An Introduction to Diophantine Equations: A Problem-Based by Titu Andreescu

By Titu Andreescu

This problem-solving booklet is an advent to the learn of Diophantine equations, a category of equations during which purely integer strategies are allowed. the cloth is prepared in elements: half I introduces the reader to ordinary tools helpful in fixing Diophantine equations, similar to the decomposition approach, inequalities, the parametric procedure, modular mathematics, mathematical induction, Fermat's approach to endless descent, and the tactic of quadratic fields; half II includes entire suggestions to all routines partly I. The presentation gains a few classical Diophantine equations, together with linear, Pythagorean, and a few better measure equations, in addition to exponential Diophantine equations. a few of the chosen workouts and difficulties are unique or are awarded with unique solutions.

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

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Note that for (x1 , y1 , z1 ) = (1, 3, 7) and (x2 , y2 , z2 ) = (14, 39, 42) we have x21 + y12 + z12 = 59 and x22 + y22 + z22 = 592 . Define now (xn , yn , zn ), n ≥ 3, by xn+2 = 59xn , yn+2 = 59yn , zn+2 = 59zn , for all n ≥ 1. Then 2 2 x2k+2 + yk+2 + zk+2 = 592 (x2k + yk2 + zk2 ); 2 2 hence x2k + yk2 + zk2 = 59k implies x2k+2 + yk+2 + zk+2 = 59k+2 . Remark. We can write the solutions as (x2n−1 , y2n−1 , z2n−1 ) = (1 · 59n−1 , 3 · 59n−1 , 7 · 59n−1 ) and (x2n , y2n , z2n ) = (14 · 59n , 39 · 59n , 42 · 59n ), n ≥ 1.

Completing the cube, we obtain x3 − 3xy 2 + y 3 = 2x3 − 3x2 y − x3 + 3x2 y − 3xy2 + y 3 = 2x3 − 3x2 y + (y − x)3 = (y − x)3 − 3(y − x)(−x)2 + (−x)3 . This shows that if (x, y) is a solution, then so is (y − x, −x). 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.

Let (xn )n≥0 and (yn )n≥0 be two sequences defined recursively as follows: xn+2 = 3xn+1 − xn , x0 = 1, x1 = 4, yn+2 = 3yn+1 − yn , y0 = 1, y1 = 2. 1. Prove that x2n − 5yn2 = −4 for all nonnegative integers n. 52 Part I. Diophantine Equations 2. Suppose that a, b are two positive integers such that a2 − 5b2 = −4. Prove that there exists a nonnegative integer k such that xk = a and yk = b. (Vietnamese Mathematical Olympiad) Solution. We first prove by induction on k that for k ≥ 0, we have (xk+1 , yk+1 ) = For k = 0, (4, 2) = 12+10 4+6 2 , 2 3xk + 5yk xk + 3yk , 2 2 3+5 1+3 2 , 2 .

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