Is a^2 - b^2 divisible by 8 when a and b are odd integers?

[annoymage]

Homework Statement

if a and b are odd integer, then 8 l (a²-b²)

Homework Equations

n/a

The Attempt at a Solution

if a=b, clearly, 8 l (a²-b²)
if not,
now, I'm not sure how to continue

should i varies b, and make a fixed, then varies a, and make b fixed,
is that really the way to show, for all odd integer a and b?

 

[HallsofIvy]

$a^2- b^2= (a+ b)(a- b)$

If a and b are both odd we can write a= 2n+1 and b= 2m+ 1 for some integers m and n. The a+ b= 2(m+ n)+ 2= 2(m+n+1) is even and a- b= 2(m-n) is also even.

That would be enough to show that $a^2- b^2$ is divisible by 4 but not enough to show it is divisible by 8.

Of course, if a- b were divisible by 4 itself, then since a+ b is even, $a^2- b^2= (a- b)(a+ b)$ would be divisible by 8.

Suppose a- b= 2(m- n) were not divisible by 4. That means that m- n must be an odd number: m- n= 2k+ 1 so that m= n+ 2k+ 1 and then m+ n= n+ 2k+ 1= 2n+ 2k+ 1= 2(m+k)+ 1, an odd number. But them m+n+ 1= 2(m+k)+ 1+ 1= 2(m+ k)+ 2= 2(m+k+1), and even number.

That is, one of a- b and a+ b is even and the other a multiple of 4 so that their product, $a^2- b^2$ is a multiple of 8.