Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

According to this Wikipedia article, we know that an integer $n\; (\geq 2)$ is prime if and only if the polynomial congruence relation

$$ (x - a)^n \equiv (x^n - a) \pmod{n} $$

holds for all integers $a$ coprime to $n$ (or even just for some such integer $a$, in particular for $a = 1$).

Then the Fermat Little Theorem says, $n$ is a probable prime if

$$ a^n \equiv a \pmod{n} $$

Inside the article it said $x$ should never be substituted by a number, so If we do so, then for primality testing of a number like $211$ we would have something like this as an example :

$$ (2 + 3)^{211} \equiv 2^{211} + 3 \pmod{n} $$

So can't we use the above equation for some reduction in modular exponentiation?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

If you have $n=211$, you can use it, but you have to know your $n$ is prime. And then you would have $$(2+3)^{211} \equiv 2^{211} + 3 \pmod{211}$$

For a prime $p$ it is always true that: $$ (x+y)^p \equiv x^p + y^p\pmod{p} $$

But if you are looking for the explanation of the test, you should not do that. The part that says you shouldn't substitute $x$ means, that you should be looking at the polynomial $(x-a)^n$ which will be determined by its coefficients, i.e. if $f=(x-a)^n$ then it can be written also as a sum $f=\sum_{i=0}^{n} a_{i}x^{n}$ (if you expand the product), and you should be checking if $a_i \equiv 0 \pmod{n}$ for all $1<i<n$ and $a_0\equiv a \pmod{n}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.