Question

Suppose A and B are both divisible by M. Prove that A^2-B^2 is divisible by M.

Answer 1

A is divisible by M :
A= M*k, k is some integer
B is divisible by M:
B = M*h , h is some integer
Now
A^2 - B^2
= (M*k)^2 - (M*h)^2
= M^2*k^2 - M^2*h^2
= M^2 * (k^2 - h^2)
= M * [M(k^2-h^2)]

so A^2 - B^2 = M*(some integer)

therefore A^2 - B^2 is divisible by M

Answer 2

you can make a stronger statement, A^2 - B^2 is divisible by M^2

Answer 3

or you could say if A is divisible by M and B is divisible by M then also \(A-B\) is divisible by M and \(A+B\) is as well