Odd Integer and Multiple of Four

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SUMMARY

The discussion focuses on proving that any odd integer \( n \) can be expressed as either one greater than a multiple of 4 or one less than a multiple of 4. The proof begins by expressing \( n \) in the form \( 2a + 1 \), where \( a \) is an integer. Two cases are considered: when \( a \) is even and when \( a \) is odd. The conclusion is drawn by analyzing the forms of integers modulo 4, specifically demonstrating that odd integers can only fall into the categories of \( 4m - 1 \) or \( 4m + 1 \).

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Homework Statement



Suppose that n is an odd integer. Prove that n is either one greater than a multiple of 4 or one less than a multiple of 4.

Homework Equations



N/A

The Attempt at a Solution



I realize that this is going to be a direct proof. However, I am stumped on where to go from here.
 
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We know that the if n is a odd integer it is in form of 2a+1.

Do two cases:
Case 1 a is even so a = 2b for some b
Case 2: a is odd so a = 2b+1 for some b (hint: express 3 in terms of 4)
 
Last edited:
A slightly different way: any integer must be of one of these forms:
a) 4n
b) 4n+ 1
c) 4n+ 2
d) 4n+ 3
for some n. Both 4n= 2(2n) and 4n+2= 2(2n+1) are even. Can you show that 4n+ 3= 4m- 1 for some number m?
 

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