Largest even number

Note that $a^2\equiv 1 \pmod{8}$ for any odd $a$ ; check $a=1,3,5,7$ . Thus $a^2\equiv b^2 \pmod{8}$ , meaning that 8 divides $a^2-b^2$ . The example $a^2=1$ and $b^2=9$ shows that $\boxed{8}$ is maximal.

Let $m,n$ be the odd distinct integers.

We can write $m = 2p + 1$ , $n = 2q + 1$ for $p \not = q$ and $p,q \in \mathbb{Z}$ .

$m^2-n^2 = (2p + 1)^2 - (2q + 1)^2 = 4p^2 + 4p - 4q^2 - 4q$

$\implies 4(p^2 + p - q^2 - q)$

$\implies 4(p - q)(p + q + 1)$

Remember that $p,q \in \mathbb{Z}$ . We can use parity rules here to determine the largest even integer dividing the difference of the squares of two odd distinct numbers.

• Case 1: $p$ and $q$ are even. We can then see that the factor $p - q$ must be even (even ± even = even).
• Case 2: $p$ and $q$ are odd. We can then see that the factor $p-q$ must be even (odd ± odd = even).
• Case 3: One of $p,q$ is odd and the other even. We can then see that the factor $p+q+1$ must be even $\qquad\enspace\enspace$ (odd ± even = odd $\rightarrow$ odd + 1 = even).

Since at least one of the factors are even in all cases, we can factor out an additional $2$ and get $4 \cdot 2 = \boxed{8}$ .