For any non-zero integer n and prime p, let vp(n)=d if pd divides n and pd+1 doe

Question

For any non-zero integer n and prime p, let vp(n)=d if pd divides n and pd+1 does not divide n. For instance, v2(4)=2, v3(6)=1 and v5(6)=0. Notice that we have n=?p pvp(n), where p is running through primes. Let m and n be two non-zero integers.
1. Prove that vp(mn) = vp(m)+vp(n).
2. Prove that vp(m+n) = min{vp(m),vp(n)}.
3. Prove that m|n if and only if for any prime p we have vp(m) = vp(n).
4. Prove that vp(gcd(n,m)) = min{vp(n),vp(m)}.
5. Let lcm(m,n) be the least common multiple of m and n. Prove that vp(lcm(m,n)) = max{vp(m),vp(n)}.
6. Prove that |mn|=lcm(m,n).gcd(m,n).
7. Prove that the number d(n) of the positive divisors of n is equal to ?p(vp(n)+1).
8. Prove that d(n) is odd if and only if n is a perfect square.
9. Prove that d(mn)=d(m)d(n) if gcd(m,n)=1.

Explanation / Answer

way too much for 350 points .Please break the question :)

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