Exercise 3.6.6 (Number of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$)

Question

Suppose that $n>1$. Show that the number of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$ is $r(n)/4$.

Richard Ganaye · Accepted Answer

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\begin{proof} As in Problem 5, there are as many primitive solutions of $x^2 + y^2 = n$ in the four quadrants. If $(a,0)$ (or $(0,b)$) is a solution on the $x$ or $y$ axes, they are not primitive solutions, because $a \wedge 0 = |a| >1$ (since $n>1$). So the number $r(n)$ of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$ is given by
$$r(n) = 4 \, \mathrm{Card}\, \{(x,y) \in \Z^2 \mid x^2 + y^2 = n,\ x\wedge y = 1,\ x>0,\ y>0 \}.$$
The number of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$ is $r(n)/4$.

\end{proof}

Exercise 3.6.6 (Number of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$)

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Exercise 3.6.6 (Number of ordered pairs $(x,y)$ of relatively prime positive integers for which $x^2 + y^2 = n$)

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