For any prime p show that τ(p!) = 2τ ((p - 1)!), σ(p!) = (p + 1)τ((p - 1)!)...?
For any prime p show that..
(a) τ(p!) = 2τ ((p - 1)!)
(b) σ(p!) = (p + 1)τ((p - 1)!)
(a) φ(p!) = (p - 1)τ ((p - 1)!):
Where τ(n) = # of pos divisors of n
σ(n) = sum of pos divisors of n
φ(n)= # of pos integers <= n that are relatively prime to n
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Note to poster: You forgot to change the functions in the last two parts...
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Since p is prime, all factors less than p are relatively prime to p.
So, gcd(p, (p-1)!) = 1.
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(a) τ(p!) = τ(p (p-1)!)
............= τ(p) τ((p-1)!), since τ is multiplicative
............= 2 τ((p-1)!), since the only positive divisors of p are 1 and p.
(b) σ(p!) = σ(p (p-1)!)
............= σ(p) σ((p-1)!), since τ is multiplicative
............= (p+1) σ((p-1)!), since the only positive divisors of p are 1 and p.
(c) φ(p!) = φ(p (p-1)!)
............= φ(p) φ((p-1)!), since τ is multiplicative
............= (p-1) φ((p-1)!), since the only positive divisors of p are 1 and p.
I hope this helps!