Exercise 3.6.7 (Number of ordered pairs $(x,y)$ of integers for which $0

Question

Suppose that $n$ is neither a perfect square nor twice a perfect square. Show that the number of ordered pairs $(x,y)$ of integers for which $0 < x < y$ and $x^2 + y^2 = n$ is $R(n)/8$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}} \begin{proof} In the solution of Problem 5, we prove that if $n$ is not a perfect square, then $$\mathrm{Card}\ T = R(n)/4,$$ where $$T = \{(x,y) \in \Z^2 \mid x^2 + y^2 = n, \ x>0, y>0\}.$$ Suppose moreover that $n$ is not twice a perfect square, so that there is no solution $(x,y)$ such that $x=y$. Then $T = T_1 \cup T_2$, where \begin{align*} T_1 & = \{(x,y) \in \Z^2 \mid x^2 + y^2 = n,\ 0 < x < y\},\ T_2 & = \{(x,y) \in \Z^2 \mid x^2 + y^2 = n,\ 0 < y < x\}. \end{align*} Consider the maps $$ \varphi \left\{ \begin{array}{ccl} T_1 & o & T_2\ (x,y) & \mapsto &(y,x) \end{array} ight. \qquad \varphi \left\{ \begin{array}{ccl} T_1 & o & T_2\ (x,y) & \mapsto &(y,x) \end{array} ight. $$ Then $\psi \circ \varphi = 1_{T_1},\ \varphi \circ \psi = 1_{T_2}$, thus $\varphi$ is bijective, which proves $|T_1| = |T_2|$. Therefore $|T_1| = |T|/2 = R(n)/8$. The number of ordered pairs $(x,y)$ of integers for which $0 < x < y$ and $x^2 + y^2 = n$ is $R(n)/8$. \end{proof}

Exercise 3.6.7 (Number of ordered pairs $(x,y)$ of integers for which $0 < x < y$ and $x^2 + y^2 = n$)

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Exercise 3.6.7 (Number of ordered pairs $(x,y)$ of integers for which $0 < x < y$ and $x^2 + y^2 = n$)

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