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Solution

suppose that n = 1 hence (1) = 1 which is odd hence we can say that (n) is odd for n=1

now assume n 2 has the prime factorization n = p₁^a₁ . . . p_r^a_r (the pi ’s are distinct primes and each ai 1).

Then (n) = (p₁^a₁) . . . (p_r^a_r).

If p is an odd prime then p^a is odd for all a 0

so we can say that (p^a) = p^a p^a-1 is even for all a 1.

So for (n) to be odd n must not be divisible by any odd prime.

Hence n = 2^a for some a 1.

But (n) = 2^a 2^a-1 is even if a 2.

Hence we must have a = 1 or n = 2. So (n) is odd iff n = 1, 2.