證明：原式=（n+1)(n+2)(n+3)(n+4)+1
=（n+1))(n+4)(n+2)(n+3)+1
=(n^2+5n+4)(n^2+5n+6)+1
設(shè)n^2+5n=t,t式自然數(shù)
∴原式=(t+4)(t+6)+1
=t^2+10t+24+1
=t^2+10t+25
=(t+5)^2
=(n^2+5n+5)^2
∴代數(shù)式（n+1)(n+2)(n+3)(n+4)+1是一個(gè)完全平方數(shù)