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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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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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Ahh I see. Thank you.
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