Question

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

I've tried a few different things, but the algebra always gets out of hand. Any help would be appreciated!

Answer 1

Let r be real number, I will show that there are two numbers a and b so
that
\[
a^2 + b^2 + 4 a b= r
\]
Solve the above equation for a in terms of b and r using the quadratic formula.
You get
\[
a= -2b-\sqrt{3 b^2+r}\\
a= -2b+\sqrt{3 b^2+r}\\

\]
choose b so that \( b^2 + r >0\) and take one of the solutions above for a.
This choice of a and b gives that
\[
a^2 + b^2 + 4 a b= r
\]
This proves that A is equal to all real numbers and therefore it closed under multiplication.

Answer 2

Ah, I left out an important bit - a and b are integers.

Answer 3

This fundamentally changes the problem :(. Sorry about that.

Answer 4

Can they be negative?

Answer 5

Yes.

Answer 6

Here is a solution to your problem

Answer 7

Let
\[
w=(a^2 + b^2 + 4 a b)( c^2 +d^2+4 cd)\\
w=a^2 c^2 + 4 a^2 c d + a^2 d^2 + 4 a b c^2 + 16 a b c d + 4 a b d^2 +
b^2 c^2 + 4 b^2 c d + b^2 d^2\\
\]
We need to find x and y so that
\[
w= x^2 + y^2 + 4 x y
\]
Use the quadratic formula to find x in terms of w and y.

Answer 8

You find
\[

x=-2 y\pm\sqrt{w+3 y^2}
\]
If we can choose y so that \( w + 3 y^2\) is a perfect square, then we will be done
Take
y=b c + a d + 4 b d
z=a c + 2 b c + 2 a d + 7 b d
then
\[
w + 3 y^2= z^2
\]
Hence one can take
\[ x =-2y \pm z\\
x=-2(b c + a d + 4 b d) \pm z=\\-2(b c + a d + 4 b d)\pm ( a c + 2 b c + 2 a d + 7 b d)\\
x= ac - bd \\or\\
x=-a c - 4 b c - 4 a d - 15 b d

\]
For both cases, you can verify
that
\[
x^2 + y^2 + 4 x y =\\a^2 c^2 + 4 a b c^2 + b^2 c^2 + 4 a^2 c d + 16 a b c d + 4 b^2 c d +
a^2 d^2 + 4 a b d^2 + b^2 d^2

\]
This shows that your set is closed