# Problem-Solving and Selected Topics in Number Theory: In the by Michael Th. Rassias

By Michael Th. Rassias

This booklet is designed to introduce one of the most very important theorems and effects from quantity conception whereas checking out the reader’s realizing via rigorously chosen Olympiad-caliber difficulties. those difficulties and their ideas give you the reader with a chance to sharpen their abilities and to use the idea. This framework publications the reader to a simple comprehension of a few of the jewels of quantity concept The ebook is self-contained and carefully offered. numerous features can be of curiosity to graduate and undergraduate scholars in quantity idea, complex highschool scholars and the lecturers who educate them for arithmetic competitions, in addition to to students who will take pleasure in studying extra approximately quantity concept. **Michael Th. Rassias** has obtained a number of awards in mathematical challenge fixing competitions together with gold medals on the Pan-Hellenic Mathematical Competitions of 2002 and 2003 held in Athens, a silver medal on the Balkan Mathematical Olympiad of 2002 held in Targu Mures, Romania and a silver medal on the forty fourth foreign Mathematical Olympiad of 2003 held in Tokyo, Japan.

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**Problem-Solving and Selected Topics in Number Theory: In the Spirit of the Mathematical Olympiads**

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D ⎞ ⎛ But μ(d)g d|n n = d (1) f (λ)⎠ . ⎝μ(d) · (2) λ| n d d|n At this point, we are going to express (2) in an equivalent form, where there will be just one sum at the left-hand side. In order to do so, we must ﬁnd a common condition for the sums d|n and λ| n . The desired d condition is λd|n. Hence, we get μ(d)g d|n n = d μ(d)f (λ). λd|n 18 2 Arithmetic functions Similarly, ⎞ ⎛ μ(d)⎠ = ⎝f (λ) · d| n λ λ|n μ(d)f (λ). λd|n Thus, ⎞ ⎛ d|n n = μ(d)g d μ(d)⎠ . ⎝f (λ) · (3) d| n λ λ|n However, by the previous theorem μ(d) = 1 if and only if d| n λ n = 1, λ and in every other case the sum is equal to zero.

If m = m1 m2 · · · mk , then the system of linear equations a1 x ≡ b1 (mod m1 ) a2 x ≡ b2 (mod m2 ) .. ak x ≡ bk (mod mk ) has a unique solution modulo m. Proof. At ﬁrst, we shall prove that the system of linear congruences has a solution modulo m and afterwards we shall prove the uniqueness of that solution. Set ri = m/mi . Then, it is obvious that gcd(ri , mi ) = 1 and thus, the linear congruence ri x ≡ 1 (mod mi ) has a unique solution. If ri denotes that solution, we have ri ri ≡ 1 (mod mi ), for i = 1, 2, .

In case d | c, the diophantine equation ax + by = c has no solutions. 1 Basic theorems 41 Proof. Case 1. If d | c, then there exists an integer k for which c = kd. But, because of the fact that d is the greatest common divisor of a and b, by Bezout’s Lemma we know that there exist integers k1 , k2 such that d = k1 a + k2 b and thus c = kk1 a + kk2 b. Hence, there is at least one pair of integers x0 = kk1 , y0 = kk2 which is a solution of the diophantine equation. In order to prove that there exist inﬁnitely many solutions and speciﬁcally of the form b x = x0 + n, d a y = y0 − n, d we set (x, y) to be an arbitrary solution of the diophantine equation.