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Number theory Difficulty 7.0 National olympiad, round 2 Prove it

Let aa and bb be positive integers with gcd(a,b)=1\gcd(a, b)=1. Show that every integer greater than ababab-a-b can be expressed in the form ax+byax+by, where x,yN0x, y \in \mathbb{N}_{0}.

Solution

1. Understanding the Problem:
We are given two positive integers aa and bb with gcd(a,b)=1\gcd(a, b) = 1. We need to show that every integer greater than ababab - a - b can be expressed in the form ax+byax + by, where x,yN0x, y \in \mathbb{N}_0. This is a classic result in number theory known as the Frobenius Coin Problem for two variables.

2. Frobenius Number for Two Variables:
The Frobenius number g(a,b)g(a, b) for two coprime integers aa and bb is given by:
g(a,b)=abab g(a, b) = ab - a - b
The theorem states that any integer greater than g(a,b)g(a, b) can be expressed as ax+byax + by for non-negative integers xx and yy.

3. Proof for Two Variables:
- Since gcd(a,b)=1\gcd(a, b) = 1, by the Extended Euclidean Algorithm, there exist integers x0x_0 and y0y_0 such that:
ax0+by0=1 ax_0 + by_0 = 1
- For any integer n>ababn > ab - a - b, we can write nn as:
n=abab+kfor some integer k>0 n = ab - a - b + k \quad \text{for some integer } k > 0
- We need to show that nn can be written in the form ax+byax + by with x,yN0x, y \in \mathbb{N}_0. Consider:
n=a(b1)+b(a1)+k n = a(b-1) + b(a-1) + k
- We can rewrite nn as:
n=a(b1+k1)+b(a1+k2) n = a(b-1 + k_1) + b(a-1 + k_2)
where k1k_1 and k2k_2 are non-negative integers such that k1+k2=kk_1 + k_2 = k. This shows that nn can be expressed as ax+byax + by for non-negative integers xx and yy.

4. Extension to Three Variables:
- For three pairwise relatively prime positive integers a,b,a, b, and cc, we need to show that ax+by+cz=nax + by + cz = n has non-negative integer solutions (x,y,z)(x, y, z) for integers n>abcabbcca+a+b+cn > abc - ab - bc - ca + a + b + c.
- The Frobenius number for three variables is not as straightforward as for two variables. However, a similar result can be conjectured:
g(a,b,c)=abcabbcca+a+b+c g(a, b, c) = abc - ab - bc - ca + a + b + c
- The proof for three variables involves more complex combinatorial arguments and is beyond the scope of elementary number theory. However, it is known that for three pairwise relatively prime integers, there exists a bound beyond which every integer can be expressed as a non-negative linear combination of a,b,a, b, and cc.

5. **Generalization to kk Variables:**
- For kk pairwise relatively prime positive integers, the problem becomes even more complex. The Frobenius number for kk variables does not have a simple closed-form expression.
- However, it is known that for any set of kk pairwise relatively prime integers, there exists a bound beyond which every integer can be expressed as a non-negative linear combination of these integers.

\blacksquare

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.