Question

If A is the set of all numbers a^2+4ab+b^2, prove that A is closed under multiplication.

No idea which direction to take. Any help would be appreciated!

Answer 1

take two arbitrary polynomials of the above form, then multiply them together, the resulting polynomial should be of the above form as well. Since all three of the polynomials are fully arbitrary up to the above form, this result should hold for all polynomials of the above form. QED, I suppose. If you want to get fancy, you can use the induction hypothesis form and factor into prime polynomials, but that doesn't seem necessary.

Answer 2

That's what I tried, but it got pretty messy pretty quickly. I also tried changing around the form of the polynomial to the following two forms to no avail:
\[a^2+4ab+b^2=(a+b)^2+2ab\]\[a^2+b^2=2(a+b)^2-(a^2+b^2)\]

Answer 3

I was kicking around trying this by induction, but I'm not sure how to set it up if I have two arbitrary members of A.