Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd

Question

$x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.

My Proof Starts:

Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus, \begin{align} x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\ &= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\ &= 4k^2 + 4k - 4l^2 - 4l \end{align}

My two concerns:

1) Is this correct so far?

2) How would I deal with the “$8\;\mid$” part?

Older posts about the same problem: If a and b are odd integers, then $8\mid (a^2-b^2)$ and Number Theory $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd — Martin Sleziak, Oct 09 '16 at 03:55
@Martin Yes, this should've been closed as a duplicate. Bit late now, we have good answers. — Deepak, Oct 09 '16 at 04:14
@Deepak Well, closing solution verification questions is a bit more nuanced, we should leave some time for the OP to receive feedback on their attempt before closing (as discussed here.) I do not think that the fact that there are answers changes whether or not a post is duplicate, at most it can change the decision in which direction to close. (I do not think it is the case here.) If further discussion is needed, let's move it to chat — Martin Sleziak, Oct 09 '16 at 04:27
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. — Martin Sleziak, Oct 09 '16 at 05:13
The title is used when creating list of related questions and also the list of similar questions which SE software offers to you when asking question. So one advantage of having a descriptive title is that this increases the likelihood that you find similar older questions from those lists. — Martin Sleziak, Oct 09 '16 at 05:15
Oops my bad. I'll do better next time. Thank you for the tips. @MartinSleziak — LizW, Oct 10 '16 at 23:00

score 8 · Accepted Answer · answered Oct 08 '16 at 19:31

8

All is correct; now the last expression can be written $$ 4\bigl(k(k+1)-l(l+1)\bigr) $$ and you just have to prove that $k(k+1)-l(l+1)$ is even.

Hint: Can you tell whether $m(m+1)$ is even, for an integer $m$?

answered Oct 08 '16 at 19:31

egreg

238,574

score 4 · Answer 2 · answered Oct 08 '16 at 18:59

$8 \mid x$ if and only if $x\equiv 0\mod 8$. Then you have $4k^2 + 4k - 4l^2 - 4l$ If $k$ is odd, $k^2$ is too. Then $4k^2 + 4k \equiv 0 \mod 8$ if $k$ is even then so is $k^2$ then $4k^2 + 4k \equiv 0\mod 8$. A similar argument for $l$ will finish your proof.

score 3 · Answer 3 · answered Oct 08 '16 at 18:52

3

You're correct so far. What you need to finish is this

Hint: $4u^2+4u = 8v$

Solution:

$ 4u^2+4u=4(u+1)u=8\binom{u+1}{2}$

answered Oct 08 '16 at 18:52

lhf

216,483

I'm not quite sure what you mean, I'm only at beginner's level for proofs. So if you can explain a bit more, please. Thank you for confirming what I have so far. – LizW Oct 08 '16 at 18:56
1

One of the two numbers $u, (u+1)$ must be even. So $4u(u+1)$ means that $u(u+1)$ is even, and hence has a factor of $2$, – amWhy Oct 08 '16 at 19:30
You're welcome, @LizW! – amWhy Oct 08 '16 at 21:56

score 2 · Answer 4 · answered Oct 08 '16 at 19:35

if $K$and $l$ are even, then $K=2K_1$ and $l=2K_2$
then $4K^2+4K-4l^2-4l=16K_1^2+8K_1-16K_2^2-8K_2$ which is clearly divisble by 8
now if $K$and $l$ are odd, then $K=2K_1+1$ and $l=2K_2+1$
then $4K^2+4K-4l^2-4l=4(4K_1^2+4K_1+1)+8K_1+4 -4(4K_2^2+4K_2+1)-8K_2+4= 16K_1^2+16K_1+8K_1+8-16K_2^2-16K_2-8K_2-8$ which is clearly divisble by 8

if $K$ is even and $l$ is odd or $K$ is odd and $l$ is even, it is the same calculation, try it!

Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd

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