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 n 2 = 8m + 1 Let n be some arbitrary odd integer, let’s say 7. If 8m +1= n 2 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 nn2n2 m8m + 1 -7496 1101 3919 5253 7496 9811081 1112115121 1316921169

6. Proof: Formal Restatement:  odd integers n,  an integer m such that n 2 = 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: n 2 = 8m + 1 Since n = 4q + 1: n 2 = (4q + 1) 2 = (4q + 1)(4q + 1) = 16q 2 + 8q + 1 = 8(2q 2 + q) + 1 Let m = 2q 2 + q. Then m is an integer since 2 and q are integers and sums and products of integers are integers. Thus, substituting, n 2 = 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: n 2 = 8m + 3 Since n = 4q + 3: n 2 = (4q + 3) 2 = (4q + 3)(4q + 3) = 16q 2 + 24q + 9 = 16q 2 + 24q + (8 + 1) = 8(2q 2 + 3q + 1) + 1 Let m = 2q 2 + 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, n 2 = 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, n 2 = 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. n 2 = (2q + 1) 2 = 4q 2 + 4q + 1 = 4(q 2 + q) + 1 Though the above allows us to represent n 2 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 n 2 = 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.