An Introduction to Diophantine Equations by Titu Andreescu

By Titu Andreescu

This problem-solving e-book is an advent to the research of Diophantine equations, a category of equations within which in basic terms integer options are allowed. The presentation gains a few classical Diophantine equations, together with linear, Pythagorean, and a few larger measure equations, in addition to exponential Diophantine equations. the various chosen workouts and difficulties are unique or are provided with unique suggestions. An advent to Diophantine Equations: A Problem-Based procedure is meant for undergraduates, complicated highschool scholars and lecturers, mathematical contest contributors ― together with Olympiad and Putnam rivals ― in addition to readers drawn to crucial arithmetic. The paintings uniquely offers unconventional and non-routine examples, rules, and methods.

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Additional info for An Introduction to Diophantine Equations

Example text

243 = (p + 3) 2 , whose only integer solution = 7. Example 3. Prove that the equation x 5 - y2 = 4 has no solutions in integers. (Balkan Mathematical Olympiad) Solution. We consider the equation modulo 11. Since (x 5 ) 2 = x 10 or 1 (mod 11) for all x, we have x 5 = -1, 0 or 1 (mod 11). So x5 - =0 4 is either 6, 7 or 8 modulo 11. However, the square residues modulo 11 are 0, 1, 3, 4, 5, or 9, so the equation has no integral solutions. Example 4. Determine all primes p for which the system of equa- tions has a solution in integers x, y.

Prove that for each integer n > 3 the equation has infinitely many solutions in positive integers. Solution. An infinite family of solutions is given by Example 5. Let a, b be positive integers. Prove that the equation has infinitely many positive integral solutions (x, y, z). {Dorin Andrica) Solution. We will use the following auxiliary result: Lemma. If A, B are relatively prime positive integers, then there exist positive integers u, v such that Au- Bv = 1 23 (1) Proof. Consider the integers (2) 1 ·A, 2 ·A, ...

K. That is, if there where an n for which P(n) was true, you could construct a sequence n > n1 > n2 > . . all of which would be greater thank, but for the nonnegative integers, no such descending is possible. Two special cases of FMID are particularly useful in the study of diophantine equations. FMID Variant 1: There is no sequence of nonnegative integers n1 > n2 > ... In some situations it is convenient to replace FMID Variant 1 by the following equivalent form: If no is the smallest positive integer n for which P(n) is true, then P(n) is false for all n < n0 .

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