# Homework Help: When will there exist solutions to ax + (a+2)y = c?

1. Jun 3, 2015

### Shackleford

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

When will there exist solutions to ax + (a+2)y = c?

2. Relevant equations

GCD

3. The attempt at a solution

Of course, this is a linear Diophantine equation. I know that a solution will exist when gcd(a, (a+2)) | c.

gcd(a, (a+2)) = d ⇒ ∃ x,y ∈ ℤ ∋ xa + y(a+2) = d

a = 0(a+2) + a
a+2 = (a) + 2
a = β(2) + r

2. Jun 3, 2015

### Staff: Mentor

gcd(a, (a+2)) has a shorter expression.
Not sure what the last three lines are saying.

3. Jun 3, 2015

### Shackleford

I just started the algorithm for finding the GCD, but I don't think that I actually need to find an expression.

Hm. Well, I know that a and a+2 are both odd or even depending on a.

4. Jun 3, 2015

### Ray Vickson

Have you forgotten to tell us whether or not a, c, x, y are positive (non-negative?) integers?

5. Jun 3, 2015

### Shackleford

Partially. a, b ∈ ℕ and x, y ∈ ℤ and 1 ≤ d ≤ min{a,b}.

6. Jun 4, 2015

### geoffrey159

If you want to solve mu+nv = d when m and n are relatively prime, you should be able to find a specific solution with Bezout theorem. With that specific solution, you should be able to find a general solution (u,v) without too much difficulty.

The thing is that ax +(a+2) y = c has not that form, or at least not always. Why ? Mfb gave you a good hint. Think about it

7. Jun 4, 2015

### Staff: Mentor

I think you should, as the final expression is very easy. It is a bit like leaving 4+5 in a final answer. It is not wrong in terms of mathematics, but you should replace it by 9.