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Consider the set $S$ of all integers of the form $x^2+y^2+4xy$, where $x$ and $y$ are integers. How could one prove the set $S$ is closed under multiplication? I have tried the bashy brute force method, but to no avail. Perhaps someone could help?

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2 Answers

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$$x^2+4xy+y^2=(7x+2y)^2-3(4x+y)^2$$

$$(a+b\sqrt3)(c+d\sqrt3)=(ac+3bd)+(ad+bc)\sqrt3$$

$$(a^2-3b^2)(c^2-3d^2)=(ac+3bd)^2-3(ad+bc)^2$$

EDIT: The 1st formula can be used to show that integers of the form $x^2+4xy+y^2$ are the same as integers of the form $r^2-3s^2$. The 3rd formula shows that the set of integers of the form $r^2-3s^2$ is closed under multiplication. The 2nd formula is just something I have to write down to get me to the 3rd formula.

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Simply awesome !! +1 – Iyengar Jan 30 '12 at 5:07
@Abhinash: Note that if $m$ and $n$ are integers, and $(x,y)$ is the solution of the system $7x+2y=m$, $4x+y=n$, then $x$ and $y$ are integers, since $7x+2y-2(4x+y)=-x$. – André Nicolas Jan 30 '12 at 6:02
Notice that your first formula is not the simplest one. And sorry for posting a similar answer without acknowledging the similarity beforehand. If at your request, I can delete it then. Thanks and regards. – awllower Mar 29 at 18:50
@awllower, no need to delete. I'm annoyed only with myself, for not finding the simpler formula. Your comment is sufficient acknowledgement of similarity. – Gerry Myerson Apr 1 at 12:04
Thanks for your generosity and appreciation. :D – awllower Apr 1 at 12:10
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Hint I:
$X^2+4XY+Y^2=(X+2Y)^2-3Y^2$.
Hint II:
Bramagupta's identity: $(p^2-3q^2)(r^2-3s^2)=(pr\pm 3qs)^2-3(ps\mp qr)^2$, for $p$, $q$ integers.
If this suffices not, tell me, or if there are mistakes. Thanks in advance.

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P.S. I used Conway's river method to derive the first hint. By that method, it is also easy to see that the transformation $(X,Y)\to (X+3Y,-Y)$ takes the form to a reduced form $X^2+2XY-2Y^2$. Moreover, we can even show that this is the only equivalent reduced form to the original one. Just for the information's sake... – awllower Mar 30 at 19:11

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