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Mathematics 11 Online
OpenStudy (anonymous):

Could some kind person check my proof that gcd(n!, n!+1)=1?

OpenStudy (anonymous):

\[n!+1=1\mod1\]\[n!+1=1\mod2\]\[n!+1=1\mod3\]...\[n!+1=1\mod n\] Therefore \[n!+1\] shares no factors with \[n!\] QED

OpenStudy (anonymous):

Remove the first line of the proof, though.

OpenStudy (ash2326):

How do you get this? N!+1 mod 3= 1

OpenStudy (ash2326):

N=2 2!+1=3 3 mod 3=0

OpenStudy (anonymous):

If N>3, then N! is a multiple of 3, so N!=0mod3

OpenStudy (anonymous):

Thus N!+1=1mod3, of course

OpenStudy (ash2326):

Then It's fine, your proof seem good

OpenStudy (anonymous):

Thanks.

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