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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

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I have updated your post to LaTeX. Please see that the updates are correct. – Jeel Shah Jan 20 '14 at 12:48
@hardmath Fixed! Thanks for the catch! – Jeel Shah Jan 20 '14 at 13:01
up vote 6 down vote accepted

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).

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Thank you so much, now I can finally sleep with this homework done. – skatter Jan 20 '14 at 12:54
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 1 at 18:16


Using Brahmagupta-Fibonacci Identity,

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

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

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But how does that help to make it of the form $a^2 + b^2 + 4ab$? – Airdish Mar 3 at 6:57
@TheOddbodNumber, Write $$a^2+4ab+b^2=(a+2b)^2-3b^2$$ and $m=3$ – lab bhattacharjee Mar 3 at 7:01
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 3 at 7:04
@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 3 at 7:17
Yes that's what I did, the expression I gave above is reduced. – Airdish Mar 3 at 7:19

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