- #1

Math100

- 779

- 220

- 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 ##.

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 ##.