Download Theorem 3.4.3

Theorem 3.4.3
The square of any odd integer has the
form 8m + 1 for some integer m.
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.
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.
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
Example
n
n2
m
8m + 1
-7
49
6
49
1
1
0
1
3
9
1
9
5
25
3
25
7
49
6
49
9
81
10
81
11
121 15
121
13
169 21
169
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
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.
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.
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.
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.]
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?
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
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.