# Solve the equation in real numbers.

### Solve the equation in real numbers.

Solve the equation in real numbers.
$1! + 2! +3! + ... + x!=y^2$

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### Re: Solve the equation in real numbers.

I'm assuming you mean $x$ is a positive integer and $y$ is an integer.
(The sum $1!+2!+3!+\ldots+x!$ is ambiguous when x is not a positive integer,
and if $y$ is allowed to be a real then $y = \pm \sqrt{1!+2!+3!+\ldots +x!}$ are the solutions)

Assuming $x$ and $y$ are positive integers the solution is given below.
Spoiler: show
If $x\geq 4$, then $1!+2!+3!+\ldots +x! \equiv 1!+2!+3!+4! \equiv 3 \bmod 5$
$y$ can always be written in the form $y = 5n+r$ for some integer $n$ and $r = 0,1,2,3,4$ (simply divide $y$ by $5$, and take the quotient as $n$ and the remainder as $r$).
This means $y^2 = 25n^2 + 10nr + r^2 \equiv r^2 \bmod 5$. No value of $r$ gives $y^2\equiv 3 \bmod 5$, so $y^2\ne 1!+2!+3!+\ldots +x!$ when $x \geq 4$.

When $x\leq 3$, there are two solutions $x=1,3$ and $y=1,3$ respectively.

R. Baber.

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