Competition problem — Number theory

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Number theory Difficulty 6.0 National olympiad Prove it

Example 2 Let $n$ be a positive integer, prove that $(n!+1,(n+1)!+1)=1$ .

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Solution

Prove we have the equation
$(n!+1)(n+1)-((n+1)!+1)=n$

Let $d=(n!+1,(n+1)!+1)$ , then by (1) we know $d \mid n$ .
Furthermore, since $d \mid n$ , hence $d \mid n!$ , combining with $d \mid(n!+1)$ we know $d \mid 1$ , so $d=1$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.