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Please notice that this post is not a duplicate from the ones posted by @Lil or @LizW since I want to use another result.

Proof:

Lets prove the next statement first:

If $n$ is an odd integer, then $n$ is equal to $4k+1$ or $4k+3$

Any integer $n$ divided by $4$ has remainder 0, 1, 2 or 3 using the quotient remainder theorem. Therefore any integer can be written as $4k, 4k+1, 4k+2$ and $4k+3$ for some integer $k$. Clearly, $4k$ and $4k+2$ are even numbers. Therefore, all odd integer numbers are $4k+1$ or $4k+3$.

Using this result, if $a$ and $b$ are odd integers then, $a=4m+1$ and $b=4n+1$ for some integers $m$ and $n$.

By algebra, $a^2=(4m+1)^2$ and $b^2=(4n+1)^2$

Then,

$\begin{align*}a^2-b^2=&(4m+1)^2-(4n+1)^2\\ =&16m^2+8m+1-16n^2-8n-1\\ =&8(2m^2+m-2n^2-n)\end{align*}$

This shows that $8\mid (a^2-b^2)$

Should I prove it for $4k+3$?

Grouper
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    This is excellent in the case that they both have that same form. But, as you've pointed out, they might not. There are actually 4 cases here, and you may be able to reduce it to 3. – Randall Aug 29 '17 at 01:53
  • I already reduced it to two, no? Because $4k$ and $4k+2$ are clearly even. My doubt goes if I had to also prove it for $4k+3$. I don't see the point because is the same reasoning,. – Grouper Aug 29 '17 at 01:59
  • oh, I see your point now... Should i prove it for each of the cases then? – Grouper Aug 29 '17 at 02:02
  • Right. $a$ can be either $4k+1$ or $4k+3$, and the same goes for $b$. You can probably make the two cases where $a$ and $b$ are of "different" forms into only one case by arguing there's not much difference in $8$ dividing $a^2-b^2$ or $b^2-a^2$. – Randall Aug 29 '17 at 02:09
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    You can get by without casework. Write $a = 2m + 1, b = 2n+1,$ then you should be able to show that $a^2 - b^2 = 4(m-n)(m+n +1)$. Now, since $(m-n )+ (m + n + 1) = 2m +1$ is odd, one of $m-n$ and $m+n+1$ must be even, and you're done. – stochasticboy321 Aug 29 '17 at 02:10

2 Answers2

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Hint:  if $a$ is an odd integer, then $a=4m\pm1\,$, so $a^2=16m^2\pm 8m+1=8(2m^2\pm m)+1\,$. Therefore $\,a^2-b^2\,$ is the difference of two numbers that each are a multiple of $\,8\,$ plus $\,1\,$.

dxiv
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If $a$ and $b$ are odd integers, then for some integers $m$ and $n$,

$a^2 - b^2 = (2m+1)^2 - (2n+1)^2 = 4(m^2 + m - n^2 - n) = 4(m(m+1) - n(n+1))$

Since $2 \mid m(m+1)$ and $2 \mid n(n+1)$, then $2 \mid (m(m+1) - n(n+1))$

It follows that $8 \mid a^2-b^2$.

Steven Alexis Gregory
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