Proofs

What does (2n+1)^2 equal?

Re: Proofs


Let

n=p*q*..*r (I)

be the (unique) factorization of the number n where p, q,..,r be primes, then the square of n is

n*n= (p*p)*(q*q)*..*(r*r) (II)

and if n is odd, then each of the factors p, q,..,r is different from 2, and hence n*n does not have the factor 2 in the (unique) factorization (II).

Proof
Proof:

Assume n is odd. Then n=2m+1 for some integer m. Then we have that n^2 = (2m+1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1. Since
2m^2 + 2m is an integer, then n^2 is odd. So if n is odd, then n^2 is odd. (End Proof)

Proof
Proof:

Assume n is odd. Then n=2m+1 for some integer m. Then we have that n^2 = (2m+1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1. Since
2m^2 + 2m is an integer, then n^2 is odd. So if n is odd, then n^2 is odd. (End Proof)

Proof
Proof:

Assume n is odd. Then n=2m+1 for some integer m. Then we have that n^2 = (2m+1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1. Since
2m^2 + 2m is an integer, then n^2 is odd. So if n is odd, then n^2 is odd. (End Proof)

unless i'm mistaken, you can slightly alter either of the proofs given for the square of an odd number to the product of any two odd numbers, and show that the result is always odd, which is a more general result.

You're not mistaken:
(2m+1)(2n+1) = 4mn+2(m+n)+1 = 2(2mn+m+n)+1