Find all the positive integer

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

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

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

• 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....
– user87543
Dec 2 '13 at 10:18
• and i do not understand what is $y^z$ in your second equation...
– user87543
Dec 2 '13 at 10:19
• 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 ?
– Joel
Dec 2 '13 at 10:23
• 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....
– user87543
Dec 2 '13 at 10:26
• $1!+2!+3!+\cdots+ x!=y^2\\ 2!+3!+\cdots+ x!=y^2-1 \\ 2!+3!+\cdots+ x!=(y-1)(y+1)$ x! is multiple of 2 when x>1
– Joel
Dec 2 '13 at 10:29

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