Verified answer

All three of these are pretty easy given the formula for φ(n).

If n's prime factorization is p1^e1 * ... * pk^ek, then φ(n) = (p1-1)*p1^{e1-1} * ... * (pk-1)*pk^{ek-1}

To prove "If φ(n)|(n-1), prove that n is a square-free integer", we'll prove the contrapositive.

That is, we'll prove that if n is NOT a square-free integer, then φ(n) does not divide n-1.

If n is not square-free, then there is at least one prime in its factorization that has a power of 2 or more. In the formula for φ(n), then, that prime appears in its factorization. Let p be this prime. Then we have that p|n and p|φ(n) and φ(n)|(n-1). So p|(n-1) as well. Thus, there is some k such that pk = (n-1) so n = pk+1 so the remainder when dividing n by p is 1. But if p|n then the remainder would be 0.

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By the formula, φ(2^k) = 1*2^{k-1}, and 2^{k-1}|2^k (the quotient is 2).

By the formula, φ(2^k3^j) = 1*2^{k-1}*2*3^{j-1} = 2^k3^{j-1}, which divides 2^k3^j (quotient is 3)

enable m=p^2 n=q^2 so m+n=p^2+q^2 yet whilst m+n be a appropriate sq. then m+n=r^2 ==> p^2+q^2=r^2 so p,q,r might desire to be Pythagorean triplets eg p=3 q=4 r=5 in gen p=3k q=4k and r=5k ok is +ve integer yet yet another set is p=12,q=5 and r=13 and there is infinite such answer subsequently it rather is actual. IVAN