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Inhaltsverzeichnis

Frontmatter

Diophantine Equations

Frontmatter

I.1. Elementary Methods for Solving Diophantine Equations

Abstract
Given the equation f(x 1, x 2, …, xn) = 0, we write it in the equivalent form
$$f_1(x_1, x_2, \ldots , x_n)f_2(x_1, x_2, \ldots , x_n) \cdots f_k(x_1, x_2, \ldots, x_n) = a,$$
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

I.2. Some Classical Diophantine Equations

Abstract
An equation of the form
$$a_1x_1 + \cdots + a_n x_n = c,$$
(2.1.1)
where a1, a2, …, a n , b are fixed integers, is called a linear Diophantine equation. We assume that n ≥ 1 and that coefficients a1, …, a n are all different from zero.
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

I.3. Pell-Type Equations

Abstract
In 1909, A. Thue proved the following important theorem: Let f = a n z n +a n -1z n-1+…+a 1 z+a 0 be an irreducible polynomial of degree ≥ 3 with integral coefficients. Consider the corresponding homogeneous polynomial
$$F(x,y)=y^nf\left(\frac{x}{y}\right)=a_nx^n+a_{n-1}x^{n-1}y+\cdots+a_1xy^{n-1}+a_0y^n.$$
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

I.4. Some Advanced Methods for Solving Diophantine Equations

Abstract
A field is a set k equipped with two commutative binary operations, addition and multiplication, such that
  • (k, +) is an abelian group under addition;
  • every nonzero element of k has a multiplicative inverse, and (k *, ·) is an abelian group under multiplication, where k * = k \ {0k};
  • 0k ≠ 1k;
  • the distributive law holds: (a + b)c = ac + bc for all a, b, c Є k.
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

Solutions to Exercises and Problems

Frontmatter

II.1. Solutions to Elementary Methods for Solving Diophantine Equations

Abstract
1. Solve the following equation in integers x, y :
$$x^{2} + 6xy + 8y^{2} + 3x + 6y = 2.$$
Solution. Write the equation in the form
$$(x + 2y)(x + 4y) + 3(x + 2y) = 2\quad{\rm or}\quad(x +2y)(x +4y + 3) = 2.$$
We obtain the solutions (0,−1), (3,−2), (3,−1), (6,−2).
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

II.2. Solutions to Some Classical Diophantine Equations

Abstract
Solution. Working modulo 3, we have y ≡ 1 (mod 3); hence y = 1 + 3s, s ∈ ℤ. The equation becomes 6x − 15z = −9 − 30s, or equivalently, 2x − 5z = −3 − 10s. Passing to modulo 2 yields z ≡ 1 (mod 2), i.e., z = 1 + 2t, t ∈ ℤ and x = 1 -5s + 5t. Hence the solutions are
$$(x,y,z)=(1 -5s + 5t,1 + 3s,1 +2t),\quad s,t \in \mathbb{Z}.$$
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

II.3. Solutions to Pell-Type Equations

Abstract
1. Find all positive integers n such that \(\frac{{n}({n+1})}{3}\) is a perfect square. (Dorin Andrica) Soloution. Let \(\frac{{n}({n+1})}{3} = y^2,\) which is equivalent to
$${(2n + 1)}^2 - 12y^2 = 1.$$
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

II.4. Solutions to Some Advanced Methods in Solving Diophantine Equations

Abstract
1.Solve the equation
$$x^2 + 4 = y^n,$$
where n is an integer greater than 1. Solution. For n = 2, the only solutions are (0, 2) and (0, –2). For n = 3, we have seen in Example 4 that the solutions are (2, 2), (–2, 2), (11, 5), and (–11,5). Lef now n ≥ 4. Clearly, for n even, the equation is not solvable, since no other squares differ by 4. For n odd, we may assume without loss of generality that n is a prime p ≥ 5. Indeed, if n = q k , where q is an odd prime, we obtain an equation of the same type: x 2 + 4 = (y k ) q .
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu

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