By Titu Andreescu
This problem-solving ebook is an creation to the learn of Diophantine equations, a category of equations within which in basic terms integer suggestions are allowed. the fabric is equipped in elements: half I introduces the reader to effortless tools priceless in fixing Diophantine equations, akin to the decomposition process, inequalities, the parametric technique, modular mathematics, mathematical induction, Fermat's approach to countless descent, and the tactic of quadratic fields; half II comprises whole strategies to all routines partly I. The presentation good points 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 offered with unique solutions.
An advent 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 drawn to crucial arithmetic. The paintings uniquely provides unconventional and non-routine examples, principles, and techniques.
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Extra resources for An Introduction to Diophantine Equations: A Problem-Based Approach
It follows that gcd(mn, m + n) = 1. , d = k(m + n), k ∈ Z+ . The solutions to the equation are given by x = km(m + n), y = kn(m + n), z = kmn, where k, m, n ∈ Z+ . Remark. 3 The Parametric Method 23 then a + b is a square. Indeed, k = 1, a = m(m + n), b = n(m + n), and hence a + b = (m + n)2 . (2) If a, b, c are positive integers satisfying 1 1 1 + = , a b c then a2 + b2 + c2 is a square. Indeed, a2 + b2 + c2 = k2 m2 (m + n)2 + n2 (m + n)2 + m2 n2 = k2 (m + n)4 − 2mn(m + n)2 + m2 n2 2 = k 2 (m + n)2 − mn .
Solution. (a) For (i), consider the family xk = m(m2 + n2 )k , yk = n(m2 + n2 )k , zk = 2k + 1, k ∈ Z+ . For (ii), consider the family xk = |m2 − n2 |(m2 + n2 )k−1 , yk = 2mn(m2 + n2 )k−1 , zk = 2k, k ∈ Z+ . (b) Since 22 + 32 = 13, we can take m = 2, n = 3 and obtain the families of solutions xk = 2 · 13k , xk = 5 · 13k−1 , yk = 3 · 13k , zk = 2k + 1, yk = 12 · 13k−1 , zk = 2k, k ∈ Z+ ; k ∈ Z+ . Remarks. (1) Taking into account Lagrange’s identity (a2 + b2 )(c2 + d2 ) = (ac − bd)2 + (ad + bc)2 , we can generate an inﬁnite family of solutions by deﬁning recursively the sequences (xk )k≥1 , (yk )k≥1 as follows: ⎧ ⎨ x k+1 = mxk − nyk , ⎩ y = nx + my , k+1 k k 22 Part I.
Diophantine Equations 2. Find all pairs (x, y) of positive integers for which x2 − y! = 2001. (Titu Andreescu) 3. Prove that the equation x3 + y 4 = 7 has no solution in integers. 4. Find all pairs (x, y) of positive integers satisfying the equation 3x − 2y = 7. 5. Determine all nonnegative integral solutions (x1 , x2 , . . , x14 ) if any, apart from permutations, to the Diophantine equation x41 + x42 + · · · + x414 = 15999. (8th USA Mathematical Olympiad) 6. Find all pairs (x, y) of integers such that x3 − 4xy + y 3 = −1.
An Introduction to Diophantine Equations: A Problem-Based Approach by Titu Andreescu