Solve the equation in real numbers.

[Math Tutor]

Solve the equation in real numbers.
[tex]1! + 2! +3! + ... + x!=y^2[/tex]

[Guest]

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

Assuming [tex]x[/tex] and [tex]y[/tex] are positive integers the solution is given below.

Spoiler: show

If [tex]x\geq 4[/tex], then [tex]1!+2!+3!+\ldots +x! \equiv 1!+2!+3!+4! \equiv 3 \bmod 5[/tex]
[tex]y[/tex] can always be written in the form [tex]y = 5n+r[/tex] for some integer [tex]n[/tex] and [tex]r = 0,1,2,3,4[/tex] (simply divide [tex]y[/tex] by [tex]5[/tex], and take the quotient as [tex]n[/tex] and the remainder as [tex]r[/tex]).
This means [tex]y^2 = 25n^2 + 10nr + r^2 \equiv r^2 \bmod 5[/tex]. No value of [tex]r[/tex] gives [tex]y^2\equiv 3 \bmod 5[/tex], so [tex]y^2\ne 1!+2!+3!+\ldots +x![/tex] when [tex]x \geq 4[/tex].

When [tex]x\leq 3[/tex], there are two solutions [tex]x=1,3[/tex] and [tex]y=1,3[/tex] respectively.

R. Baber.