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Suppose the smallest of the integers is n.

Then the product of the four consecutive integers is n*(n+1)*(n+2)*(n+3)

=(n2+3n)(n2+3n+2) = n4+6n3+11n2+6n

So product +1 = n4+6n3+11n2+6n+1 which can be factorised as follows:

n4+3n3+n2 +3n3+9n2+3n + n2+3n+1

=[n2+3n+1]2

Thus, one more that the product of four consecutive integers is a perfect square.

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