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I wonder if someone could shed some light in the following question

Let $(x,y)$ denote the greatest common divisor of $x$ and $y$, and let $x_1,y_1,x_2,y_2$ be integers.

Is the following statement true?

If $(x_1,y_1)=(x_2,y_2)=(x_1^2+y_1^2,x_2^2+y_2^2)=1$, then $$(x_1x_2\pm y_1y_2, x_1y_2\mp x_2y_1)=1$$

If not, what further hypotheses are necessary to guarantee the claim?

Thanks in advance, Guillermo

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2 Answers 2

HINT $\ \: $ Interpret it in Gaussian integers $\rm\: \mathbb Z[i]\:.\: $ Define $\rm\ a + b\ i\ $ to be primitive if it has no rational prime divisor, i.e. $\rm\ (a,b) = 1\:.\:$ Then the statement says that the product of two primitive Gaussian integers is primitive if they have coprime norms. Put $\rm\ \ \alpha = x_1+y_1\ i,\ $ $\rm\ \beta = x_2+y_2\ i\:.\ $ Suppose $\rm\ p\:|\:\alpha\beta\ $ for a prime $\rm\: p\in \mathbb Z\:.\:$ If $\rm \: p\:$ is prime in $\rm\: \mathbb Z[i]\: $ then $\rm\ p\:|\:\alpha\ $ or $\rm\ p\:|\:\beta\ $ contra $\ldots\ $ Else $\rm\: p = \pi\pi'\: $ for a prime $\rm\: \pi \in \mathbb Z[i]\:$ hence $\rm\ \pi\pi'|\:\alpha\beta\ \Rightarrow\ \ldots$

Alternatively, if you are not familiar with Gaussian integer arithmetic then you may employ the Brahmagupta–Fibonacci identity for composition of squares. It is a consequence of the fact that the norm is multiplicative: $\rm\ \ \ N(\alpha)\ N(\beta)\ =\ N(\alpha\beta)\ $ i.e. $\rm\ \alpha\alpha'\beta\beta' = (\alpha\beta) (\alpha\beta)'\: $ which, rationally, is

$$\rm (x_1^2 + y_1^2)\ (x_2^2 + y_2^2)\ =\ (x_1x_2\pm y_1y_2)^2 + ( x_1y_2\mp x_2y_1)^2 $$

The upper $\pm$ signs are from $\rm\ N(\alpha)\ N(\beta')\ = \ N(\alpha\beta')\:$.

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+1 slick! –  Timothy Wagner Nov 28 '10 at 16:49

@Bill: Thanks, I see that $\pi\pi'|\alpha\beta$ implies that $p^2|N(\alpha)N(\beta)$. Since $(N(\alpha),N(\beta))=1$ we have that $p^2$ divides $N(\alpha)$ or $N(\beta)$, say $p^2|N(\alpha)$. Thus, $p$ must divide $\alpha$, a contradiction.

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