Question

S is the set of numbers in the form a^2+4ab+b^2 where a and b are integers. If x and y are members of S how can you prove that xy is also in the set? This problem is giving me algebraic meltdown.

Answer 1

nice name

Answer 2

I know that: if x=a^2+4ab+b^2 and y=s^2+4st+t^2 then:
xy=a^2 s^2+4 a^2 s t+a^2 t^2+4 a b s^2+16 a b s t+4 a b t^2+b^2 s^2+4 b^2 s t+b^2 t^2
but I can't see how to make this look the correct format.

Answer 3

Divide the sets into two disjoint subsets A and B such that A consists entirely of (x+y)^2+xy 's and B consists entirely of xy's>It is clear that A +B=S.

Answer 4

That is both the sets are required to complete the set S

Answer 5

I am not getting any cogent results,

Let say, \[ x = \left(a^2+4 a b+b^2\right) \]and \[ y= \left(m^2+4 m n+n^2\right) \]

Then \( xy= a^2 m^2+4 a b m^2+b^2 m^2+4 a^2 m n\) \(+16 a b m n+4 b^2 m n +a^2 n^2+4 a b n^2+b^2 n^2 \)

Now this is quite close to \( (a m+b n){}^2 +4 (a n+b m)(a m+b n)+(a n+b m){}^2\) but we will still need \(12 a b m n\).

Answer 6

I don't follow Aron's solution I'm afraid. Could you possibly make it a bit clearer?