# The Diophantine Frobenius Problem by Jorge L. Ramírez Alfonsín

By Jorge L. Ramírez Alfonsín

In the course of the early a part of the final century, Ferdinand Georg Frobenius (1849-1917) raised he following challenge, often called the Frobenius challenge (FP): given really major optimistic integers a1,...,an, locate the most important normal quantity (called the Frobenius quantity and denoted by means of g(a1,...,an) that's not representable as a nonnegative integer blend of a1,...,an, .

At first look FP may possibly glance deceptively really expert. but it vegetation up time and again within the so much unforeseen areas and has been super important in investigating many various difficulties. a couple of equipment, from numerous parts of arithmetic, were utilized in the desire of discovering a formulation giving the Frobenius quantity and algorithms to calculate it. the most purpose of this e-book is to spotlight such equipment, principles, viewpoints and functions to a broader viewers.

Similar mathematics books

The Mathematics of Paul Erdos II (Algorithms and Combinatorics 14)

This is often the main complete survey of the mathematical lifetime of the mythical Paul Erd? s, the most flexible and prolific mathematicians of our time. For the 1st time, the entire major parts of Erd? s' learn are lined in one undertaking. as a result of overwhelming reaction from the mathematical neighborhood, the undertaking now occupies over 900 pages, prepared into volumes.

Additional info for The Diophantine Frobenius Problem

Example text

2 [100] There is no ﬁnite set of polynomials {h1 , . . , hn } such that for each choice of a1 , a2 , a3 there is some i such that hi (a1 , a2 , a3 ) = g(a1 , a2 , a3 ). Proof. H = n i=1 (hi (X1 , X2 , X3 ) − Y ) would vanish on A. An explicit general formula for computing g(a1 , a2 , a3 ) can be found. Let L1 , L2 and L3 be the smallest positive integers such that there exist integers xij ≥ 0, 1 ≤ i, j ≤ 3, i = j with L1 a1 = x12 a2 + x13 a3 , L2 a2 = x21 a1 + x23 a3 , L3 a3 = x31 a1 + x32 a2 .

1, we need the following Proposition. 2 Let bi for each i = 1, . . , n and ¯bi for each i = 1, . . , n be the integers as deﬁned in Procedure A. Then, g(¯b1 , . . , ¯bn+1 ) = 4g(b1 , . . , bn ) + 1. Proof. Let g be an integer such that g > 4g(b1 , . . , bn ) + 1. Let g = g − ¯bn+1 where ≡ g (mod 2). If = 0 then g = g > 4g(b1 , . . , bn ) + 1 > 2g(b1 , . . , bn ). Otherwise, if = 1 then g = g − ¯bn+1 > 4g(b1 , . . , bn ) + 1 − (2g(b1 , . . , bn ) + 1) = 2g(b1 , . . , bn ). Hence, g > 2g(b1 , .

An = dG d d G(a1 , . . , an ) = dG n−1 holds. Notice that G(a1 , . . , an ) = i=1 ai xi with xi > 0 (this follows n−1 ai xi + an xn from the fact that we can write an + G(a1 , . . , an ) = i=1 n−1 with xi > 0 and thus G(a1 , . . , an ) = i=1 ai xi + an (xn − 1) that contradicts the deﬁnition of G unless xn = 1). Let ai = dai , i = 1, . . , n − 1. Hence, n−1 G= n−1 ai xi = d i=1 ai xi and G is divisible by d, say G = dG . 2) i=1 It is clear that G cannot be expressed as a linear combination of n−1 ai yi a1 , .