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Section5.4Using the Chinese Remainder Theorem

We will here present a completely constructive proof of the CRT (Theorem 5.3.1). That is, we will not just prove it can be done, we will show how to get a solution to a given system of linear congruences.

Keep in mind that this is a procedure that works. It may have a number of steps, but its power is not to be underestimated. After some careful examples, we'll see some other uses.

Subsection5.4.1Constructing simultaneous solutions

Remember that we are trying to solve the system of equations x\equiv a_i (mod n_i). It is important to confirm that all n_i are coprime in pairs. Then the following steps will lead to a solution. You will find basically this proof in any text; I use the notation in [C.1.1].

  1. First, let's call the product of the moduli n_1 n_2 \cdots n_k=N.

  2. Take the quotient N/n_i and call it c_i. It's sort of a “complement” to the ith modulus within the big product N.

  3. Now find the inverse of each c_i modulo n_i. That is, for each i, find a solution d_i such that \begin{equation*}c_i d_i\equiv 1\text{ (mod }n_i)\end{equation*}

    Notice that this is possible. You can't find an inverse modulo any old thing! But in this case, c_i is the product of a bunch of numbers, all of which are coprime to n_i, so it is also coprime to n_i, as required.

  4. For each i, multiply the three numbers a_i \cdot c_i \cdot d_i.

  5. Now we evaluate each of these products (indexed by i) modulo the various n_j. That looks bad, but most things cancel:

    • By definition, each c_j is divisible by n_i (except for c_i itself), so modulo n_i the product is \begin{equation*}a_j c_j d_j \equiv 0\text{ (mod }n_i)\; .\end{equation*}

    • The product \begin{equation*}a_i c_i d_i \equiv a_i \cdot 1\equiv a_i\text{ (mod }n_i)\end{equation*}

  6. Now add all these products together to get our final answer, \begin{equation*}x=a_1 c_1 d_1+a_2 c_2 d_2+\cdots +a_k c_k d_k\; .\end{equation*} For each n_i, we can do the sum modulo n_i too; the previous step shows this sum is \begin{equation*}x\equiv 0+0+\cdots +a_i +\cdots +0\text{ (mod }n_i)\; .\end{equation*} So this is definitely a solution.

  7. Any other solution x' has to still fulfill x'\equiv a_i\equiv x (mod n_i), so n_i\mid x'-x for all moduli n_i. Since all n_i are relatively prime to each other, N\mid x'-x too (if a\mid c and b\mid c and \gcd(a,b)=1, then ab\mid c). So x'\equiv x (mod N), which means x is the only solution modulo N!

Clearly this needs an example.

Example5.4.1A first CRT example

Let's look at how to solve our original system using this method.

  • x\equiv 1 (mod 5)

  • x\equiv 2 (mod 6)

  • x\equiv 3 (mod 7)

We'll follow along with each of the steps in Sage.

First, I'll make sure I know all my initial constants.

Next, I'll write down all the c_i, the complements to the moduli, so to speak. Remember, c_i=N/n_i.

Now we need to solve for the inverse of each c_i modulo n_i. One could do this by hand. For instance, \begin{equation*}42d_1\equiv 2d_1\equiv 1\text{ (mod }5)\text{ yielding }d_1=3,\text{ since }2\cdot 3=6\equiv 1\text{ (mod }5)\; .\end{equation*} But that is best done on homework for careful practice; in the text, we might as well use the power of Sage.

Now I'll create each of the big product numbers, as well as their sum.

Of course, we don't recognize 836 as our answer. But:

Let's try some more interesting moduli for an example to do on your own. Can you follow the template?

  • x\equiv 1 (mod 6)

  • x\equiv 11 (mod 35)

  • x\equiv 3 (mod 11)

Sage can also approach this in a similar way, as we saw earlier.

Subsection5.4.2A theoretical but highly important use of CRT

Now, there are many, many useful things we can do with the CRT.

That means that any question about congruences is really a question about congruences modulo simple moduli (see Proposition 6.5.1 for a strong statement of this). We will use this fact again and again in the remainder of the text, and it is a huge reason why the Chinese Remainder Theorem is so intensely powerful.