# Division Algorithm

1. Jul 26, 2012

### MAGICMATHS

1. My difficulty is to show that if a is an integer such that 2 does not divide a and 3 does not divide a then 24 does not divide a squared minus 1

2.Is there any equation which helps?

3. My idea is that it has to be an integer such that 6 does not divide a..........therefore i have to show that 6a,6a+1,...6a+5 converted to asquared minus 1 are not divisible by 24........pls can anyone suggest if im solving it correctly.

2. Jul 26, 2012

### HallsofIvy

Staff Emeritus
Well, not "6a" because that would be divisible by both 2 and 3. And you don't mean "6a+ ...". If "2 does not divide a and 3 does not divide a" then either a= 6k+ 1 or a= 6k+ 5 for some integer k.

Now, if a=6k+ 1, $a^2- 1= 36k^2+ 12k+1-1= 36k^2- 12k= 12k(3k- 1)$. That will be divisible by 3 if either k or 3k-1 is even- and it is easy to see that they can't both be odd.

If a= 6k+ 5, $a^2-1= 36k^2+ 60k+ 25- 1= 36k^2+ 60k+ 24= 12(3k^2+ 5k+ 2)= 12(3k+ 2)(k+ 1)$. Now it is only necessary to show that 3k+2 and k+ 1 can't both be odd.

3. Jul 26, 2012

### MAGICMATHS

how can i explain that 6k is divisible by 6 when if i change it to asquared - 1 (pls refer to previous message) it becomes 36ksquared - 1........thats not easy to explain as divisible by 6!:(

4. Jul 26, 2012

### HallsofIvy

Staff Emeritus
The hypothesis is "a is an integer such that 2 does not divide a and 3 does not divide a" so $(6k)^2- 1$ is not relevant.