Prove that (n) is an odd integer if and only if n is a perfect square or twice a

Question

Prove that (n) is an odd integer if and only if n is a perfect square or twice a perfect square.

(hint: if p is an odd prime, then 1+P+P2+...+Pk is odd only when k is even)

Use proof notation and show all steps for lifesaver rating.

Ref: elementary number theory, 7th edition, burton

ch6-7

Explanation / Answer

Since s(n) = (1 + p_1 + (p_1)^2 + ... + (p_1)^(a_i)) ... (1 + p_k + (p_k)^2 + ... + (p_k)^(a_k)), s(n) is odd Each (1 + p_i + (p_i)^2 + ... + (p_i)^(a_k)) is odd p_i = 2, or for primes p_i > 2, each a_i is even (that is, a_i = 2 r_i for some integer r_i) [consider the number of terms in (1 + p_i + (p_i)^2 + ... + (p_i)^(a_k)) when p_i is odd...] n is a perfect square, or n is twice a perfect square. I hope this helps!

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