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Find all the positive integers (x,y), such that

a) $1!+2!+3!+\cdots+ x!=y^2$

b)$1!+2!+3!+\cdots+x!=y^z$

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    $\begingroup$ please post only one question at a time and please change your title and even try writing down what all you have tried to solve this problem.... $\endgroup$ – user87543 Dec 2 '13 at 10:18
  • $\begingroup$ and i do not understand what is $y^z$ in your second equation... $\endgroup$ – user87543 Dec 2 '13 at 10:19
  • $\begingroup$ a) I worked like that : $2\cdot\left(1!+\frac{3!}{2}+\frac{4!}{2} \cdots\frac{x!}{2}\right) = (y-1)(y+1)$ $y=3 \ \ or \ \ y = 1$ (3,3),(1,1) Is that right ? But how to prove this to all postive integers ? $\endgroup$ – Joel Dec 2 '13 at 10:23
  • $\begingroup$ where did you got $2$ from? where did you got $(y+1)(y-1)y=3$ from :O you should be more careful in writing.... $\endgroup$ – user87543 Dec 2 '13 at 10:26
  • $\begingroup$ $1!+2!+3!+\cdots+ x!=y^2\\ 2!+3!+\cdots+ x!=y^2-1 \\ 2!+3!+\cdots+ x!=(y-1)(y+1)$ x! is multiple of 2 when x>1 $\endgroup$ – Joel Dec 2 '13 at 10:29
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For the first one, note that $y^2\equiv 0,1,4\pmod{5}$. Now, if $x\geq 5$ then $$1!+2!+\cdots +x!\equiv 1+2+6+24\equiv 3\pmod{5}.$$ Hence if $x\geq 5$ then there are no soloutions.

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For part b, assuming $z$ is also a positive integer,for $x\ge 27$, first note that the last digit of summation is 3, so the last digit of $y$ must be $3$ or $7$ and $z=4k+1$ or $z=4k+3$

$1!+2!+\cdots +x!\equiv 9 \pmod{27}$

But $y^{4k+3}$ or $y^{4k+1}$ cannot be $27t+9$. So all answers, if any, must be less than $27$. This can be checked by a computer, and there is no other $y^z$ equal to the sum but $3^2$ and trivial $1^z$

Note that:

$y^{4} mod 27 \in \{0,1,4,7,10,13,16,19,22,25\}\\ y^{3} mod 27 \in \{0,1,8,10,17,19,26\} $

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