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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