Find the values of ## n\geq 1 ## for which ## 1+2+3+\dotsb +n ##.

[Math100]

Homework Statement

Find the values of $n\geq 1$ for which $1!+2!+3!+\dotsb +n!$ is a perfect square.

Relevant Equations

None.

Let $f(n)=1!+2!+3!+\dotsb +n!$ for $n\in\mathbb{N}$.
Then every $k!$ is even except $1!$.
This means $f(n)$ is odd for $n>1$.
Since each $k!$ is divisible by $5$ for $k\geq 5$,
it follows that $f(4)\pmod {5}\equiv (1+2+6+24)\pmod {5}\equiv 3\pmod {5}$.
Thus, $f(n)\equiv 3\pmod {2}$ and $f(n)\equiv 3\pmod {5}$ for $n>3$, which implies that $f(n)\equiv 3\pmod {10}$.
Note that $f(1)=1^{2}, f(2)=3$ and $f(3)=3^{2}$, where $f(2)=3$ isn't a perfect square.
Therefore, the values of $n\geq 1$ for which $1!+2!+3!+\dotsb +n!$ is a perfect square are $n=1, 3$.

 

[anuttarasammyak]

This is just a checking of your result.

for n $\geq$ 8 f(n)$\equiv 0$(mod 9)
So f(n) is written as $f(n)=3^2 A$
for n $\geq$ 9 f(n)$\equiv 3$(mod 10)
So the first place of A is 7. Any square number does not have 7 in its first place. So f(n) for n $\geq$ 9 is not square number.

for n $\geq$ 1 f(n)$\equiv 1$(mod 2)
for n $\geq$ 2 f(n)$\equiv 0$(mod 3)
for n $\geq$ 3 f(n)$\equiv 1$(mod 4)
for n $\geq$ 4 f(n)$\equiv 3$(mod 5)
for n $\geq$ 5 f(n)$\equiv 3$(mod 6)
for n $\geq$ 6 f(n)$\equiv 5$(mod 7)
for n $\geq$ 7 f(n)$\equiv 1$(mod 8)
for n $\geq$ 8 f(n)$\equiv 0$(mod 9)
for n $\geq$ 9 f(n)$\equiv 3$(mod 10)

 

[andrewkirk]

Therefore, the values of $n\geq 1$ for which $1!+2!+3!+\dotsb +n!$ is a perfect square are $n=1, 3$.

You have not proven that. You have only shown that 1 and 3 are such values and 2 is not. You need to prove there are no values higher than 3. You could try using your mod results to do that.

 

[Delta2]

You have not proven that. You have only shown that 1 and 3 are such values and 2 is not. You need to prove there are no values higher than 3. You could try using your mod results to do that.

Yes, true but he has prove that $f(n)=3 \pmod {10}$ for $n>3$ and it is kind of trivial to show that $a^2 \pmod {10}\neq 3$ for any natural number $a$, so $f(n)\neq a^2$ for any natural number $a$..

 

[fresh_42]

Homework Statement:: Find the values of $n\geq 1$ for which $1!+2!+3!+\dotsb +n!$ is a perfect square.
Relevant Equations:: None.

Let $f(n)=1!+2!+3!+\dotsb +n!$ for $n\in\mathbb{N}$.

I would deal with the small cases by hand:

$f(1)=1$ and $f(3)=9$ are perfect squares, $f(2)=3$ and $f(4)=33$ are not.

Then every $k!$ is even except $1!$.
This means $f(n)$ is odd for $n>1$.
Since each $k!$ is divisible by $5$ for $k\geq 5$,
it follows that $f(4)\pmod {5}\equiv (1+2+6+24)\pmod {5}\equiv 3\pmod {5}$.

A typo? It follows $f(n) \equiv f(4) \equiv 3 \pmod 5.$ Now conclude with @Delta2 's proposal in post #4:

$a^2 \equiv c \in \{0,1,4\} \pmod 5$ so $f(n)\neq a^2$ for any integer $a.$

Thus, $f(n)\equiv 3\pmod {2}$ and $f(n)\equiv 3\pmod {5}$ for $n>3$, which implies that $f(n)\equiv 3\pmod {10}$.
Note that $f(1)=1^{2}, f(2)=3$ and $f(3)=3^{2}$, where $f(2)=3$ isn't a perfect square.
Therefore, the values of $n\geq 1$ for which $1!+2!+3!+\dotsb +n!$ is a perfect square are $n=1, 3$.