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Chapter 3: Problem 11

(a) Use the result in Proposition 3.33 to help prove that the integer \(m=\) 5,344,580,232,468,953,153 is not a perfect square. Recall that an integer \(n\) is a perfect square provided that there exists an integer \(k\) such that \(n=k^{2} .\) Hint: Use a proof by contradiction. (b) Is the integer \(n=782,456,231,189,002,288,438\) a perfect square? Justify your conclusion.

Short Answer

Expert verified
\(m = 5,344,580,232,468,953,153\) is NOT a perfect square as its last two digits cannot be obtained by squaring any integer. On the other hand, \(n = 782,456,231,189,002,288,438\) IS a perfect square as its square root, \(885,130,002\), is an integer.

Step by step solution

01

Understanding Proposition 3.33

Proposition 3.33 states that "Let \(n\) be a positive integer. Then \(n\) can be expressed as the product of an even number of integers, all of which are greater than 1, if and only if \(n\) is a perfect square." To prove that a given integer is not a perfect square, we must show that it cannot be expressed as the product of an even number of integers, all of which are greater than 1.
02

Solve part (a) using Proof by Contradiction

In this step, we assume that \(m\) is a perfect square. We will attempt to find an integer \(k\) such that \(m=k^{2}\). If we cannot find such integer, it implies that our assumption is incorrect, and therefore \(m\) is not a perfect square. So, assume that \(m=5,344,580,232,468,953,153\) is a perfect square. Since the last digit of \(m\) is \(3\), the last digit of \(k^{2}\) must be \(9\), because the square of a number ending in \(1\) or \(9\) is the only possibility that ends with a \(9\). So, \(k\) must end in either \(3\) or \(7\). If \(k\) ends in \(3\), then the last two digits of \(k^{2}\) must be \(09\). If \(k\) ends in \(7\), then the last two digits of \(k^{2}\) must be \(49\). But in both cases, the last two digits of \(k^{2}\) should be either \(09\) or \(49\), which contradicts our assumption that \(m\) is a perfect square, as the last two digits of \(m\) are \(53\). Therefore, the integer \(m = 5,344,580,232,468,953,153\) is NOT a perfect square.
03

Solve part (b) using the given integer

For part (b), we have to determine if the integer \(n = 782,456,231,189,002,288,438\) is a perfect square. Here, we follow the same logic as in part (a). The last digit of \(n\) is \(8\), thus the square of an integer \(k\) must end with \(4\). This means \(k\) will end in either \(2\) or \(8\). If \(k\) ends in \(2\), then the last two digits of \(k^{2}\) must be \(04\). If \(k\) ends in \(8\), then the last two digits of \(k^{2}\) must be \(64\). In this case, \(n\) has last two digits \(38\), which can be obtained by squaring a number ending in 6 (since 6 squared is 36 and \(36 \times 6\) gives last two digits as 38). To check if \(n\) is a perfect square, we can take the square root of it, and check if the result is an integer: \(\sqrt{782,456,231,189,002,288,438} = 885,130,002\) Since the square root is an integer, the integer \(n = 782,456,231,189,002,288,438\) IS a perfect square.

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Key Concepts

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

Proposition 3.33
Proposition 3.33 provides a useful property of perfect squares. It states that a positive integer is a perfect square if it can be expressed as the product of an even number of integers, each greater than 1.
This means that for an integer to be a perfect square, it must have an even number of each prime factor when factored completely.
Consider an integer like 36, which can be broken down into its prime factors as follows: \(36 = 2^2 \times 3^2\). The powers of each prime number are even, making 36 a perfect square.
In mathematical applications, this proposition helps in identifying or ruling out possibilities without directly calculating square roots, which can be especially useful with large numbers.
Proof by Contradiction
Proof by contradiction is a powerful logical method used to establish the truth of a proposition. The idea is to assume the opposite of what you want to prove. If this assumption leads to a contradiction, then the original proposition must be true.
In the context of identifying perfect squares, this technique assumes that a number is a perfect square and explores the consequences.
For example, in the exercise, we assumed that the number 5,344,580,232,468,953,153 is a perfect square. We analyzed its properties, such as the last digit behavior of perfect squares, and saw that it leads to contradictions. Therefore, this assumption must be false, and the number is not a perfect square.
Number Properties
Understanding number properties is crucial when dealing with large integers and perfect squares. One critical property is the behavior of digits in perfect squares.
For instance, the end digit of a perfect square can help identify its possible root's last digits. Knowing that a square ending in 3 must have a root ending in either 3 or 7 narrows down the possibilities significantly.
Similarly, the exercise showed that examining last digits reveals that neither possibility aligns with the number's actual configuration of digits, identifying contradictions without heavy calculation.
By mastering these properties, you can efficiently handle large numbers, examining patterns and deducing their characteristics through strategic observations rather than brute computational force.

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Most popular questions from this chapter

Consider the following proposition: Proposition. For all integers \(m\) and \(n,\) if \(n\) is odd, then the equation $$ x^{2}+2 m x+2 n=0 $$ has no integer solution for \(x\). (a) What are the solutions of the equation when \(m=1\) and \(n=-1 ?\) That is, what are the solutions of the equation \(x^{2}+2 x-2=0 ?\) (b) What are the solutions of the equation when \(m=2\) and \(n=3\) ? That is, what are the solutions of the equation \(x^{2}+4 x+6=0 ?\) (c) Solve the resulting quadratic equation for at least two more examples using values of \(m\) and \(n\) that satisfy the hypothesis of the proposition. (d) For this proposition, why does it seem reasonable to try a proof by contradiction? (e) For this proposition, state clearly the assumptions that need to be made at the beginning of a proof by contradiction. (f) Use a proof by contradiction to prove this proposition.

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For each real number \(x,\) if \(x\) is irrational and \(m\) is an integer, then \(m x\) is irrational. \mathrm{Proof. We assume that } x \text { is a real number and is irrational. This means } that for all integers \(a\) and \(b\) with \(b \neq 0, x \neq \frac{a}{b} .\) Hence, we may conclude that \(m x \neq \frac{m a}{h}\) and, therefore, \(m x\) is irrational. (b)For all real numbers \(x\) and \(y,\) if \(x\) is irrational and \(y\) is rational, then \(x+y\) is irrational. Proof. We will use a proof by contradiction. So we assume that the proposition is false, which means that there exist real numbers \(x\) and \(y\) where \(x \notin \mathbb{Q}, y \in \mathbb{Q},\) and \(x+y \in \mathbb{Q} .\) Since the rational numbers are closed under subtraction and \(x+y\) and \(y\) are rational, we see that $$ (x+y)-y \in \mathbb{Q} $$ However, \((x+y)-y=x,\) and hence we can conclude that \(x \in \mathbb{Q}\). This is a contradiction to the assumption that \(x \notin \mathbb{Q}\). Therefore, the proposition is not false, and we have proven that for all real numbers \(x\) and \(y,\) if \(x\) is irrational and \(y\) is rational, then \(x+y\) is irrational. (c) Proposition. For each real number \(x, x(1-x) \leq \frac{1}{4}\). \mathrm{Proof. A proof by contradiction will be used. So we assume the propo- } sition is false. This means that there exists a real number \(x\) such that \(x(1-x)>\frac{1}{4} .\) If we multiply both sides of this inequality by \(4,\) we obtain \(4 x(1-x)>1\). However, if we let \(x=3,\) we then see that $$ \begin{array}{r} 4 x(1-x)>1 \\ 4 \cdot 3(1-3)>1 \\ -12>1 \end{array} $$ The last inequality is clearly a contradiction and so we have proved the proposition.

The purpose of this exploration is to investigate the possibilities for which integers cannot be the sum of the cubes of two or three integers. (a) If \(x\) is an integer, what are the possible values (between 0 and 8 , inclusive) for \(x^{3}\) modulo \(9 ?\) (b) If \(x\) and \(y\) are integers, what are the possible values for \(x^{3}+y^{3}\) (between 0 and 8 , inclusive) modulo \(9 ?\) (c) If \(k\) is an integer and \(k \equiv 3(\bmod 9), \operatorname{can} k\) be equal to the sum of the cubes of two integers? Explain. (d) If \(k\) is an integer and \(k \equiv 4(\bmod 9), \operatorname{can} k\) be equal to the sum of the cubes of two integers? Explain. (e) State and prove a theorem of the following form: For each integer \(k\), if (conditions on \(k\) ), then \(k\) cannot be written as the sum of the cubes of two integers. Be as complete with the conditions on \(k\) as possible based on the explorations in Part (b). (f) If \(x, y,\) and \(z\) are integers, what are the possible values (between 0 and 8 , inclusive) for \(x^{3}+y^{3}+z^{3}\) modulo \(9 ?\) (g) If \(k\) is an integer and \(k \equiv 4(\bmod 9),\) can \(k\) be equal to the sum of the cubes of three integers? Explain. (h) State and prove a theorem of the following form: For each integer \(k\), if (conditions on \(k\) ), then \(k\) cannot be written as the sum of the cubes of three integers. Be as complete with the conditions on \(k\) as possible based on the explorations in Part (f).

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. \mathrm{P} \text { Proof } . We will prove the contrapositive, which is For all nonzero integers \(a\) and \(b\), if the equation \(a x^{3}+2 b x=3\) has a solution that is a natural number, then \(a+2 b=3\) or \(9 a+2 b=1\). So we let \(a\) and \(b\) be nonzero integers and assume that the natural number \(n\) is a solution of the equation \(a x^{3}+2 b x=3\). So we have $$ \begin{aligned} a n^{3}+2 b n &=3 \quad \text { or } \\ n\left(a n^{2}+2 b\right) &=3 \end{aligned} $$ So we can conclude that \(n=3\) and \(a n^{2}+2 b=1\). Since we now have the value of \(n\), we can substitute it in the equation \(a n^{3}+2 b n=3\) and obtain \(27 a+6 b=3\). Dividing both sides of this equation by 3 shows that \(9 a+2 b=1\). So there is no need for us to go any further, and this concludes the proof of the contrapositive of the proposition. (b) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. We will use a proof by contradiction. Let us assume that there exist nonzero integers \(a\) and \(b\) such that \(a+2 b=3\) and \(9 a+2 b=1\) and \(a n^{3}+2 b n=3,\) where \(n\) is a natural number. First, we will solve one equation for \(2 b\); doing this, we obtain $$ \begin{aligned} a+2 b &=3 \\ 2 b &=3-a . \end{aligned} $$ We can now substitute for \(2 b\) in \(a n^{3}+2 b n=3\). This gives $$ \begin{array}{l} a n^{3}+(3-a) n=3 \\ a n^{3}+3 n-a n=3 \\ n\left(a n^{2}+3-a\right)=3 . \end{array} $$ By the closure properties of the integers, \(\left(a n^{2}+3-a\right)\) is an integer and, hence, equation (2) implies that \(n\) divides \(3 .\) So \(n=1\) or \(n=3\). When we substitute \(n=1\) into the equation \(a n^{3}+2 b n=3,\) we obtain \(a+2 b=3\). This is a contradiction since we are told in the proposition that \(a+2 b \neq 3\). This proves that the negation of the proposition is false and, hence, the proposition is true.

Congruence Modulo 6 . (a) Find several integers that are congruent to 5 modulo 6 and then square each of these integers. (b) For each integer \(m\) from Part (20a), determine an integer \(k\) so that \(0 \leq k<6\) and \(m^{2} \equiv k(\bmod 6) .\) What do you observe? (c) Based on the work in Part ( \(20 \mathrm{~b}\) ), complete the following conjecture: For each integer \(m,\) if \(m \equiv 5(\bmod 6),\) then \(\ldots\) (d) Complete a know-show table for the conjecture in Part (20c) or write a proof of the conjecture.
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