设这四个连续整数依次为:n-1,n,n+1,n+2,则
(n-1)n(n+1)(n+2)+1,
=[(n-1)(n+2)][n(n+1)]+1
=(n2+n-2)(n2+n)+1
=(n2+n)2-2(n2+n)+1
=(n2+n-1)2.
故四个连续整数的积加上1是一个整数的平方.
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