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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

This is annoying me.

Prove that for n>3, 1! + 2! + 3! + ... + n! is never square.

I've managed to prove it by considering modulo 5. All squares are either 1 or 4 (mod 5). n! = 0 (mod 5) for n≥5. Therefore the sum = 1! + 2! + 3! + 4! = 3 (mod 5), and so it can't be square.

However, I first attempted it by considering modulo 4. All squares are either 0 or 1 (mod 4). But the sum is odd, and so the square would have to be 1 (mod 4). n! = 0 (mod 4) for n≥4. Therefore the sum = 1! + 2! + 3! = 1 (mod 4). It could therefore be square??

Thanks.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's not a contradiction.

The modulo 5 proof perfectly shows that none of those sums can be square.

(Side-note: squares can be 0 mod 5 as well as 1 and 4, but that doesn't break the argument)

The modulo 4 proof fails to show this, but it doesn't prove that a square sum of factorials exists either so we're alright.

Why did the vector cross the road?

It wanted to be normal.

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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

Ahh I see. Thank you.

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