Question

All type of linear congruence problem solution higher power

Answer 1

A linear congruence is the problem of finding an integer x satisfying

ax ≡ b (mod m)

for specified integers a, b, and m. This problem could be restated as finding xsuch that

the remainder when ax is divided by m is b,a*x % m == b in C code,ax – b is divisible by m, orin base m, ax ends in b.

Two solutions are considered the same if they differ by a multiple of m. (It’s easy to see that x is a solution if and only if x + km is a solution for all integers k.)

For example, we may want to solve 7x ≡ 3 (mod 10). It turns out x = 9 will do, and in fact that is the only solution. However, linear congruences don’t always have a unique solution. For example, 8x ≡ 3 (mod 10) has no solution; 8x is always an even integer and so it can never end in 3 in base 10. For another example, 8x ≡ 2 (mod 10) has two solutions, x = 4 and x = 9.

In this post, we answer two questions.

How many solutions does ax ≡ b(mod m) have?How can we compute them?

The answer to the first question depends on the greatest common divisor of a and m. Let g = gcd(a, m). If bis not divisible by g, there are no solutions. If b is divisible by g, there are g solutions.

So if g does divide b and there are solutions, how do we find them? The brute force solution would be to try each of the numbers 0, 1, 2, …, m-1 and keep track of the ones that work. That works in theory, but it is impractical for large m. Cryptography applications, for example, require solving congruences where m is extremely large and brute force solutions are impossible.

First, suppose a and m are relatively prime. That is, assume g = gcd(a, m) = 1. We first put the congruence ax ≡ b(mod m) in a standard form. We assume a > 0. If not, replace ax ≡ b (mod m) with –ax ≡ –b (mod m). Also, we assume a < m. If not, subtract multiples of m from auntil a < m.

Now solve my ≡ –b (mod a). This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. We can repeat this process recursively until we get to a congruence that is trivial to solve. The algorithm can be formalized into a procedure suitable for programming. The result is closely related to the Euclidean algorithm.

Now what if the numbers a and m are not relatively prime? Then first solve the congruence (a/g)y ≡ (b/g) (mod (m/g)) using the algorithm above. Then the solutions to ax ≡ b (mod m) are x = y + tm/g where t = 0, 1, 2, …, g-1.

Here are a couple examples.

First, let’s solve 7x ≡ 13 (mod 100). Since 7 and 100 are relatively prime, there is a unique solution. The algorithm says we should solve 100y ≡ -13(mod 7). Since 100 ≡ 2 (mod 7) and -13 ≡ 1 (mod 7), this problem reduces to solving 2y ≡ 1 (mod 7), which is small enough to solve by simply sticking in numbers. We find y = 4. Then x = (100*4 + 13)/7 = 59. You can verify that 7*59 = 413 so 7*59 ≡ 13 (mod 100). In general, we may have to apply the algorithm multiple times until we get down to a problem small enough to solve easily.

Now let’s find all solutions to 50x ≡ 65 (mod 105). Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. So we first solve 10x ≡ 13 (mod 21). The algorithm above says we can solve this by first solving 21y ≡ -13 (mod 10), which reduces immediately to y ≡ 7 (mod 10), and so we take y = 7. This says we can take x = (105*7 + 65)/50 = 16. The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. So the solutions are 16, 37, 58, 79, and 100.

I intend to write posts in the future about how to solve simultaneous systems of linear congruences and how to solve quadratic congruences.