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 ##.