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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 $1^2equiv 5^2$

Remark $ $ Alternatively $bmod 8!: {rm odd}^2equiv {pm1,pm3}^2equiv 1,$ so $,x,y,$ odd $,Rightarrow,x^2!-!y^2equiv 1-1equiv 0,$

Therefore $,8mid x^2-y^2, $ hence $ 24mid x^2-y^2iff 6mid x^2-y^2, $ which is true as in the hint.