Problem solving in NT (day 2 week 1)

Well, I am finally free! It's more like I was busy with classes and covering up the NT4 course, dang it is hard! Finally got hand of convolutions etc!

ISL 2002 N3: Let $p_1,p_2,\ldots,p_n$ be distinct primes greater than $3$. Show that $2^{p_1p_2\cdots p_n}+1$ has at least $4^n$ divisors.

Proof: Now consider $d\mid p_1p_2\ldots p_n$ and $d'\mid p_1p_2\ldots p_n$. Note that there is a prime $p|2^d+1|2^{p_1p_2\cdots p_n}+1$ and $p\nmid 2^{d'}+1$ by zsig for any $d, d'$. Hence there are any $2^n$ distinct such $p$ ( as there are $2^n$ prime factors of $p_1p_2\cdots p_n$).

Since there are atleast $2^n$ distinct primes dividing $p_1p_2\cdots p_n \implies $ there are at least $2^{2^n}$ primes factors which is $>>4^n$.

The exceptional case doesn't matter here.

[ISL 2000 N4]:Find all triplets of positive integers $ (a,m,n)$ such that $ a^m + 1 \mid (a + 1)^n$

Proof: Except for the exception of the Zsigmondy theorem, there exists $p$ such that $p|a^m+1$ and $p\nmid (a+1)$ but for $m>1$, we get that $a^m+1\nmid (a+1)^n$. So $m=1$.

The exception case is $(a,m,n)=(2,3,n)$ and works for $(a,m,n)=(a,1,n)$

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Sunaina thinks absurd

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