By Titu Andreescu

This problem-solving e-book is an creation to the research of Diophantine equations, a category of equations within which merely integer strategies are allowed. The presentation gains a few classical Diophantine equations, together with linear, Pythagorean, and a few greater measure equations, in addition to exponential Diophantine equations. some of the chosen workouts and difficulties are unique or are awarded with unique options. An creation to Diophantine Equations: A Problem-Based process is meant for undergraduates, complex highschool scholars and academics, mathematical contest individuals ― together with Olympiad and Putnam rivals ― in addition to readers attracted to crucial arithmetic. The paintings uniquely provides unconventional and non-routine examples, rules, and methods.

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**Additional resources for An Introduction to Diophantine Equations: A Problem-Based Approach**

**Example text**

2n−2 , 2n−2 + 1, 2n−2 2n−2 + 1 . 40 Part I. Diophantine Equations Indeed, 1 1 1 1 1 + 2 + · · · + n−2 + n−2 + n−2 n−2 2 2 2 2 +1 2 (2 + 1) n−2 1 1 2 + n−2 n−2 = 1 − n−2 + n−2 n−2 2 2 (2 + 1) 2 (2 + 1) 1 1 = 1 − n−2 + n−2 = 1. 2 2 (3) Another way to construct solutions to equation (1) is to consider the sequence am+1 = a1 · · · am + 1, a1 = 2, m ≥ 1. Then for all n ≥ 3, 1 1 1 1 + + ··· + + = 1. a1 a2 an−1 an − 1 Indeed, from the recurrence relation it follows that ak+1 − 1 = ak (ak − 1), k ≥ 1, or 1 ak+1 − 1 = 1 1 − , ak − 1 ak k ≥ 1.

For example, from 1 1 1 + + =1 2 3 6 and a = 7, we obtain the solution 2, 3, 7, 72 , . . , 7n−3 , 6 · 7n−3 , n ≥ 4, while from 1 1 1 1 + + + =1 2 3 7 42 we get 2, 3, 7, 43, 432 , . . , 43n−4 , 42 · 43n−4 , n ≥ 5. From the construction above it follows that equation (1) has inﬁnitely many families of solutions with distinct components. (6) It is not known whether there are inﬁnitely many positive integers n for which equation (1) admits solutions (x1 , x2 , . . , xn ), where x1 , x2 , . . , xn are all distinct odd positive integers.

Moreover, if m = n, then necessarily m = n = 1. 6 Fermat’s Method of Inﬁnite Descent (FMID) 51 which shows that (n − m, m) satisﬁes the same relation and 0 < n − m ≤ m. By FMID Variant 2, the transformation (m, n) → (n−m, m) must terminate after ﬁnitely many steps, and it terminates only when m = n = 1. Hence all pairs of numbers satisfying the relation are obtained from (1, 1) by applying the inverse transformation (m, n) → (n, m + n) several times: (1, 1) → (2, 1) → (3, 2) → (5, 3) → · · · . The components of all such pairs are Fibonacci numbers Fn , where the sequence (Fn )n≥0 is deﬁned by F0 = 0, F1 = 1 and Fn+1 = Fn + Fn−1 , n ≥ 1.