## anonymous 4 years ago 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.

1. anonymous

nice name

2. anonymous

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.

3. anonymous

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.

4. anonymous

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

5. anonymous

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 $$12 a b m n$$.

6. anonymous

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