A Bit More Number Theory

Notation: Let MATH be three integers . We write MATH if $\QTR{Large}{q}$ divides $\QTR{Large}{m-n}$.

Note that from the previous section we have MATH implies MATH or MATH

Lemma: Let MATH , $\QTR{Large}{p}$ prime, then there exists an integerMATH such that MATH

$\qquad $PROOF: A simple counting argument shows that there are integers MATH with MATH.

Assume $\QTR{Large}{j>k}$. We have MATH. But since MATH we have MATH.

Lemma: Let MATH then MATH.

PROOF: Let MATH be the least integer such that MATH. It suffices to show that $\QTR{Large}{m}$ divides $\QTR{Large}{p-1}$. To do this consider the set of integers.

MATH, We can divide this set into disjoint subsets of $\QTR{Large}{m}$ elements using the relationship MATH iff MATH for some $\QTR{Large}{k}$.

Observations: The subsets are of the form MATH. The other points is that $\QTR{Large}{m}$ depends on the $\QTR{Large}{a}$. For example, let

$\QTR{Large}{p=7}$ and $\QTR{Large}{a=2}$. Note MATH. On the other hand, if $\QTR{Large}{p=7}$ and $\QTR{Large}{a=3}$ , we have MATH Of course, MATH.

Theorem: Let $\QTR{Large}{p}$ and $\QTR{Large}{q}$ be primes. Let $\QTR{Large}{n=pq}$. Let $\QTR{Large}{a}$ be relatively prime to both $\QTR{Large}{p}$ and $\QTR{Large}{q}$ , hence $\QTR{Large}{n}$, then MATH.

Hence, MATH.

PROOF: Both $\QTR{Large}{p}$ and $\QTR{Large}{q}$ , divide MATH. (MATH)

Observation: Another way of looking at this is that MATH and MATH Of course, the Extended Euclidean Algorithm tells us that MATHcan be computed for any relatively prime pair MATH