By Titu Andreescu
This problem-solving publication is an advent to the research of Diophantine equations, a category of equations during which in simple terms integer recommendations are allowed. the fabric is prepared in components: half I introduces the reader to hassle-free equipment important in fixing Diophantine equations, reminiscent of the decomposition procedure, inequalities, the parametric technique, modular mathematics, mathematical induction, Fermat's approach to endless descent, and the tactic of quadratic fields; half II includes whole ideas to all routines partly I. The presentation good points 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 offered with unique solutions.
An creation to Diophantine Equations: A Problem-Based Approach is meant for undergraduates, complex highschool scholars and academics, mathematical contest contributors — together with Olympiad and Putnam rivals — in addition to readers attracted to crucial arithmetic. The paintings uniquely provides unconventional and non-routine examples, rules, and techniques.
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Extra info for An Introduction to Diophantine Equations: A Problem-Based Approach
Remark. For n = 1, we get the equation √ 2x1 = x2 , and since √ 2 is irrational, there is no solution in this case. For n = 2, we have x22 x23 + x21 x23 = 3x21 x22 , or equivalently, a2 + b2 = 3c2 . We can assume that the numbers a, b, and c are all diﬀerent from zero and that they are relatively prime, 44 Part I. Diophantine Equations meaning gcd(a, b, c) = 1. The square of an integer is congruent to 0 or 1 modulo 3, and hence both a and b are divisible by 3. Now, c is also divisible by 3 and we get a contradiction.
Note the solutions (0, k), k ∈ Z+ , and (1, 1). We will prove that there are no other solutions by using FMID on the prime 50 Part I. Diophantine Equations factors of x. Let p1 be a prime divisor of x and let q be the least positive integer such that p1 | 2q − 1. From Fermat’s Little Theorem we have p1 | 2p1 −1 − 1, and therefore q ≤ p1 − 1 < p1 . Let us prove now that q | x. If it didn’t, then x = kq + r, with 0 < r < q, and 2x − 1 = 2kq 2r − 1 = (2q )k · 2r − 1 = (2q − 1 + 1)k · 2r − 1 ≡ 2r − 1 (mod p1 ).
Prove that the following system of equations has inﬁnitely many solutions in positive integers: ⎧ ⎨ ax − yz − c = 0, ⎩ bx − yt + d = 0. (Titu Andreescu) 28 Part I. Diophantine Equations 6. Find all triples (x, y, z) of integers such that xy(z + 1) = (x + 1)(y + 1)z. 7. Solve in integers the equation x2 + xy = y 2 + xz. 8. Prove that the equation x3 + y 3 + z 3 + w3 = 2008 has inﬁnitely many solutions in integers. (Titu Andreescu) 9. Prove that there are inﬁnitely many quadruples (x, y, z, w) of positive integers such that x4 + y 4 + z 4 = 2002w .