设四个连续整数为：(n-1),n,(n+1),(n+2)
(n-1)*n*(n+1)*(n+2) +1
= [(n+1)*n] * [(n-1)*(n+2)] +1
= [n^2+n] * [n^2+n-2] +1
= (n^2+n)^2 - 2(n^2+n) + 1
= (n^2+n+1)^2
根号{ (n-1)*n*(n+1)*(n+2) +1} = n^2+n+1
n为整数,n^2+n+1为整数,得证
试说明四个连续正整数的乘机与1的和的算数平方根仍是一个整数
设四个连续整数为：(n-1),n,(n+1),(n+2)
(n-1)*n*(n+1)*(n+2) +1
= [(n+1)*n] * [(n-1)*(n+2)] +1
= [n^2+n] * [n^2+n-2] +1
= (n^2+n)^2 - 2(n^2+n) + 1
= (n^2+n+1)^2
根号{ (n-1)*n*(n+1)*(n+2) +1} = n^2+n+1
n为整数,n^2+n+1为整数,得证