Lisa Jacobsen's Analytic theory of continued fractions III PDF

By Lisa Jacobsen

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E. excluding the triangle (3, 4, 5)), all other heights cannot be integers. (11) (2y − 1)x Indeed, = ∆ = yt gives (2y−1)x = 2yt (here x = h2y−1 for simplicity). Since 2 2yt is integer only if (2y − 1)|t = 3u. e. (2y − 1)(2y + 1) = 3(4u2 + 1) = 4(3u2 ) 3. Therefore (2y − 1)|3u implies (2y − 1)|3u2 , so we must have (2y − 1)|3, implying (2y + 1)z 2yt y = 2 (y > 1). For h2y+1 we have similarly = ∆ = yt, so z = , where 2 2y + 1 (2y + 1)|t = 3u ⇔ (2y + 1)|3 ⇔ y = 1 (as above). Therefore z = h2y+1 cannot be integer in all CH-triangles.

Therefore x must be odd and y even. By (1) (with d = 1) one can write x 2 = a2 + b 2 , y 2 = 2ab, z = a2 − b 2 (4) with a > b, (a, b) = 1 having distinct parities. Let a = even, b = odd. Since y 2 = 2ab, y is even, let y = 2Y . From ab = 2Y 2 and a = 2a one gets a b = Y 2 so a = p2 , b = q 2 , with 39 (p, q) = 1. We have obtained: a = 2p2 , b = q 2 , so x2 = (2p2 )2 + (q 2 )2 , y = 2pq. (5) Applying (1) once again to (5) we get x = r 2 + s2 , 2p2 = 2rs, q 2 = r 2 − s2 (6) with r > s, (r, s) = 1 having distinct parities.

D. Th´erond, L’Hypoth`ese de Fermat pour les exposants n´egatifs, L’Enseignement Math. 4. 61 3 On the equations a b c a b c + = and + = x y d x y z 1. The equation a b c + = x y d (1) where a, b, c, d are given positive integers, and x, y unknown positive integers occurs many times for particular values of a, b, c, d. e. x = kX, y = kY with (X, Y ) = 1. Then bdX = Y (ckX − ad), implying Y |bdX. Since (X, Y ) = 1, Euclid’s theorem yields Y |bd. The numbers b, d being given, there exist a finite number of such values of Y .

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