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For a proof of the Fermat Little Theorem in the case where \(a\) and \(p\) are relatively prime, consider the remainders of the numbers \(1, a, a^{2}, \ldots\) on division by \(p .\) These remainders must ultimately repeat (why?), and so \(a^{n+r} \equiv a^{r}(\bmod p)\) or \(a^{r}\left(a^{n}-1\right) \equiv 0(\bmod p)\) or \(a^{n} \equiv 1(\bmod p) .\) (Justify each of these alternatives.) Take \(n\) as the smallest positive integer satisfying the last congruence. By applying the division algorithm, show that \(n\) divides \(p-1\).

Short Answer

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Question: Prove Fermat's Little Theorem for the case where \(a\) and \(p\) are relatively prime: if \(a\) and \(p\) are relatively prime, then \(a^{p-1}\equiv 1(\bmod p)\). Solution: First, list the remainders of the numbers \(1, a, a^2, a^3, \ldots\) when divided by \(p\). Next, find a relation between \(a^n\) and \(a^r\), and identify the smallest \(n\) that satisfies the congruence \(a^n\equiv 1(\bmod p)\). Then, use the division algorithm to find the relationship between \(n\) and \(p-1\), and find a new congruence. Finally, use the new congruence to prove that \(n\) divides \(p-1\), thus proving Fermat's Little Theorem.

Step by step solution

01

List the remainders

We start by listing the remainders of the numbers \(1, a, a^2, a^3, \ldots\) when divided by \(p\). Since there are only \(p-1\) possible residues mod \(p\), by the Pigeonhole Principle, these remainders must ultimately repeat.
02

Find a relation between \(a^n\) and \(a^r\)

Let \(r\) be the smallest positive integer such that the remainder of \(a^r\) is the same as that of \(a^{n+r}\) for some positive integer \(n\). Then, we can write the congruence \(a^{n+r}\equiv a^r(\bmod p)\). Rearranging the congruence, we have \(a^r(a^n-1)\equiv 0(\bmod p)\). Notice that the last congruence implies that \(p\) divides \(a^r(a^n-1)\).
03

Identify the smallest \(n\) satisfying the congruence

Let \(n\) be the smallest positive integer satisfying the congruence \(a^n\equiv 1(\bmod p)\). Since \(a\) and \(p\) are relatively prime, we know that \(a^r\) is not divisible by \(p\). It follows that \(p\) must divide \((a^n-1)\).
04

Use the division algorithm

Now, we will use the division algorithm to find the relationship between \(n\) and \(p-1\). We can write \(p - 1 = n \cdot q + r\) for some non-negative integers \(q\) and \(r\), where \(0\leq r < n\).
05

Find a new congruence with the given remainder information

Raise both sides of the congruence from Step 3 to the power of \(q\), we have, \[(a^n)^q \equiv 1^q(\bmod p).\] Rewriting the congruence yields, \[a^{n\cdot q} \equiv 1(\bmod p).\]
06

Use the congruence from Step 5 to prove \(n\) divides \(p-1\)

We have \(a^{n\cdot q} \equiv 1(\bmod p)\). By multiplying both sides by \(a^r\), we obtain: \[a^{n\cdot q+r} \equiv a^r(\bmod p).\] Notice that \(n\cdot q+r=p-1\). Thus, the congruence becomes: \[a^{p-1} \equiv a^r(\bmod p).\] But, since \(n\) is the smallest positive integer satisfying the congruence \(a^n\equiv 1(\bmod p)\), it follows that \(p-1=n\cdot q\). Therefore, \(n\) divides \((p-1)\). Thus, we have proved Fermat's Little Theorem for the case where \(a\) and \(p\) are relatively prime: if \(a\) and \(p\) are relatively prime, then \(a^{p-1}\equiv 1(\bmod p)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Congruences
A fundamental concept in number theory that is quite pervasive, specifically in problems like Fermat's Little Theorem, is that of congruences. At its most basic, a congruence relation describes an equivalence between numbers with respect to a given modulus.

For instance, when we say that two numbers are congruent mod p, such as in the expression \(a \equiv b \pmod{p}\), we mean that when a and b are divided by p, they leave the same remainder. This kind of relationship is essential to understand because it tells us about the cyclical patterns that numbers exhibit when modular arithmetic is applied.

In the step-by-step solution for the theorem, congruences were crucial to establish the repeating nature of remainders when powers of a number are divided by a prime number p. The property that powers of a number eventually reach a cycle modulo p is what allows us to deduce the special case of Fermat's Little Theorem.
Pigeonhole Principle
The Pigeonhole Principle is an intuitive but surprisingly powerful tool in mathematics, particularly in combinatorics and number theory. The principle itself is simple: if there are more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon.

How does this apply to Fermat's Little Theorem? In our case, we can think of the 'pigeons' as the remainders when powers of a number a are divided by p, and the 'pigeonholes' as the possible remainders (0 to p-1). Since for prime numbers p there are exactly p-1 non-zero remainders possible, and we are considering the sequence \(1, a, a^2, \ldots\), we are bound to have a repeat remainder since we’re calculating more powers of a than there are remainders available.

Understanding this principle is key to rationalizing why we would expect such a pattern of repeat remainders in the sequence, which thereby forms part of the foundation for proving the theorem.
Division Algorithm
The division algorithm is a theorem in number theory that gives us a precise description of the division process. It states that given any two integers a (the dividend) and b (the divisor), with b > 0, there exist unique integers q (the quotient) and r (the remainder) such that \(a = bq + r\) and \(0 \leq r < b\).

This concept underpins much of arithmetic and is especially useful in proofs like that of Fermat's Little Theorem. In our exercise, the division algorithm was employed to show the relationship between n and p-1. The specific use of it helps us to formalize the step where we assert that n, the smallest integer satisfying \(a^n \equiv 1 \pmod{p}\), actually divides p-1 outright, a crucial leap in our proof.
Number Theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions. Situated at the heart of number theory are concepts such as primes, divisibility, and the abstractions that arise from considering numbers themselves as mathematical objects.

The entirety of Fermat's Little Theorem unfolds within the realm of number theory and requires a solid grasp of how prime numbers interplay with modular arithmetic. Much of the elegance of number theory, and indeed problems like the one considered for Fermat's Little Theorem, comes from its capacity to reveal deep truths through relatively simple equations and logical expressions.

Understanding number theory is crucial for delving deeper not only into this theorem but also into a myriad of other questions about the structure and properties of the number systems that govern arithmetic operations in mathematics.

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Most popular questions from this chapter

The best-known quotation from Descartes is "I think, therefore I am," from the Discourse on Method. The context is Descartes' resolve only to accept those ideas that are selfevidently true. There is a well-known joke based on this quote: Descartes goes into a restaurant. The waiter asks him, "Would you like tonight's special?" He replies, "I think not" and disappears. Comment on the logical validity of this joke. \(^{44}\)

To solve the fourth-degree equation \(x^{4}-p x^{2}-q x-\) \(r=0\), Descartes considered the cubic equation in \(y^{2}\). \(y^{6}-2 p y^{4}+\left(p^{2}+4 r\right) y^{2}-q^{2}=0 .\) If \(y\) is a solution, show that the original polynomial factors into two quadratics: \(r_{1}(x)=x^{2}-y x+\frac{1}{2} y^{2}-\frac{1}{2} p-\frac{q}{2 y}, r_{2}(x)=x^{2}+y x+\) \(\frac{1}{2} y^{2}-\frac{1}{2} p+\frac{q}{2 y}\), each of which can be solved. Apply this method to solve the equation \(x^{4}-17 x^{2}-20 x-6=0\) Note that the corresponding equation in \(y, y^{6}-34 y^{2}+\) \(313 y^{2}-400=0\), has the solution \(y^{2}=16\)

Show that the odds against at least one 1 appearing in a throw of three dice is \(125: 91\). (This answer was stated by Cardano.)

There are 12 balls in an urn, 4 of which are white and 8 black. Three blindfolded players, \(A, B, C\) draw a ball in. turn, first \(A\), then \(B\), then \(C\). The winner is the one who first draws a white ball. Assuming that each (black) ball is replaced after being drawn, find the ratio of the chances of the three players.

This problem illustrates one of Descartes' machines (Fig. 14.21). Here \(G L\) is a ruler pivoting at \(G\). It is linked at \(L\) with a device \(C N K L\) that allows \(L\) to be moved along \(A B\), always keeping the line \(K N\) parallel to itself. The intersection \(C\) of the two moving lines \(G L\) and \(K N\) determines a curve. To find the equation of the curve, begin by setting \(C B=y, B A=x\), and the constants \(G A=a\), \(K L=b\), and \(N L=c\). Then find \(B K, B L\), and \(A L\) in terms of \(x, y, a, b\), and \(c\). Finally, use the similarity relation \(C B: B L=G A: A L\) to show that the equation is $$ y^{2}=c y-\frac{c}{b} x y+a y-a c $$ Descartes stated, without proof, that this curve is a hyperbola. Show that he was correct.

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