I want to prove that for all normal number (M ) Takes place of (m!+1,(m+1)!+1)=1 is true
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What does "Takes place of (m!+1,(m+1)!+1)=1 is true" mean? – lulu Nov 21 '20 at 15:56
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it mean that GCD for (m!+1,(m+1)!+1)=1 is true for any M we try – Mohamad Abo allo Nov 21 '20 at 15:57
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1For that matter, what do you mean by "normal number"? normal numbers have a precise definition, but I doubt you have that definition in mind. – lulu Nov 21 '20 at 15:58
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i mean M>0 so if we try to use 1 or 2 we always will get the answer of GCD equal to 1 – Mohamad Abo allo Nov 21 '20 at 15:59
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So I think you just mean "natural number". Anyway, I have posted something below. I think it is relevant, but I am not sure. – lulu Nov 21 '20 at 16:03
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I think (but am not sure) that you are asking: for $n\in \mathbb N$ show that $\gcd(n!+1,(n+1)!+1)=1$
To see that, suppose we had a counterexample $n$. Then let $p$ be a prime which divided $\gcd(n!+1,(n+1)!+1)$. It would then follow that $p$ divided the difference $(n+1)!+1-n!-1=(n+1)!-n!=n!\times n$. But if $p\,|\,n!\times n$ then $p\,|\,n!$ in which case we certainly have $p\,\nmid (n!+1)$ so we have a contradiction, and we are done.
lulu
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