# Proof Using Modular Arithmetic

• Oct 18th 2009, 03:29 PM
ilikecandy
Proof Using Modular Arithmetic
show that if an integer n has at least two distinct odd prime divisors then there exists k<φ(n) such that
a^k ≡ 1(mod n)
for every a relatively prime to n.

What I have so far: The expression a^k ≡ 1(mod n) looks similar to Euler's Generalization of Fermat's Little Theorem. Also, n can be expressed as xp1p2, where x is a positive integer, and p1 and p2 are odd distinct prime numbers.

I know this isn't much, but I don't to where to go afterwards to complete the proof.
• Oct 19th 2009, 12:11 AM
NonCommAlg
Quote:

Originally Posted by ilikecandy
show that if an integer n has at least two distinct odd prime divisors then there exists $k < \varphi(n)$ such that $a^k \equiv 1 \mod n,$ for every a relatively prime to n.

What I have so far: The expression a^k ≡ 1(mod n) looks similar to Euler's Generalization of Fermat's Little Theorem. Also, n can be expressed as xp1p2, where x is a positive integer, and p1 and p2 are odd distinct prime numbers.

I know this isn't much, but I don't to where to go afterwards to complete the proof.

$(*)$: suppose $\gcd(u,v)=1, \ r \mid t,$ and $s \mid t.$ if $\ a^r \equiv 1 \mod u$ and $a^s \equiv 1 \mod v,$ then $a^t \equiv 1 \mod uv.$

now let $n=p^{\alpha} q^{\beta}m,$ where $p \neq q$ are some odd primes and $\gcd(m,p)=\gcd(m,q)=1.$ let $t=\text{lcm}(\varphi(p^{\alpha}), \varphi(q^{\beta})) < \varphi(p^{\alpha})\varphi(q^{\beta}),$ where $\varphi$ is the Euler's totient function.

let $k=t \varphi(m) < \varphi(n)$ and suppose $\gcd(a,n)=1.$ by $(*)$ we have $a^t \equiv 1 \mod p^{\alpha}q^{\beta}.$ we also have $a^{\varphi(m)} \equiv 1 \mod m.$ so, again by $(*),$ we get: $a^k \equiv 1 \mod n. \ \Box$

Remark: i suggest you read about the Carmichael $\lambda$ function a little bit to learn about the minimum value of that $k$ in your problem.