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Not sure where to get started on this on. I started listing numbers for n starting at 1 but the numbers get very big very fast and I cannot find a pattern. Is there a better way of doing this or what is the pattern?

Andreas Caranti
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user63342
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3 Answers3

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We have:

$$ n! + (n+1)! + (n+2)! = n!(1 + (n+1) + (n+1)(n+2)) = n!(n+2)^2 $$

So the problem reduces to finding $n$ so that $n!$ is a perfect square. For $n = 0$ or $1$ this is obvious. For $n > 1$, it turns out such $n$ doesn't exist. To see this, consider the largest prime $p$ in $\{1, 2, \ldots, n\}$. $n!$ only contains one factor of $p$ by Bertrand's postulate, so it cannot be a perfect square.

Also check out this question for a detailed proof that $n!$ is not a perfect square for $n > 1$.

Community
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Ayman Hourieh
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$$n! + (n+1)! + (n+2!) = n! \cdot (1+n+1+(n+1)\cdot (n+2)) = n! \cdot (n+2)^2$$ $(n+2)^2$ is clearly square. So $n! + (n+1)! + (n+2!)$ is only a square, if $n!$ is squared. This happens only when $n=0$ or $n=1$.

Jakube
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Since $n! + (n+1)! + (n+2)! = n! \left(1 + (n+1) + (n+1)(n+2) \right) = n! \left(n+2\right)^2$ you are looking for such $n \geqslant 1$ that $n!$ is a perfect square with the only solution here being $n=1$.

Indeed, assuming there exists such $n>1$ that $n!$ is a perfect square. Let $p$ be the largest prime not exceeding $n$. Clearly $p$ divides $n!$, but it divides it only once.

Sasha
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