Problem
NT-B2-M01-P013 Neighbouring Factorials
#13
★★★★☆ Level 4 of 5
Prove that \( \gcd(n!+1,(n+1)!+1)=1 \) for every positive integer \(n\).
Subtract \((n+1)(n!+1)\) from the second number.
Let \(d\) divide both numbers. Then \(d\mid (n+1)!+1-(n+1)(n!+1)=-n\). Hence \(d\mid n\), so \(d\mid n!\). But \(d\mid n!+1\), therefore \(d\mid1\). Thus \(d=1\).
This is useful before the valuations module: factorials are handled here by gcd, not by prime exponents.