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What is $x$, if $1! + 2! + 3! + \cdots + (x-1)! + x! = k^2$ and $k$ is an integer?

Using trial and error, it is obvious that for $x < 4$, the given equation has solutions only for $x = 1,k = \pm1$ and $x=3, k = \pm 3$. Where to go from here?

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    $\begingroup$ A more general question is handled here $\endgroup$
    – lulu
    Oct 4, 2020 at 22:27

1 Answer 1

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It can be proved that for $x \ge 4$, no solutions exist. The expressions

$$\begin{aligned}1! + 2! + 3! + 4! &= 33 \\ 1! + 2! + 3! + 4! + 5! &= 153 \\ 1! + 2! + 3! + 4! + 5! + 6! &= 873 \\ 1! + 2! + 3! + 4! + 5! + 6! + 7! &= 5913\end{aligned}$$ end with the digit $3$. Now, for $x \ge 4$ the last digit of the sum $1! + 2! + 3! + \cdots + x!$ is equal to 3 and therefore, this sum cannot be equal to a square of a whole number $k$ (because the square of a whole number cannot end with $3$).

Therefore, $1$ and $3$ are the only solutions for $x$.

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    $\begingroup$ You have given particular samples only. How can we gaurantee that in general "for x≥4 the last digit of the sum 1!+2!+3!+⋯+x! is equal to 3" $\endgroup$
    – Mick
    Oct 5, 2020 at 5:37
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    $\begingroup$ @Mick For $n \ge 5$, $n!$ ends with the digit 0. $\endgroup$ Oct 5, 2020 at 7:52
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    $\begingroup$ I see what you mean now but if that comment has been included in your post, the deduction will be much clearer. $\endgroup$
    – Mick
    Oct 5, 2020 at 17:10

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