For any set S = {a, a+1, ..., a+5} where 6|a, 24|($x^2$ - $y^2$) for distinct odd integer x and y in set S if and only if one of x and y is congruent to 1 modulo 6 and the other is congruent to 5 modulo 6
I think I have one half of the biconditional right but I am very stuck on where how to complete it
If one of x and y is congruent to 1 mod 6 and the other to 5 mod 6 then
$x^2 - y^2$ = $(a+1)^2 - (a+5)^2 = a^2+2a+1 - (a^2 +10a + 25)$ = -8a - 24
using 6k = a I have
-48k -24 = 24 (2k+1) (-1)
So, if one of x and y is congruent to 1 mod6 and the other to 5 mod6, 24|($x^2 - y^2$)
Next I know I need to show the reverse that if 24|($x^2 - y^2$)then x and y must be congruent to 1 mod 6 and 5 mod 6
I'm pretty sure I need a few cases but I'm stuck. After getting the first part the second should come to me and its something I should know but I just can't come up with it. Any hints are appreciated
Hint $\,x,y$ odd $\,\Rightarrow\,\bmod 6\!:\ x,y \equiv 1,3,5,\,$ and the only distinct pair with equal square is $1^2\equiv 5^2$
Remark $ $ Alternatively $\bmod 8\!:\ {\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1\,$ so $\,x,y\,$ odd $\,\Rightarrow\,x^2\!-\!y^2\equiv 1-1\equiv 0\,$
Therefore $\,8\mid x^2-y^2,\ $ hence $\ 24\mid x^2-y^2\iff 6\mid x^2-y^2,\ $ which is true as in the hint.