# Abstract algebra question chapter 1.2

1. Sep 3, 2009

### blairebear

1. The problem statement, all variables and given/known data

prove that if 2 does not divide a then 24 divides a^2-1

2. Relevant equations

I know that if 2 does not divide a then a is odd.

I proved that the square for all odd integers are of the form 8K+1

I also proved the square of any integer is either in the form 3k, 3k+1 or 3k+2. I know 3k and 3k+2 are even so the odd are in the form 3k+1

3. The attempt at a solution

Can I simply say that because both 8 and 3 are in this form then 8*3 is in this form?

2. Sep 3, 2009

### blairebear

That is not what I mean to say. based on before I know that 8 divides a^2-1 and 3 divides a^2-1. Can I then say that since both 3 and 8 divide a^2-1 then 24 divides a^2-1

3. Sep 3, 2009

### blairebear

Ok, I just realized that 3k and 3k+2 are not necessarily even and 3k+1 is not necessarily odd. So now I am even more confused

4. Sep 3, 2009

### blairebear

OK, I am going to say this is not true. When a=3 then a^2 is in the form 3k. so I have disproved it I think.
However the next part says if 2 or 3 do not divide a. That eliminates every third number. I also read my proof wrong and it is either 3k or 3k+1 not 3k+2 so eliminating all the times when 3 divides a will leave me the other two options and for each of those a^2 is in the form 3k+1. OK, I need to write it up but I think I got it.

5. Sep 3, 2009

### Hurkyl

Staff Emeritus
If this hypothesis is true, then this conclusion is true, by the Chinese Remainder Theorem.

6. Sep 3, 2009

### HallsofIvy

Or, since 3 and 8 have no prime factors in common, yes, any number that is divisible by both 3 and 8 is divisible by 3(8)= 24.