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?
Proving the multiplicativity of a binary quadratic form
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$\begingroup$
quadratic-forms
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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0Thanks $f$or your generosity and appreciation. :D – 2013-04-01
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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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0P.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... – 2013-03-30