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I am trying to prove a version of the Chinese Remainder Thoerem using the pigeonhole principle.

The theorem that was provided:

If $n$ and $m$ are relatively prime, then for all integers $0 ≤a< n$ and $0 ≤ b < m,$ there is an integer solution to the equation set $x\equiv a\pmod n$ and $x\equiv b\pmod m$.

Note that those are equal signs and not congruency, but in this case, I don't think that should impact it. I started with trying to rearrange the statements to maybe make more sense of them (so it became $x-nc=a$ which became $x=a+nc$ where $c$ is whatever constant needed to satisfy the equation due to the definition of modulo), however, from here I'm stuck.

Specifically with the pigeonhole theorem, I'm unsure how to use it in this case. I was also given the hint to consider the sequence $b, b+m, b+2m, ..., b+(n-1)m,$ but unsure how to consider it.

Thanks for any help you can give.

Thomas Finley
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user2494584
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  • That finite sequence is a complete residue class modulo $n$. So $b+km \text{ mod } n = a$ for some $k$, but by construction equals $b$ modulo $m$ – David P Feb 10 '14 at 00:28
  • By the Theorem here, $,\gcd(n,m)=1\Rightarrow f(x) = mx+b \pmod{n},$ is $1$-to-$1$ so onto, by Pigeonhole. – Bill Dubuque Jun 21 '21 at 21:47

2 Answers2

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From the hint:

$$b, b+m, b+2m,..., b+(n-1)m$$

forms a complete residue class modulo $n$. To see this, first consider the sequence

$$0,m, 2m, 3m, ..., (n-1)m$$

Since $(m,n)=1$, we have $xm \equiv ym \text{ mod n} \iff x \equiv y \text{ mod n}$, since $m$ has a mutiplicative inverse modulo $n$. Hence each of these elements in the sequence are unique. Adding $b$ does nothing more than cyclically permute the residue class.

By the pigeonhole principle, for any $0\le a < n$, we must have $b+km=a \text{ mod n}$ for some $k\in\{0,...,n-1\}$. Then $x=b+km$ satisfies the property you want.

David P
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This is a particular and a "partial" case of the Chinese Remainder Theorem. The problem also appeared in the book "Topics in Algebra " by I.N Herstein in the chapter of Ring Theory. But we can do it in a much elegant way. Here is a basic solution to the problem:

We consider the numbers $a,a+2m,a+3m,\cdots,a+(n-1)m.$

We note that this is a complete residue system congruent modulo $n.$

Definition: A set $S$ of integers is called a complete residue system congruent modulo $n$ (, where $n\in\Bbb Z$) if every $y\in \Bbb Z$ is congruent to one and only one numbers from the set $S.$ For example, $S=\{0,1,2,...,n-1\}$ is a complete residue system congruent modulo $n$ as, ever integer $y$ leaves a remainder $r\in S$ when divided by $n$ i.e $y=nq+r$ so, $y\equiv r(\in \Bbb Z)\pmod n.$ Also, as this $r$ is unique( for a number cannot leave two different remainders!) so, this proves that $S$ is a complete residue system. Also, this set $S$ is not unique. The set, $S'=\{1,2,...,n\}$ may also serve as a complete residue system modulo $n.$ (Why?)

Lemma: A set of $n$ integers is called a complete residue system congruent modulo $n$ iff no two of those $n$ numbers are congruent to each other modulo $n.$ (Prove!)

(For a more detailed discussion on these above facts please refer to Chapter -2 of the book, "An Introduction to the Theory of Numbers" by Niven,Zuckerman and Montgomery.)

Ok, so we come back to the problem. The set $a,a+2m,a+3m,\cdots,a+(n-1)m $ is a complete residue system congruent modulo $n.$ This is because, it satisfies the lemma above.

To prove this, we need to show that no two of these $n$ numbers are congruent modulo $n.$ Let us assume the contrary, for the sake of contradiction.

Say, $p=a+k_1m,q=b+k_2m\in \{a,a+2m,a+3m,\cdots,a+(n-1)m\}$ and $p\neq q$(i.e $k_1\neq k_2$). If $p\equiv q\pmod n$ then it means that $$a+k_1m\equiv a+k_2m\pmod n\implies n|m(k_1-k_2)\implies n|(k_1-k_2),$$ as $\gcd (a,m)=1.$ But $k_1,k_2\in\{0,1,...,n-1\}$ so this pissible only if, $k_1=k_2$, a contradiction.

So, indeed, $a,a+2m,a+3m,\cdots,a+(n-1)m $ is a complete residue system congruent modulo $n.$ So, each of these $n$ numbers leaves a remainder of $\{0,1,2,...,n-1\}$ in some order.

Given, $a,b\in \Bbb Z$ so, $b\equiv a+km(=x)\pmod n,$ where $k\in\{0,1,2,...,n-1\}.$ This means, $x=a+km=b+nk'$ where $k'\in\Bbb Z.$

Now, we observe, that, $x=a+km$ is a solution to $x\equiv a\pmod m$ and $x\equiv b\pmod n.$ This completes the proof.

Thomas Finley
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