At what n is the number $1!+2!+3!+4!+...+n!$ a square of another integer number? All in all I have found $2$ values: $1$ and $3$. I think they are the only ones, but the only thing left is to prove it. How?
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6Consider it modulo $5$. – Daniel Fischer Jan 12 '17 at 14:57
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@DanielFischer iam much into this problem,can someone show without modulo? – Jan 12 '17 at 15:00
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Also related: http://math.stackexchange.com/q/33863/9464 – Jan 12 '17 at 15:12
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those are the same questions! – mdave16 Jan 12 '17 at 15:46
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Let $1!+2!+3!+\cdots \cdots +n! = y^2$
$\star$ if $n=1\;,$ then $1=y^2$ (True)
$\star$ if $n=2\;,$ then $1+2=3=y^2$ (False)
$\star$ if $n=3,$ then $1+2+6=9=y^2$ (True)
$\star$ if $n=4,$ then $1+2+6+24=33=y^2$(False)
for $n\geq 5,$ then L .H .S end with $3$ and we now that square of any integer does not have last digit $3$
so we have only $n=1,n=3$ for which $y^2$ is a perfect square quantity
DXT
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1Along the lines of Daniel's comment, this amounts to "considering it modulo 10". – Ben Grossmann Jan 12 '17 at 15:05
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@Help , since for $n\geq 5$, $10 | n!$ so $n!$ ends in a 0, so the sum will end in a 3, the last digit of $1! + 2! + 3! + 4! $. – Giuseppe Jan 12 '17 at 15:37
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@idliketodothis you can prove it using modular arithmetic, but if you don't know that, then you can make a simple check-by-check case using the last digit of number. – Xam Jan 12 '17 at 15:58
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@idliketodothis If $y$ ends in $1$, $y^2$ ends in $1$. If $y$ ends in $2$, $y^2$ ends in $4$. If $y$ ends in $3$, $y^2$ ends in $9$. If $y$ ends in $4$, $y^2$ ends in $6$. Etcetera. You'll see that there's no possible value of $y$ for which $y^2$ ends in $3$. – Akiva Weinberger Jan 12 '17 at 16:42
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@idliketodothis In fact, squares can only end in the digits $0$, $1$, $4$, $5$, $6$, or $9$. – Akiva Weinberger Jan 12 '17 at 16:42