Let \(m,n \in \mathbb{Z}\). If \(m\) and \(n\) are both odd, then \(m^2+n^2\) is not prefect square

 By way of contradiction, assume \(m^2+n^2=p^2\) where \(\exists m,n,p \in \mathbb{Z}\). Since \(m\) and \(n\) are both odd \(m\) can be written a \(m=2k+1\) and \(n\) can be written as \(n=2j+1\). Therefore, \[\begin{align} m^2+n^2=(2k+1)^2+(2j+1)^2 \\=(4k^2+4k+1)+(4j^2+4j+1) \\=2(2k^2+2k)+1+2(2j^2+2j)+1 \\=2(2k^2+2k)+2(2j^2+2j)+2 \end{align}\].

Since \(2(2k^2+2k)\) and \(2(2j^2+2j)\) must be even, \(p^2\) must be even and therefore \(p\) must also be even (previously proved). This means \(p\) can be written as \(p=2\ell\). Thus, \(m^2+n^2=p^2\) can be written as \[\begin{align}2(2k^2+2k)+2(2j^2+2j)+2=4\ell^2 \\ (2k^2+2k)+(2j^2+2j)+1=2\ell^2 \end{align}\] 

Since \((2k^2+2k)+(2j^2+2j)+1\) is odd, and \(2\ell^2\) is even this is a contradiction, so \(m^2+n^2\) cannot be a perfect square if \(m\) and \(n\) are odd.

Two integers have the same parity if and only if their sum is even

(i) Let \(m\) and \(n\) be two integers having the same parity. Then \(m\) and \(n\) are either both even or both odd.

Case 1. (\(m\) and \(n\) both even.) If \(m\) and \(n\) are both even, then \(m=2k\) and \(n=2j\) for some integers \(k,j\). Then \[m+n=2k+2j=2(k+j)\] and hence \(m+n\) is even.

Case 2. (\(m\) and \(n\) both odd). If \(m\) and \(n\) are both odd, then \(m=2k+1\) and \(n=2j+1\) for some integers \(k,j\). Then \[m+n=(2k+1)+(2j+1)=2(k+j+1)\] and hence \(m+n\) is even.

(ii) We will prove the reverse implication, that two integers have the same parity if their sum is even, by contraposition. Hence, suppose that \(n\) and \(m\) have opposite parity. Without loss of generality we may assume that \(m\) is even and \(n\) is odd. Then \(m=2k\) and \(n=2j+1\) for some integers \(k,j\). So \[m+n=2k+(2j+1) = 2(k+j)+1,\] which proves that \(m+n\) is odd. Thus by contraposition we have proved that if their sum is even, two integers have the same parity.

An integer is even if and only if its square is even

(i) Let \(n\) be an even integer. Then \(n=2k\) for some integer \(k\). So \[n^2 = (2k)^2 = 4k^2 = 2 (2k^2) \] and since \(2k^2\) is an integer, \(n^2\) also satisfies the definition of an even integer.

(ii) To prove that \(n\) is even if \(n^2\) is even, we proceed by contraposition. Hence, suppose that \(n\) is not even and therefore odd. Then \(n=2k+1\) for some integer \(k\). So \[n^2=(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1\] and thus is also odd. Thus by contraposition, we have proved that if \(n^2\) is even, \(n\) is even.