Read e-book online An Introduction to Diophantine Equations: A Problem-Based PDF

By Titu Andreescu

This problem-solving publication is an creation to the research of Diophantine equations, a category of equations during which in simple terms integer options are allowed. the cloth is equipped in elements: half I introduces the reader to basic tools beneficial in fixing Diophantine equations, resembling the decomposition process, inequalities, the parametric process, modular mathematics, mathematical induction, Fermat's approach to endless descent, and the strategy of quadratic fields; half II comprises whole recommendations to all workouts partially I. The presentation positive aspects a few classical Diophantine equations, together with linear, Pythagorean, and a few larger measure equations, in addition to exponential Diophantine equations. some of the chosen routines and difficulties are unique or are offered with unique solutions.

An creation to Diophantine Equations: A Problem-Based Approach is meant for undergraduates, complex highschool scholars and lecturers, mathematical contest members — together with Olympiad and Putnam opponents — in addition to readers drawn to crucial arithmetic. The paintings uniquely offers unconventional and non-routine examples, rules, and techniques.

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

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Xn are all distinct odd positive integers. A simple parity argument shows that in this case n must be odd. There are several known examples of such integers n. For instance, if n = 9, we have 1 1 1 1 1 1 1 1 1 + + + + + + + + = 1; 3 5 7 9 11 15 33 45 385 42 Part I. Diophantine Equations if n = 11, 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + = 1; 3 5 7 9 15 21 27 35 63 105 135 if n = 15, 1 1 1 1 1 1 1 1 1 + + + + + + + + 3 5 7 9 15 21 35 45 55 1 1 1 1 1 1 + + + + + = 1; + 77 165 231 385 495 693 and if n = 17, 1 1 1 1 1 1 1 1 1 + + + + + + + + 3 5 7 9 15 21 35 45 55 1 1 1 1 1 1 1 1 + + + + + + + = 1.

P (n0 + s − 1) are true; • For all k ≥ n0 , P (k) is true implies P (k + s) is true. Then P (n) is true for all n ≥ n0 . Mathematical Induction (strong form): Suppose that • P (n0 ) is true; • For all k ≥ n0 , P (m) is true for all m with n0 ≤ m ≤ k implies P (k + 1) is true. Then P (n) is true for all n ≥ n0 . This method of proof is widely used in various areas of mathematics, including number theory. 5 The Method of Mathematical Induction 37 to show how mathematical induction works in studying Diophantine equations.

The two solutions are distinct, since y − x = x and −x = y lead to x = y = 0. Similarly, x3 − 3xy 2 + y 3 = x3 − 3x2 y + 3xy2 − y 3 + 2y 3 + 3x2 y − 6xy 2 = (x − y)3 + 3xy(x − y) − 3xy 2 + 2y 3 = (−y)3 − 3(−y)(x − y)2 + (x − y)3 , so (−y, x − y) is the third solution to the equation. We use these two transformations to solve the second part of the problem. Let (x, y) be a solution. Since 2891 is not divisible by 3, 32 Part I. Diophantine Equations x3 + y 3 is not divisible by 3 as well. So either both x and y give the same residue modulo 3 (different from 0), or exactly one of x and y is divisible by 3.

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