If both integers $x$ and $y$ can be represented as $a^2 + b^2 + 4ab$, prove that $xy$ can also be represented like this ...

Question

There is a set $Q$ which contains all integral values that can be represented by $$a^2 + b^2 + 4ab$$, where $a$ and $b$ are also integers. If some integers $x$ and $y$ exist in this set, prove that $xy$ does too.

I really have no idea how I can go about solving this. I tried simple multiplication of the two assuming one to be $(a^2 + 4ab + b^2)$ and other as $(c^2 + 4cd + d^2)$ but ultimately it leads to a long equation I can make no tail of :/

Any help whatsoever would be greatly appreciated

I have updated your post to LaTeX. Please see that the updates are correct. — Jeel Shah, Jan 20 '14 at 12:48

score 7 · Accepted Answer · answered Jan 20 '14 at 12:52

7

Since $a^2+b^2+4ab=(a+2b)^2-3b^2$, your numbers are exactly the numbers of the form $x^2-3y^2$. Now $x^2-3y^2$ is the norm of the algrebraic number $x+y\sqrt{3}$, so you have the identity

$$ (x^2-3y^2)(u^2-3v^2)=(xu+3yv)^2-3(xv+yu)^2 $$

(multiplicativity of norms).

answered Jan 20 '14 at 12:52

Ewan Delanoy

61,600

Thank you so much, now I can finally sleep with this homework done. – skatter Jan 20 '14 at 12:54
3

To make the resulting identity explicit in terms of $a, b, c, d$, if $f(x,y) = x^2 + 4xy + y^2$, then $$f(ac-bd,ad+4bd+bc) = f(a,b) f(c,d).$$ – heropup Jan 20 '14 at 13:13
But how is that last expression of the form $a^2 + 4ab + b^2$? – Airdish Mar 01 '16 at 18:16

score 1 · Answer 2 · answered Jan 20 '14 at 17:07

1

Generalization:

Using Brahmagupta-Fibonacci Identity,

$$(a^2+nb^2)(c^2+nd^2)=(ac\pm nbd)^2+n(ad\mp bc)^2$$

$$n=-m\implies (a^2-mb^2)(c^2-md^2)=(ac\mp mbd)^2-m(ad\mp bc)^2$$

answered Jan 20 '14 at 17:07

lab bhattacharjee

274,582

But how does that help to make it of the form $a^2 + b^2 + 4ab$? – Airdish Mar 03 '16 at 06:57
@TheOddbodNumber, Write $$a^2+4ab+b^2=(a+2b)^2-3b^2$$ and $m=3$ – lab bhattacharjee Mar 03 '16 at 07:01
I tried it and this is what it came to: $(a+2b)^2(c+2d)^2 + 9b^2d^2-3d^2(a+2b)^2-3b^2(c+2d)^2$ – Airdish Mar 03 '16 at 07:04
@TheOddbodNumber, $${(a+2b)^2-3b^2}{(c+2d)^2-3d^2}={(a+2b)(c+2d)\pm3cd}^2-3{(a+2b)c\mp b(c+2d)}^2=\cdots$$ – lab bhattacharjee Mar 03 '16 at 07:17
Yes that's what I did, the expression I gave above is reduced. – Airdish Mar 03 '16 at 07:19
Is it not $((a+2b)(c+2d) \mp 3cd)^2$? You've written $\pm$. – Airdish Mar 03 '16 at 07:21

If both integers $x$ and $y$ can be represented as $a^2 + b^2 + 4ab$, prove that $xy$ can also be represented like this ...

2 Answers2

Linked