# Proving product for odd integers is odd using FTA

I know that the fundamental theorem of arithmetic gives a unique (except, for ordering) prime factorization of any natural $\gt 1$. Now, any odd number will not have 2 as a prime in that factorization. If I have to prove using the fundamental theorem of arithmetic, that product of two odd integers is odd; does it suffice to say that the product's prime-factorization does not have a prime value of 2.

• You are right. And it is not necessary to write neither the factorization nor the form $2n+1$. A proof is also to say that the product does not contains the prime $2$. – Piquito Nov 10 '17 at 12:25

The FTA does indeed suffice, but the claim follows perhaps more easily by considering $$(2n+1)(2k+1)= 2n \cdot 2k +2k +2n +1 \\ = 4nk+2k+2n+1 \\ = 2(2nk+k+n)+1$$