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# Theorem 3.4.3

Theorem 3.4.3. The square of any odd integer has the form 8 m + 1 for some integer m. Introduction. Experimentation Most people do not attempt to prove an unfamiliar theorem by simply writing down a formal proof.

## Theorem 3.4.3

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1. Theorem 3.4.3 The square of any odd integer has the form 8m + 1 for some integer m.

2. Introduction Experimentation • Most people do not attempt to prove an unfamiliar theorem by simply writing down a formal proof. • It is often necessary to experiment with various approaches until one is found that works. • Formal proofs are like the result of a mystery where all of the loose ends are neatly tied up.

3. Introduction Division into Cases & Alternative Representations for Integers • One useful strategy for proving a theorem is to divide the problem into multiple cases. • All of the individual cases together describe all of the ways to represent a given set of numbers. • Any given number will fit only one of the cases, but all cases must be considered in order to account for any arbitrary value. • Alternative representations of integers provide an opportunity to use the division into cases as a viable strategy.

4. Example n2 = 8m + 1 Let n be some arbitrary odd integer, let’s say 7. If 8m +1= n2 8m + 1 = 49 8m = 48 m = 6

5. Example An interesting observation: each consecutive value of m can be calculated using a triangle series. m = 0 m = 0 + 1 m = 0 + 1 + 2 m = 0 + 1 + 2 + 3 m = 0 + 1 + 2 + 3 + 4 m = 0 + 1 + 2 + 3 + 4 + 5 m = 0 + 1 + 2 + 3 + 4 + 5 + 6

6. Proof: Formal Restatement:  odd integers n,  an integer m such that n2 = 8m + 1. Suppose n is a particular but arbitrarily chosen odd integer. By the quotient-remainder theorem, n can be written in one of the forms: 4q or 4q + 1 or 4q + 2 or 4q + 3 Since n is odd, n must have one of the forms: 4q + 1 or 4q + 3 We can therefore divide this proof into two cases.

7. Case 1 (n = 4q + 1 for some integer q): We must find an integer m such that: n2 = 8m + 1 • Since n = 4q + 1: n2 = (4q + 1)2 = (4q + 1)(4q + 1) = 16q2 + 8q + 1 = 8(2q2 + q) + 1 • Let m = 2q2 + q. Then m is an integer since 2 and q are integers and sums and products of integers are integers. Thus, substituting, n2 = 8m + 1, where m is an integer.

8. Case 2 (n = 4q + 3 for some integer q): We must find an integer m such that: n2 = 8m + 3 • Since n = 4q + 3: n2 = (4q + 3)2 = (4q + 3)(4q + 3) = 16q2 + 24q + 9 = 16q2 + 24q + (8 + 1) = 8(2q2 + 3q + 1) + 1 • Let m = 2q2 + 3q + 1. Then m is an integer since 1, 2, 3, and q are integers and sums and products of integers are integers. Thus, substituting, n2 = 8m + 1, where m is an integer.

9. Conclusion Cases 1 and 2 show that given any odd integer, whether of the form 4q + 1 or 4q + 3, n2 = 8m + 1 for some integer m. [This is what we needed to show.]

10. Experimentation Revisited The text presents the following approach as an example of an attempt that doesn’t quite prove our theorem : Since n is odd, n can be represented as 2q + 1 for some integer q. n2 = (2q + 1)2 = 4q2 + 4q + 1 = 4(q2 + q) + 1 Though the above allows us to represent n2 as 4m + 1, it doesn’t appear to help us represent it as 8m + 1. Or does it?

11. Related Homework Problems Problem 24 asks to prove that any two consecutive integers is even. Problem 25 then asks to write a new proof that n2 = 8m + 1 based on the initial failed attempt on the previous slide. It turns out that problem 24 is instrumental in providing a second solution to our proof. Section 3.4 Problems 24 – 27, 34 – 41

12. Exam Question One definition of a “perfect square” is a number whose square root is an integer. • Prove that the product of any four consecutive integers + 1 is a perfect square.

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