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Suppose $n|a^2-1$ Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$
Suppose $n|a^2-1$ where $a>1$ and n is odd. Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$.
Part 2
Show that if $a<n-1$ then this gives a nontrivial factorization of n
What I did:
I found the gcd$(a-...
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$\operatorname{lcm}(n,m,p)\times \gcd(m,n) \times \gcd(n,p) \times \gcd(n,p)= nmp \times \gcd(n,m,p)$, solve for $n,m,p$?
$\newcommand{\lcm}{\operatorname{lcm}}$
I saw this in the first Moscow Olympiad of Mathematics (1935), the equation was :
$$\lcm(n,m,p)\times \gcd(m,n) \times \gcd(n,p)^2 = nmp \times \gcd(n,m,p)$$
My ...
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