Download PDF by Titu Andreescu: An Introduction to Diophantine Equations

By Titu Andreescu

This problem-solving booklet is an advent to the research of Diophantine equations, a category of equations during which in simple terms integer options are allowed. 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. a few of the chosen workouts and difficulties are unique or are offered with unique suggestions. An advent to Diophantine Equations: A Problem-Based procedure is meant for undergraduates, complicated highschool scholars and academics, mathematical contest members ― together with Olympiad and Putnam rivals ― in addition to readers attracted to crucial arithmetic. The paintings uniquely offers unconventional and non-routine examples, rules, and strategies.

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The following examples are meant to show how mathematical induction works in studying diophantine equations. Example 1. Prove that for all integers n > 3, there exist odd positive integers x, y, such that 7x 2 + y 2 = 2n. (Bulgarian Mathematical Olympiad) 32 Solution. We will prove that there exist odd positive integers Xn, Yn such that 7x~ + y~ = 2n, n > 3. For n = 3, we have integer n > 3 we have odd integers Xn, Yn satisfying 7x~ X3 = Y3 = 1. Now suppose that for a given + y~ = 2n. We shall exhibit a pair (xn+b Yn+d of odd positive integers such that 7x~+l + Y~+l = 2n+l.

Prove that there are infinitely many quadruples of positive integers (x, y, u, v) such that xy + 1, xu+ 1, xv + 1, yu + 1, yv + 1, uv + 1 are all perfect squares. 4. The Modular Arithmetic Method In many situations simple modular arithmetic considerations are employed in proving that certain diophantine equations are not solvable or in reducing the range of their possible solutions. Example 1. Show that the equation (x + 1) 2 + (x + 2) 2 + · · · + (x + 2001) 2 = y2 is not solvable. Solution. Let x = z - 1001.

Is p If p = 3, then q5 < 27, If q = 3, we obtain p3 - which is not possible. 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.

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