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#1 h Jul 25, 2024, 2:23 PM

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Find all positive integer $m$ and $n$ for which:
$1! +2! +......+n! = m^2$

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ATGY

#2 h Jul 25, 2024, 2:29 PM • 1 Y

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For $n \geq 5$ , $1! + \dots + 5! + ... + n!$ , we have $5! + ... + n! \equiv 0 \mod{10}$ , so $1! + \dots + n! \equiv 1! + ... + 4! \equiv 3\mod{10}$ , which is impossible, since $3$ is not a qr mod 10.

checking $n = 1, 2, 3, 4$ gives $(1, 1)$ and $(3, 3)$ as the only solutions

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#4 h Jul 27, 2024, 12:47 PM

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Thanks you!

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#5 h Jul 27, 2024, 6:02 PM

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Checking $mod \ 5$ works also. Famous problem. Try this one:
Solve in $\mathbb{N}$ $1!+2!+\ldots+x! = y^z.$

This post has been edited 2 times. Last edited by UrkeBurke, Jul 27, 2024, 6:04 PM
Reason: typo

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frc

#6 h Aug 22, 2024, 4:25 AM

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Set $z\geq 2$ .
When $x\geq 8$ , $1!+\ldots+x!\equiv 1!+\ldots+8!\equiv 18 \pmod{27}$ .So $3\mid y \Rightarrow v_3(y^z)\geq z$ But $v_3(1!+\ldots+x!)=2$ ,so $z=2$ ,which is solved already.
When $x<8$ ,it's easy to get the solutions:(1,1,m),(3,3,2)

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#7 h Aug 22, 2024, 7:43 AM

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Matricy wrote:

Find all positive integer $m$ and $n$ for which:
$1! +2! +......+n! = m^2$

See the last digit for $n>=5$ in the rest casework

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#8 h Saturday at 1:21 PM

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For $n \geq 5$ , $\sum_{i =1}^{n} i! \equiv 3 \pmod 5$ , thus $\sum_{i=1}^{n} = m^2$ is not possible.
Now checking for $n<5$ gives that only $n =1$ works and hence, $(m,n) = (1,1)$ .

This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Saturday at 1:21 PM

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