# Difference between diophantine equations $ax + by = c$ and $ax - by = c$

I had just solved this problem of Sierpinski

Prove that the sequence $$2^n - 3, \; n = 2,3,4, \dots$$ contains infinitely many terms divisible by $$5$$ and $$13$$ but no terms divisible by $$5\cdot 13$$.

For showing the existence of infinite terms divisible by $$5$$ and $$13$$ I basically applied Fermat's little theorem and some congruences.

Now for showing whether a term exists such that it is divisible by $$65$$ I thought of using the fact that if $$a|b$$ and $$c | b$$ and $$gcd(a,c) = 1$$ then $$ac|b$$. But for that, I would require to show that there is a common term divisible by both $$5$$ and $$13$$.

I finally found that values of $$n$$ satisfying $$3 + 4k, \; k \in \mathbb{N}$$ are divisible by $$5$$ and values of $$n$$ satisfying $$4 + 12m, \; m \in \mathbb{N}$$ are divisible by $$13$$. So now I have to prove that no $$k$$ and $$m$$ exist such that $$3 + 4k = 4 + 12m$$. This comes out to be:

$$4k - 12m = 1$$

Which is a linear diophantine equation. Now $$gcd(4,12) \not\mid 1$$ so no solution exists.

But I feel that doing it this way is wrong as $$k, m \in \mathbb{N}$$ but due to the $$-$$ (minus) sign in the equation negative values of $$k$$ and $$m$$ are also being considered.

So now coming to my main question, when are diophantine equations $$ax + by = c$$ and $$ax - by = c$$ equivalent and what are their differences?

P.S I did solve this without using the diophantine part by proving that no positive integers $$k$$ and $$m$$ exist such that $$k - 3m = \frac{1}{4}$$.

• There is no reason why you can't use $k, m \in \mathbb Z$. – steven gregory May 3 at 23:08

No need to worry about signs $$4k-12m$$ is a multiple of $$\gcd(4,12)=4$$, while $$1$$ is not.
Edit: Let$$a,b>0$$.
Then, $$ax+by=c$$ is practically the same as $$ax-by=c$$ if $$x,y\in\mathbb{Z}$$ because if $$(x',y')$$ is a solution of the first one, then $$(x',-y')$$ is a solution for the other one.
Now, if you are just looking for $$x,y\in\mathbb{N}$$, then $$ax+by=c$$ will have a finite number of solutions, while $$ax-by=c$$ may have infinite solutions (note that if $$(x',y')$$ is a solution then $$(x'+kb,y'+ka)$$ is a solution)