# Thread: [SOLVED] Prove: For all integers a and b, if a|b then a^2|b^2?

1. ## [SOLVED] Prove: For all integers a and b, if a|b then a^2|b^2?

I dont really know where to get start on this, all I can really figure out is:

a|b <=> b = a (d)

a^2|b^2 <=> b^2 = a^2 (d)

therefore there is some common d that is in a|b and a^2|b^2

I kinda doubt this is enough for a proof or if it really even proves anything really.

Can someone help me out here?

2. Originally Posted by Kitizhi
I dont really know where to get start on this, all I can really figure out is:

a|b <=> b = a (d)

a^2|b^2 <=> b^2 = a^2 (d)

therefore there is some common d that is in a|b and a^2|b^2

I kinda doubt this is enough for a proof or if it really even proves anything really.

Can someone help me out here?
from $\displaystyle a|b$ implies $\displaystyle b=ad$ for some integer $\displaystyle d$ we have $\displaystyle b^2=(ad)^2 = a^2d^2$ and $\displaystyle d^2$ is also an integer.. thus $\displaystyle a^2|b^2$..