6.1. Sums of Two Squares
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Proof. If α has pure symmetric period, then α = $q0 ; q1 , q2 , . . . , q!−1 % = $q!−1 ; q!−2 , . . . , q1 , q0 % = −1/α" , where the last equality comes from Corollary 5.6 on page 230. In other words, N (α) = αα" = −1, so part (a) implies part (b). If we have that N (α) = (P 2 − D)/Q2 = −1, then
D = P 2 + Q2 ,
so part (b) implies part (c). If D = P 2 + Q2 , then N (α) = (P 2 − D)/Q2 = −1, so α = −1/α" . Therefore, by Corollary 5.6, α has pure symmetric period. Thus, part (c) implies part (a), and the logical circle is complete. ! Example 6.3 Let α = (P + Since
√
D)/Q = (9 +
α > 1, and − 1 < α" = (9 −
√
√
145)/8.
145)/8 < 0,
then α is a reduced quadratic irrational. Also,
α = $2; 1, 1, 1, 2%, so α has pure symmetric period and N (α) = (81 − 145)/64 = −1, and
145 = 12 + 122 = 82 + 92 ,
the only two (primitive) representations by Theorem 6.3 on page 247 since r2 (D) = 2 = 2d−1 given that 145 = 5 · 29. Remark 6.4 Theorem 6.4 on the facing page is one of the prettiest results to emerge from the connection between continued fractions and sums of two squares. Note that the representation given in Theorem 6.4 need not be primitive. For instance, if √ α = (5 + 50)/5,