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User:Temperal/The Problem Solver's Resource6

< User:Temperal
Revision as of 10:17, 11 January 2009 by Xpmath (talk | contribs) (→‎Solving Linear Congruences)
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Beginner/Intermediate Number Theory

This section covers number theory, specifically Fermat's Little Theorem, Wilson's Theorem,Euler's Totient Theorem, and Quadratic residues.

To use this page, we recommend knowing the basics of Linear congruence, Modular arithmetic, and have a grasp of basic number theory needed for the AMC 10 and 12.

Definitions

  • $n\equiv a\pmod{b}$ if $n$ is the remainder when $a$ is divided by $b$ to give an integral amount. Also, this means b divides (n-a).
  • $a|b$ (or $a$ divides $b$) if $b=ka$ for some integer $k$.
  • $\phi$ is the greek letter phi. $\phi(n)$ is the number of integers less than or equal to m that are at the same time relatively prime to n. If the prime factorization of n is $p_1^{e_1}p_2^{e_2}...p_n^{e_n}$, $\phi(n)=n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)...\left(1-\frac{1}{p_n}\right)$.

Special Notation

Occasionally, if two equivalent expressions are both modulated by the same number, the entire equation will be followed by the modulo.

$(a_1, a_2,...a_n)$ refers to the greatest common factor of $a_1, a_2, ...a_n$ and $[a_1, a_2, ...a_n]$ refers to the lowest common multiple of $a_1, a_2,...a_n$.

Properties

For any number there will be only one congruent number modulo $m$ between $0$ and $m-1$.

If $a\equiv b \pmod{m}$ and $c \equiv d \pmod{m}$, then $(a+c) \equiv (b+d) \pmod {m}$.

  • $a \pmod{m} + b \pmod{m} \equiv (a + b) \pmod{m}$
  • $a \pmod{m} - b \pmod{m} \equiv (a - b) \pmod{m}$
  • $a \pmod{m} \cdot b \pmod{m} \equiv (a \cdot b) \pmod{m}$

Fermat's Little Theorem

For a prime $p$ and a number $a$ such that $(a,b)=1$, $a^{p-1}\equiv 1 \pmod{p}$. A frequently used result of this is $a^p\equiv a\pmod{p}$.

Example Problem 1

Find all primes p such that $p|2^p+1$.

Solution

Firstly, p=2 clearly does not work. Now, as all other primes are odd, $(2, p)=1$ and hence $2^p\equiv2\pmod{p}$. After adding one, we have $3\equiv0\pmod{p}$ since p divides $2^p+1$. However, that means p must divide 3, so the only prime possible is 3. Indeed, $2^3+1=9$ is a multiple of 3.

Wilson's Theorem

For a prime $p$, $(p-1)! \equiv -1 \pmod p$.

Example Problem 2

Let $a$ be an integer such that $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{23}=\frac{a}{23!}$. Find the remainder when $a$ is divided by $13$.

Solution

After multiplying through by $23!$, we know that every term on the left-hand-side will be divisible by 13 except for $\frac{23!}{13}$. We wish to find the remainder when $23\cdot22\cdot21...\cdot1$ is divided by 13. From Wilson's Theorem, we know that $12!\equiv-1\pmod{13}$ so we consider (mod 13). Thus, the remainder is $10\cdot9\cdot8...\cdot1\cdot12!\equiv10!\cdot{-1}\pmod{13}$ which comes out to be 7. Thus, our answer is 7.

Euler's Phi Theorem

If $(a,m)=1$, then $a^{\phi{m}}\equiv1\pmod{m}$, where $\phi{m}$ is the number of relatively prime numbers lower than $m$. This is mostly a generalization of Fermat's Little Theorem, although much more useful.

Quadratic Residues

An integer n is a quadratic residue (mod m) if and only if there exists an integer p such that $p^2\equiv n\pmod{m}$. Some useful facts are that all quadratic residues are $0$ or $1\pmod{4}$ and $0$, $1$, or $4$ $\pmod{8}$. All cubic residues (mod 9) are 0, 1, or -1.

Example Problem 3

Does there exist an integer such that its cube is equal to $3n^2 + 3n + 7$, where n is an integer? (IMO longlist 1967)

Solution

Consider $3n^2 + 3n + 7$ (mod 9), and n (mod 3). If n is divisible by 3, $3n^2 + 3n$ is clearly divisible by 9. If n is congruent to 1 (mod 3), $3n^2 + 3n$ is congruent to 6 (mod 9). If n is congruent to 2 (mod 3), then $3n^2 + 3n\equiv3(n)(n + 1)\pmod{9}$. As n+1 is divisible by 3, it is congruent to 0 (mod 9). Hence, $3n^2 + 3n + 7$ is either 7 or 4 (mod 9). However, all cubes are 0,1, or -1 (mod 9), so there does not exist such an integer.

Solving Linear Congruences

As mentioned at the top, you should at least know how to solve simple linear congruences, with just one linear congruence. However, solving with two or more congruences is more complex, and many times there is not even a solution. The Chinese Remainder Theorem shows when the congruences do have a unique solution. *to be continued*

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