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Chapter 3: Problem 15

Let \(n\) be a natural number greater than 4 and let \(a\) be an integer that has a remainder of \(n-2\) when it is divided by \(n\). Make whatever conclusions you can about the remainder of \(a^{2}\) when it is divided by \(n\). Justify all conclusions.

Short Answer

Expert verified
The remainder of \(a^2\) when it is divided by \(n\) is \(4(1-n)\).

Step by step solution

01

Recall and understand the properties of modular arithmetic

In modular arithmetic, if two numbers have the same remainder when divided by a divisor, then we say that they are congruent modulo the divisor. For example, if \(a\) and \(b\) have the same remainder when divided by \(n\), then we write \(a \equiv b \pmod{n}\).
02

Use the given information to express a in terms of congruence modulo n

We are given the information that when \(a\) is divided by \(n\), it has a remainder of \(n-2\). In other words, \(a\) is congruent to \(n-2\) modulo \(n\). Therefore, we can write \(a \equiv n-2 \pmod{n}\).
03

Square both sides of the congruence

Now, we want to find the remainder when \(a^2\) is divided by \(n\). So, let's square both sides of the congruence \(a \equiv n-2 \pmod{n}\) to obtain: \[a^2 \equiv (n-2)^2 \pmod{n}\]
04

Expand the right-hand side of the congruence

\[(n-2)^2 \equiv n^2 - 4n + 4\pmod{n}\]
05

Apply the property of modular arithmetic that states if \(a \equiv b \pmod{n}\), then \(a + c \equiv b + c \pmod{n}\)

Since \(n^2\) is divisible by \(n\), it is congruent to 0 modulo \(n\). Therefore, \(n^2 - 4n + 4 \equiv -4n + 4 \pmod{n}\).
06

Make conclusion about the remainder of \(a^2\) when it is divided by \(n\)

Now, we can substitute this back into the congruence we obtained in Step 3: \[a^2 \equiv -4n + 4 \pmod{n}\] Thus, the remainder of \(a^2\) when it is divided by \(n\) is \(-4n+4\) or \(4(1-n)\).

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Key Concepts

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

Congruence Relations
Congruence relations are a fundamental concept in number theory, particularly in modular arithmetic. When we say two numbers, say \(a\) and \(b\), are congruent modulo \(n\), it simply means that they leave the same remainder when divided by \(n\). This relationship can be expressed as \(a \equiv b \pmod{n}\).
This concept allows us to think of numbers as representatives of a class which share similar divisibility properties.
  • Congruence is useful because computations of congruent numbers modulo \(n\) can simplify complicated arithmetic operations.
  • For instance, if we know \(a \equiv n-2 \pmod{n}\), then \(a\) behaves just like \(n-2\) in the modular system dictated by \(n\).
Understanding congruence helps you not only solve equations like the one in the given exercise but also allows tracking and predicting the behavior of numbers under division.
Natural Numbers
Natural numbers are those that start from 1, proceeding forward in a continuous sequence like 1, 2, 3, and so on. They are the numbers we ordinarily count with, and in this context, play a critical role.
In our exercise, we specifically focus on natural numbers greater than 4.
Natural numbers provide a foundation for arithmetic since they are integral and positive.
  • This set of numbers excludes negative numbers and fractions, focusing only on whole, positive integers.
  • In the context of modular arithmetic, every operation happening with natural numbers remains within their domain, making calculations straightforward without the complications introduced by negative or fractional values.
Understanding natural numbers is crucial when calculating remainders or setting constraints, just as in the original problem where \(n\) must be greater than 4 to make logical sense in the context of having \(n-2\) as a remainder.
Remainders
Remainders are what is "left over" after division, and they are at the heart of modular arithmetic. When you divide a number \(a\) by another number \(n\), the remainder is the part of \(a\) that is not evenly divisible by \(n\).
For example, dividing 11 by 4 leaves a remainder of 3, since 4 goes into 11 twice (which is 8), and 11 minus 8 equals 3. In modular arithmetic, this is expressed as \(11 \equiv 3 \pmod{4}\).
  • Remainders can simplify complex problems, breaking down numbers into manageable parts.
  • In our specific case, the problem gives a remainder of \(n-2\) when \(a\) is divided by \(n\), which dictates how \(a^2\)'s remainder will be ultimately calculated.
Understanding remainders allows you to see beyond surface calculations, helping predict results of seemingly complex equations with ease.
Squaring Numbers
Squaring a number, quite simply, means multiplying a number by itself. This operation increases the size and scope of numbers, turning integers into larger counterparts while maintaining congruence.
In the context of the exercise, squaring was key to understanding how \(a^2\) behaves under modular arithmetic.
  • When you square a number in the modular system, congruence relations still hold. If \(a \equiv b \pmod{n}\), then \(a^2 \equiv b^2 \pmod{n}\).
  • This principle is leveraged in our step-by-step solution where \(a^2\) becomes congruent to \((n-2)^2\), simplifying calculations by allowing modular division rules to apply smoothly.
By grasping how squaring functions in number theory, you gain more tools to tackle equations, predict outcomes, and understand deeper numerical patterns.

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Most popular questions from this chapter

Evaluation of proofs This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false. \- If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. \- If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. \- If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.(a) Proposition. If \(m\) is an even integer, then \((5 m+4)\) is an even integer. \mathrm{\\{} \text { Proof. We see that } 5 m + 4 = 1 0 n + 4 \(=2(5 n+2)\). Therefore, \((5 m+4)\) is an even integer. (b) Proposition. For all real numbers \(x\) and \(y,\) if \(x \neq y, x>0,\) and \(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\) \mathrm{\\{} P r o o f . ~ S i n c e ~ \(x\) and \(y\) are positive real numbers, \(x y\) is positive and we can multiply both sides of the inequality by \(x y\) to obtain $$ \begin{aligned} \left(\frac{x}{y}+\frac{y}{x}\right) \cdot x y &>2 \cdot x y \\ x^{2}+y^{2} &>2 x y \end{aligned} $$ By combining all terms on the left side of the inequality, we see that \(x^{2}-2 x y+y^{2}>0\) and then by factoring the left side, we obtain \((x-y)^{2}>0 .\) Since \(x \neq y,(x-y) \neq 0\) and so \((x-y)^{2}>0 .\) This proves that if \(x \neq y, x>0,\) and \(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\) (c) Proposition. For all integers \(a, b,\) and \(c,\) if \(a \mid(b c),\) then \(a \mid b\) or \(a \mid c\). \mathrm{\\{} P r o o f . ~ W e ~ a s s u m e ~ t h a t ~ \(a, b,\) and \(c\) are integers and that \(a\) divides \(b c\). So, there exists an integer \(k\) such that \(b c=k a\). We now factor \(k\) as \(k=m n,\) where \(m\) and \(n\) are integers. We then see that $$ b c=m n a . $$ This means that \(b=m a\) or \(c=n a\) and hence, \(a \mid b\) or \(a \mid c\). (d) Proposition. For all positive integers \(a, b,\) and \(c,\left(a^{b}\right)^{c}=a^{\left(b^{c}\right)}\). This proposition is false as is shown by the following counterexample: If we let \(a=2, b=3,\) and \(c=2,\) then $$ \begin{aligned} \left(a^{b}\right)^{c} &=a^{\left(b^{c}\right)} \\ \left(2^{3}\right)^{2} &=2^{\left(3^{2}\right)} \\ 8^{2} &=2^{9} \\ 64 & \neq 512 \end{aligned} $$

(a) If \(x\) and \(y\) are integers and \(x y=1,\) explain why \(x=1\) or \(x=-1\). (b) Is the following proposition true or false? For all nonzero integers \(a\) and \(b\), if \(a \mid b\) and \(b \mid a,\) then \(a=\pm b\).

Prove that for each real number \(x\) and each irrational number \(q,(x+q)\) is irrational or \((x-q)\) is irrational.

Are the following statements true or false? Justify your conclusions. (a) For each \(a \in \mathbb{Z},\) if \(a \equiv 2(\bmod 5),\) then \(a^{2} \equiv 4(\bmod 5)\). (b) For each \(a \in \mathbb{Z},\) if \(a^{2} \equiv 4(\bmod 5),\) then \(a \equiv 2(\bmod 5)\) (c) For each \(a \in \mathbb{Z}, a \equiv 2(\bmod 5)\) if and only if \(a^{2} \equiv 4(\bmod 5)\).

The Last Two Digits of a Large Integer. Notice that \(7,381,272 \equiv 72(\) mod 100\()\) since \(7,381,272-72=7,381,200\). which is divisible by 100 . In general, if we start with an integer whose decimal representation has more than two digits and subtract the integer formed by the last two digits, the result will be an integer whose last two digits are \(00 .\) This result will be divisible by 100 . Hence, any integer with more than 2 digits is congruent modulo 100 to the integer formed by its last two digits. (a) Start by squaring both sides of the congruence \(3^{4} \equiv 81\) (mod 100 ) to prove that \(3^{8}=61\) (mod 100 ) and then prove that \(3^{16}=21\) (mod 100 ). What does this tell you about the last two digits in the decimal representation of \(3^{16} ?\) (b) Use the two congruences in Part (24a) and laws of exponents to determine \(r\) where \(3^{20} \equiv r(\bmod 100)\) and \(r \in \mathbb{Z}\) with \(0 \leq r<100 .\) What does this tell you about the last two digits in the decimal representation of \(3^{20} ?\) (c) Determine the last two digits in the decimal representation of \(3^{400}\). (d) Determine the last two digits in the decimal representation of \(4^{804}\). Hint: One way is to determine the "mod 100 values" for \(4^{2}, 4^{4}, 4^{8}\), \(4^{16} \cdot 4^{32}, 4^{64},\) and so on. Then use these values and laws of exponents to determine \(r,\) where \(4^{804} \equiv r(\) mod 100\()\) and \(r \in \mathbb{Z}\) with \(0 \leq r<\) \(100 .\) (e) Determine the last two digits in the decimal representation of \(3^{3356}\). (f) Determine the last two digits in the decimal representation of \(7^{403}\).
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