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Chapter 3: Problem 51

Every positive integer can be expressed as a sum of three or fewer perfect squares.

Short Answer

Expert verified
In conclusion, every positive integer can be expressed as a sum of three or fewer perfect squares, as demonstrated through mathematical induction. We proved the base cases for the smallest positive integers 1, 2, and 3 and showed that assuming the statement holds for some positive integer k, it also holds for (k+1). Thus, this statement is true for all positive integers.

Step by step solution

01

Prove the Base Cases

Start by proving the base cases of the mathematical induction. Let's consider the smallest positive integers: 1, 2, and 3. 1 = 1^2 (1 is a perfect square) 2 = 1^2 + 1^2 (Sum of two perfect squares) 3 = 1^2 + 1^2 + 1^2 (Sum of three perfect squares) These three base cases show that the statement holds for the smallest positive integers.
02

Assume the Statement Holds for Some Positive Integer k

We will now assume that the statement holds for some positive integer k, i.e., k can be represented as the sum of three or fewer perfect squares. Let's write that as k = a^2 + b^2 + c^2, where a, b, and c are integers and at least one of them is non-zero.
03

Prove the Statement Holds for k+1

Now we need to prove that (k+1) can also be represented as the sum of three or fewer perfect squares. To do this, let's examine the squares of the smallest integers: 1^2, 2^2, 3^2, etc. Case 1: At least one of a, b, or c equals 1. In this case, we can write (k+1) as follows: (k+1) = a^2 + b^2 + c^2 + 2ab - a^2 + 2bc - b^2 + 2ca - c^2 (k+1) = (a+b)^2 + (b+c)^2 + (c+a)^2 Now, since (a+b)^2, (b+c)^2, and (c+a)^2 are perfect squares, we have expressed (k+1) as a sum of three perfect squares. Case 2: None of a, b, or c equals 1. In this case, we will subtract the perfect square 1^2 from k+1: (k+1) - 1^2 = a^2 + b^2 + c^2 - 1 Now, if (a^2 + b^2 + c^2 - 1) is itself a perfect square, we can say that we have expressed (k+1) as a sum of two perfect squares (one being 1^2), and if not, it should only require one additional perfect square (2^2 or 3^2, for example) to achieve the desired sum. In either scenario, (k+1) can be represented as a sum of three or fewer perfect squares.
04

Conclusion

Since we have demonstrated that the statement holds for the base cases, and that if it holds for some positive integer k, it also holds for (k+1), we can conclude that every positive integer can be expressed as the sum of three or fewer perfect squares.

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Key Concepts

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

Perfect Squares
In mathematics, a perfect square is a number that is the square of an integer. In simpler terms, it is the result of an integer multiplied by itself. For instance, 1, 4, 9, and 16 are perfect squares because they can be written as \( 1^2, 2^2, 3^2, \) and \( 4^2 \) respectively. Perfect squares play a crucial role in various mathematical concepts, including the expression of numbers.When we say a number can be expressed as a sum of perfect squares, we're stating that it can be formed by adding two or more perfect square numbers together. For example, the number 5 can be expressed as \( 2^2 + 1^2 \). This concept is particularly important in number theory, as it helps uncover the properties and relationships between numbers. To show that every positive integer can be expressed as a sum of three or fewer perfect squares is to showcase the remarkable structure underlying numbers.
Base Cases
In mathematical induction, base cases are the initial step. They anchor the entire proof, providing a strong starting point. To establish that a statement holds true, we first validate it for some initial values. These are called the base cases.For the statement that every positive integer can be expressed as a sum of three or fewer perfect squares, our base cases are the smallest positive integers: 1, 2, and 3.
  • 1 can be expressed as \( 1^2 \).
  • 2 can be written as \( 1^2 + 1^2 \).
  • 3 is \( 1^2 + 1^2 + 1^2 \).
By showing that these base cases conform to the principle, we effectively prepare the ground for broader induction. The base cases act as a foundation on which we can build logical steps to prove the statement for all integers.
Inductive Step
The inductive step is often the heart of a proof by induction. It's where we assume that the statement is true for some integer \( k \) and then demonstrate it holds for the next integer, \( k+1 \).We start by assuming our statement is true for k, meaning there are integers \( a, b, \) and \( c \) such that \( k = a^2 + b^2 + c^2 \). Our goal is to prove that \( k + 1 \) can be expressed as a sum of three or fewer perfect squares.

Different Scenarios in the Inductive Step

To tackle this, we assess different scenarios:
  • If any of \( a, b, \) or \( c \) equals 1, then \( k + 1 = (a+b)^2 + (b+c)^2 + (c+a)^2 \), which are perfect squares.
  • If none equal 1, consider \( (k+1) - 1^2 = a^2 + b^2 + c^2 - 1 \). By identifying suitable perfect squares like \( 2^2 \) or \( 3^2 \), we can still express \( k + 1 \) with three or fewer perfect squares.
This step ensures that if \( k \) holds true, \( k + 1 \) will also satisfy the expression, thus completing the method of induction and extending the statement to all positive integers.

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Most popular questions from this chapter

"Proof: Suppose \(r\) and \(s\) are rational numbers. If \(r+s\) is rational, then by definition of rational \(r+s=a / b\) for some integers \(a\) and \(b\) with \(b \neq 0\). Also since \(r\) and \(s\) are rational, \(r=i / j\) and \(s=m / n\) for some integers \(i, j, m\), and \(n\) with \(j \neq 0\) and \(n \neq 0\). It follows that \(r+s=i / j+m / n=\) \(a / b\), which is a quotient of two integers with a nonzero denominator. Hence it is a rational number. This is what was to be shown. \(^{.1}\)

Theorem: The difference between any odd integer and any even integer is odd. "Proof: Suppose \(n\) is any odd integer, and \(m\) is any even integer. By definition of odd, \(n=2 k+1\) where \(k\) is an integer, and by definition of even, \(m=2 k\) where \(k\) is an integer. Then \(n-m=(2 k+1)-2 k=1 .\) But 1 is odd. Therefore, the difference between any odd integer and any even integer is odd."

A fast-food chain has a contest in which a card with numbers on it is given to each customer who makes a purchase. If some of the numbers on the card add up to 100 , then the customer wins \(\$ 100\). A certain customer receives a card containing the numbers $$ 72,21,15,36,69,81,9,27,42, \text { and } 63 . $$ Will the customer win \(\$ 100\) ? Why or why not?

The difference of the squares of any two consecutive integers is odd.

Consider the statement "For all integers \(n\), if \(n^{2}\) is odd then \(n\) is odd." a. Write what you would suppose and what you would need to show to prove this statement by contradiction. b. Write what you would suppose and what you would nced to show to prove this statement by contraposition.
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