By Titu Andreescu

This problem-solving e-book is an creation to the examine of Diophantine equations, a category of equations during which merely integer suggestions are allowed. The presentation good points 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 strategies. An creation to Diophantine Equations: A Problem-Based process is meant for undergraduates, complicated highschool scholars and academics, mathematical contest individuals ― 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 methods.

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**Extra info for An Introduction to Diophantine Equations**

**Example text**

2001 (Titu Andreescu) 3. Prove that the equation has no solution in integers. 4. Find all pairs of positive integers (x, y) satisfying the equation 30 5. Determine all nonnegative integral solutions {x1, x2, . x 14) if any, apart from permutations, to the diophantine equation xf + xi + · · · + x1 4 = 15999 (8th USA Mathematical Olympiad) 6. M. - Bucharest) 7. Find all triples of nonnegative integers (x, y, z) solutions to the equation (Bulgarian Mathematical Olympiad) 8. Prove that the equation 4xy- x- y = z2 has no solution in positive integers.

More precisely; we show that if (x, y, z) is a solution of (1) with x + y8 + z8 2 > 0, then (x, y, z) = (xn, Yn, Zn), where n is the unique integer such that Define the new solution (u, v, w) by the relation so that 1 < u + V8 + w8 2 < 1 + 8 + 8 2 • We have 1 > (u+v8+w8 2 )- 1 (u + V8W + W8 2 w2 )(u + V8W 2 (u 2 1 - 2vw) + (2w 2 - + W8 2 w) uv)8 + (v 2 - 2[(u- v8) 2 + (v8- W8 2 ) 2 + (w8 2 hence lu- V81, lv8- W8 2 1, lw8 2 - uw)8 2 - u) 2 ], ul are all less than or equal to -/2 > J2. > O, so x + V8 + W8 2 > 1 + 8 + 8 2 , a contradiction.

The second and third condition imply -1 < u, v < 1, which yields only the solutions (u, v, w) = (1, 0, 0) or ( -1, 1, 0). The third solution fails the first condition, so we deduce (u, v, w) = (1, 0, 0) and conclude that as desired. Exercises and Problems 1. Prove that the equation 6(6a 2 in integers except a= b = c = n + 3b2 + c2 ) = 5n2 has no solution = 0. (Asian Pacific Mathematical Olympiad) 2. Determine a positive constant c such that the equation has exactly three solutions (x, y) in positive integers.