This paper presents a short but non-obvious and interesting theorem in Number Theory that I originally discovered while working on a problem. This theorem states that bcbcbc - b - c is the largest number which cannot be written as mb+ncmb + nc. Given all b,c,mb, c, m and nNn \in \mathbb{N}. In this article I prove the above statement and also show a problem where this theorem could be directly applied to considerably make the problem easier.

Introduction And Statement Of Result

I created the theorem I will present below while working on a problem involving conductors of natural numbers. Here we refer to the “conductor” of two natural numbers as computing the product of the two numbers and subtracting each of the original numbers from it. Simply enough, in a mathematical form we say:

Conductor(x,y)=xyxyConductor(x, y) = xy - x - y

Using the below theorem which I originally created just to solve the problem made the problem significantly easier for me and in this paper I also show how this also proved to be super useful for other problems too. I hope seeing this to be used for a wide variety of problems.

Theorem 1

The conductor of two natural numbers, bcbcbc - b - c, is the largest number which cannot be written as mb+ncmb + nc. Given all b,c,mb, c, m and nNn \in \mathbb{N}.

Proof. Without loss of generality let us assume b<cb < c

0,c,2c,3c,4c(b2)c,(b1)c0, c, 2c, 3c, 4c \ldots (b-2)c, (b-1)c

is an exhaustive set of residues for modulo bb

Let us take some rr for which:

r>(b1)cbr > (b-1)c-b

then,

rncbr \equiv nc \mid b

for some nn such that 0n(b1)0 \leq n \leq (b-1)

Now it is obvious that a number larger than bcbcbc-b-c can not be written as mb+ncmb +nc

Note: We are yet to prove that bcbcbc-b-c can not be written in this form, we have only proved for numbers larger than bcbcbc-b-c .

Let us now see for the number bcbcbc-b-c itself.

If bcbc=mb+ncbc-b-c = mb+nc

this     nb1\implies n \geq b-1

And this leads to:

mb+ncnc(b1)c>bcbcmb+nc \geq nc \geq (b-1) \cdot c > bc-b-c

Which is a contradiction.

Usage

We will now see a problem where using this theorem makes it significantly easier. This problem is from The International Mathematical Olympiad (IMO) 1983 Day 1 Problem 3 1.

Problem

Let a,ba,b and cc be positive integers, no two of which have a common divisor greater than 11. Show that 2abcabbcca2abc-ab-bc-ca is the largest integer which cannot be expressed in the form xbc+yca+zabxbc+yca+zab, where x,y,zx,y,z are non-negative integers.

Using Theorem 1 we use the same starting argument:

0,bc,2bc,3bc,4bc(a2)bc,(a1)bc0, bc, 2bc, 3bc, 4bc \ldots (a-2)bc, (a-1)bc

is an exhaustive set of residues for modulo aa

Given N>2abcabbccaN > 2abc - ab - bc - ca

we may take

xbc=N(moda)xbc = N \pmod a

with 0x<a0 \leq x < a

But:

\[\begin{align*} N - xbc &> 2abc - ab - bc - ca - (a-1)bc \\ &= abc - ab - ca \\ &= a(bc - b - c) \end{align*}\]

So,

Nxbc=ka with k>bcbcN - xbc = ka \text{ with } k > bc - b - c

Hence we can find non-negative y,zy, z so that:

k=zb+yck = zb + yc

So,

N=xbc+yca+zabN = xbc + yca + zab

Finally, we show that for N=2abcabbccaN = 2abc - ab - bc - ca we cannot find non-negative x,y,zx, y, z

so that N=xbc+yca+zabN = xbc + yca + zab

Nbc(moda)N \equiv -bc \pmod a

so we must have x1(moda)x \equiv -1 \pmod a and hence xa1x \geq a-1

Similarly, y=>b1y => b-1, and zc1z \geq c-1

Hence,

xbc+yca+zab3abcabbcca>Nxbc + yca + zab \geq 3abc - ab - bc - ca > N

Contradiction.

Data availability

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

References

  1. International Mathematical Olympiad (IMO). The international mathematical olympiad (imo) 1983 problems, 1983. URL: https://www.imo-official.org/year_info.aspx?year=1983.