By Titu Andreescu

This problem-solving e-book is an advent to the examine of Diophantine equations, a category of equations within which in basic terms integer options are allowed. 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. a number of the chosen workouts and difficulties are unique or are offered with unique strategies. An creation to Diophantine Equations: A Problem-Based strategy is meant for undergraduates, complex highschool scholars and academics, mathematical contest contributors ― together with Olympiad and Putnam opponents ― in addition to readers drawn to crucial arithmetic. The paintings uniquely provides unconventional and non-routine examples, principles, and methods.

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**Example text**

Xn = n. Remark. Substituting back into the equation yields the identity 1 + 1 · 1! + 2 · 2! + · · · + (n- 1) · (n- 1)! = n! Example 2. Solve in positive integers the system of equations (Titu Andreescu) Solution. The inequalities cannot be both true, because adding them up would yield a contradiction. So at least one of the inequalities x 2 + 3y < (x + 2) 2 and y 2 + 3x < (y+2) 2 is true. Without loss of generality, assume that x 2 +3y < (x+2) 2 . Then x 2 < x 2 + 3y < (x + 2) 2 implies x 2 + 3y = (x + 1) 2 or 3y = 2x + 1.

Thus k1 - k2 < = 0. It is not difficult to see that k · A ¢. 0 (mod B) for all k E {1, 2, ... , B - 1}. e. there exist u E {1, 2, ... , B - 1} and v E Z+ such that A · u = B · v + 1. D Remark. e. uo (and vo) is minimal. Then all solutions in positive integers to the equation {1) are given by Um = Uo + Bm, Vm = Vo + Am, m E z+ (3) Coming back to the original problem, let us consider the sequence (Yn)n~l, given by (4) 24 Clearly gcd(yn, Yn+d = 1, n E Z+· From the above Lemma, there exist sequence of positive integers (un)n~l, (vn)n~l such that From (4) we obtain buny~ + (aun- Vn)Yn + Un- 1 = 0, n E Z+ (5) Regarding (5) as a quadratic equation in Yn and taking into account that Yn E Z+, it follows that the discriminant is· a perfect square.

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. In fact, 2 2 Yn) + (1Xn =f Yn) = 2(7 2 + 2) = 2n+l 7 (Xn ± Xn Yn , 2 2 . l y one of t h e numb ers Xn + Yn and lxn-_Ynl 1s . o dd (as P rec1se 2 2 Xn sum is the larger of Xn and Yn, which is odd).