# Prove that #if n# is odd, then #n=4k+1# for some #k in ZZ# or #n=4k+3# for some #k in ZZ#?

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I'm stuck! I know I need to use cases, I'm just not sure how to work them out.

I'm stuck! I know I need to use cases, I'm just not sure how to work them out.

##### 1 Answer

Feb 18, 2017

Here's a basic outline:

**Proposition**: If

**Proof**: Let

Then, by division algorithm,

#R=0,1,2,# or#3# (remainder).

Case 1: R=0. If the remainder is#0# , then#n=4k=2(2k)# .

#:. n# is even

Case 2: R=1. If the remainder is#1# , then#n=4k+1# .

#:. n# is odd.

Case 3: R=2. If the remainder is#2# , then#n=4k+2=2(2k+1)# .

#:. n# is even.

Case 4: R=3. If the remainder is#3# , then#n=4k+3# .

#:. n# is odd.