Prove $(n-1)! + 1 = n^2$ has only one integer solution

Question

Prove $(n-1)! + 1 = n^2$ has only one integer solution, namely $5$.

I think that we can use the derivate of the gamma function to say that the LHS is growing more than the RHS from $n=5$ onwards, so it can not have solutions bigger than $5$. But computing the value of $\Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt$ and proving this is strictly bigger than $2n$ is where I get stuck.

Hint: write it as $,(n-1)!=(n-1)(n+1),$. – dxiv May 31 '22 at 07:06 — dxiv, May 31 '22 at 07:06
@dxiv, its my mistake, I didn't saw that. – VVR May 31 '22 at 07:20 — VVR, May 31 '22 at 07:20

score 5 · Accepted Answer · answered May 31 '22 at 07:04

5

Use the fact that $$(n-1)! +1 \geq (n-1)(n-2)(n-3) +1$$ for all $n > 3,$ and then compare $(n-1)(n-2)(n-3) +1$ with $n^{2}.$ Since both are polynomials, this should not be too difficult.

answered May 31 '22 at 07:04

ckefa

3,746

Prove $(n-1)! + 1 = n^2$ has only one integer solution

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