By Peter J. Cameron

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**Extra resources for A Course on Number Theory [Lecture notes]**

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We have to show that −1 < y1 < 0. Let P∗ = Qa0 − P. Then y1 = = = = = 1 √ (P + D)/Q − a0 1 √ (P + D − Qa0 )/Q 1 √ (−P ∗ + D)/Q √ P∗ + D (D − (P∗ )2 )/Q √ P∗ + D , Q∗ where Q∗ = (D − (P∗ )2 )/Q. Now Q∗ = D − (Qa0 − P)2 D − P2 = −Qa20 + 2Pa0 + , Q Q so Q∗ is an integer. Moreover, Q∗ divides D − (P∗ )2 (the quotient is just Q). So we have written y1 in the same form as y, with the same D, but maybe different P and Q. Also, we have y1 = 1/(y − a0 ), and so y1 = 1/(y − a0 ), so −1 < y1 < 0 as required.

B) If y is reduced, then 0 < P < D and 0 < Q < 2 D. √ Proof (a) We know that y = u + v d where u and v are rationals and d is squarefree. Suppose first that v is positive. Let q be the least common multiple of the denominators of u and v, and u = p/q, v = r/q. Then √ √ p+r d p + r2 d pq + q2 r2 d y= = = . q q q2 Put P = pq, Q = q2 , and D = q2 r2 d, and note that Q divides P2 − D. If u < 0, then write −y in the specified form and then replace Q by −Q. (b) Now suppose that y is reduced; recall√ that this means y > 1 and −1 < y < 0, where y is the conjugate of y (so y = (P − D)/Q).

2 If y is the value of a periodic continued fraction, then y is a quadratic irrational. Proof Let y = [a0 ; a1 , . . , am , am+1 , . . , am+k ]. Let z = [am+1 ; . √ , am+k ]. 1, z is a (reduced) quadratic irrational, say z = u + v d, where u and v are rational numbers and d is a squarefree integer. We have [a0 , . . , am , z] [a1 , . . , am , z] [a0 , . . , am ]z + [a0 , . . , am−1 ] = . [a1 , . . , am ]z + [a1 , . . , am−1 ] y = [a0 ; a1 , . . , am , z] = Let [a0 , . . , am ] = A, [a0 , .