User:Temperal/The Problem Solver's Resource6
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Beginner/Intermediate Number Theory
- if is the remainder when is divided by to give an integral amount. Also, this means b divides (n-a).
- (or divides ) if for some integer .
- is the greek letter phi. 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 , .
Occasionally, if two equivalent expressions are both modulated by the same number, the entire equation will be followed by the modulo.
refers to the greatest common factor of and refers to the lowest common multiple of .
For any number there will be only one congruent number modulo between and .
If and , then .
- , where is a positive integer that divides and .
Fermat's Little Theorem
For a prime and a number such that , . A frequently used result of this is .
Example Problem 1
Find all primes p such that .
Firstly, p=2 clearly does not work. Now, as all other primes are odd, and hence . After adding one, we have since p divides . However, that means p must divide 3, so the only prime possible is 3. Indeed, is a multiple of 3.
For a prime , .
Example Problem 2
Let be an integer such that . Find the remainder when is divided by .
After multiplying through by , we know that every term on the left-hand-side will be divisible by 13 except for . We wish to find the remainder when is divided by 13. From Wilson's Theorem, we know that so we consider (mod 13). Thus, the remainder is which comes out to be 7. Thus, our answer is 7.
Euler's Phi Theorem
If , then , where is the number of relatively prime numbers lower than . This is mostly a generalization of Fermat's Little Theorem, although much more useful.
An integer n is a quadratic residue (mod m) if and only if there exists an integer p such that . Some useful facts are that all quadratic residues are or and , , or . 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 , where n is an integer? (IMO longlist 1967)
Consider (mod 9), and n (mod 3). If n is divisible by 3, is clearly divisible by 9. If n is congruent to 1 (mod 3), is congruent to 6 (mod 9). If n is congruent to 2 (mod 3), then . As n+1 is divisible by 3, it is congruent to 0 (mod 9). Hence, 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*