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Find all natural solutions $m!=a^2n!$ .

It's clear that $m>=n$.
When $m=n$ we have solutions $(1,m,m)$.
When $m=n+1$ we have solutions $(a,a^2,a^2-1)$.
I think that when $m>n+1$ we have no solutions but I can't prove this.

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The product of consecutive integers is never a power.

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