Solve in natural

Question

$1!+2!+3!...x!=y^2$ Solve over the $N$

I see (1;1), (3;3)

$x^5+y^5+z^5=2009$ prove that there are no solutions in $Z$

Answer 1

Working in $\mod 11$ , we get $2009 \equiv 7 \mod 11$ . Now let us look at the possible remainders of $x^5$ in $\mod 11$.

$$\begin{align} n &\quad 0 \,\,\,\,1\,\,\,\,2\,\,\,3\,\,\,\,4\,\,\,\,5\,\,\,\,6\,\,\,\,7\,\,\,\,8\,\,\,\,9\,\,\,\,10 \\ n^5 & \color{#4d4}{\quad 0 \,\,\,1\,\,\,10\,\,1\,\,\,\,1\,\,\,\,1\,\,10\,\,10\,\,10\,\,1 \,\,\,10}\end{align}$$

It is easy to conclude that there is no possible way to achieve $7$ as the remainder.

Hence there is no solution for the given equation.