, since $(n!)^2$ is a perfect square. For $1 \le n \le 100$ this is the case for $n = 3, 8, 15, 24, 35, 48, 63, 80, 99$ (perfect squares minus one that are below $100$ ). $\square$

b. We rewrite the product $n! \cdot (n+1)! \cdot (n+2)! \cdot (n+3)!$ as follows:
$(n!)^2 \cdot (n+1) \cdot (n+2)! \cdot (n+3)! = (n!)^2 \cdot (n+1) \cdot ((n+2)!)^2 \cdot (n+3).$
Since $(n!)^2$ and $((n+2)!)^2$ are both perfect squares, the above product is a perfect square if and only if $(n+1)(n+3)$ is a perfect square. However, $(n+1)(n+3)$ cannot be a perfect square. Indeed, suppose that $(n+1)(n+3) = k^2$ were a perfect square. Since $(n+1)^2 < (n+1)(n+3) < (n+3)^2$ we would have $n+1 < k < n+3$ , so $k = n+2$ . This is impossible because $(n+1)(n+3) = (n+2)^2 - 1$ , which is not equal to $(n+2)^2$ . $\square$

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