# Maximum common divisor

1. May 21, 2006

### kezman

Find all a(integers) that $$(5a - 3:7a^2 - a + 1) = 1$$

I only know that

$$d|7a^2 - a + 1$$
$$d|5a - 3$$

2. May 23, 2006

### aujing

what is the meaning of "putting the 2 functions in the parathesis seperated by the colon" ??

3. May 23, 2006

### kezman

(a:b) = d
d is maximum common divisor of a and b

4. May 23, 2006

### vsage

Correct me if I'm wrong, but aren't you asking for which values of a are the polynomials (5a - 3) and (7a^2 - a + 1) relatively prime? (5a - 3 : 7a^2 - a + 1) = 1 is what you have written. That doesn't quite mesh with your thread title so I'm a little lost. For example, where did "d" come from?

5. May 23, 2006

### Hurkyl

Staff Emeritus
How do you normally find GCDs? The Euclidean algorithm, right? Have you tried it here?

6. May 24, 2006

### kezman

If (a:b) = 1 then a and b are co-prime
Also (a:b) = c then c|a and c|b
and exists d so that d|a , d|b and d|c
thats the d I try to use.