By Weng A.

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Example text

Our goal is to prove the following theorem. 10) and let c = s 2 + 1. Then, the system of Diophantine equations y 2 − bx 2 = r 2 , z 2 − cx 2 = s 2 has only trivial solutions (x, y, z) = (0, ±r, ±s). 10) belonging to the same class as one of (2q 3 + q, ±q). 2 for c = ci with 2 ≤ i ≤ 7, where Linear Forms in Logarithms 45 c2 = 4r 4 + 1, c3 = (4r 3 − 4r 2 + 3r − 1)2 + 1, c4 = (4r 3 + 4r 2 + 3r + 1)2 + 1, c5 = (8r 4 + 4r 2 )2 + 1, c6 = (16r 5 − 16r 4 + 20r 3 − 12r 2 + 5r − 1)2 + 1, c7 = (16r 5 + 16r 4 + 20r 3 + 12r 2 + 5r + 1)2 + 1 (see [27, p.

96bc and N = 1021 with any choice of r and c left, we get after two steps that n < 2. Here, one may also use that D(−1)-triples {1, b, c} cannot be extended to a quadruple for r ≤ 143000, which was verified by computer. Hence, in some cases one can avoid to use reduction at all. We are still left to deal with the cases of small indices m and n in the equation z = Vm = Wn . From [21] we know that n ≥ 3; otherwise we have only the trivial solution (corresponding with an extension with d = 1, which is no real extension, because we ask for elements in D(n)-m-tuple to be distinct).

Bujaˇci´c and A. Filipin |2m − 3n | < 2m , (m − 1) log 2 < n log 3 < (m + 1) log 2, and the problem of finding a lower bound for |2m − 3n | clearly reduces to this special case. Consider the linear form Λ = 3n 2−m − 1. 14, we get log |Λ| > −c0 (1 + log m). 87 · 108 . Hence, the following theorem is proved. 1 Let m, n be positive integers. 87·10 8 This theorem enables us to find the list of all powers of 3 that increased by 5 give a power of 2. 2 The only integer solutions to the Diophantine equation 2m − 3n = 5 are (m, n) = (3, 1), (5, 3).

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Elliptic Curves (student project) by Weng A.


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