# An equation has infinitely many positive integers solutions

We know that the Diophantine equation: $$ax + by = c$$, has infinitely many integer solutions if $$\gcd(a,b)|c$$. Now, I am asking about the case where this equation has infinitely many positive integers solutions.

My solution:

Assuming $$a>0$$ and $$b>0$$

If $$(x, y)$$ is a solution, then the other solutions have the form $$(x + kv, y − ku)$$, where $$k$$ is an arbitrary integer, and $$u$$ and $$v$$ are the quotients of $$a$$ and $$b$$ (respectively) by the greatest common divisor of $$a$$ and $$b$$.

Now, $$x+kv≥0$$ and $$y-ku≥0$$ if $$-x/v≤k≤y/u$$

However, this is not able to give infinitely many integers $$k$$.

• Or you can simply say that for $a,b >0$, the expression $ax+by$ will exceed $c$ if $x$ and $y$ are positive and are bigger than a certain threshold (e,g, if say $x \geq \frac{c}{a})$, so it cannot have infinitely many positive integer solutions. Jul 2, 2021 at 8:36

Claim
$$ax + by = c$$ has infinite number of positive solution if and only if $$a$$ and $$b$$ are of opposite signs and $$\gcd(a,b)|c$$
Proof
WLOG let $$a>0$$ and $$b<0$$. There exists a solution because $$\gcd(a,b)|c$$.
If $$(x',y')$$ is a solution then $$(x'-b,y'+a)$$ is a solution so we get a increasing ordered pair of $$(x,y)$$ that satisfies the equation. Hence there are infinite number of positive solutions for this case.

It is not possible if $$a$$ and $$b$$ are of same sign. Suppose there are infinitely many solution when $$a$$ and $$b$$ are of same sign then there exists a solution $$(x',y')$$ with $$\mid x'\mid>c$$ and $$\mid y'\mid>c$$ , but this implies $$\mid ax'+by'\mid>c$$ thus leading to a contradiction.

For example $$(a,b,c)=(rt,-rt,2rt)$$ where $$r$$ is a constant and $$t$$ is a integer variable will give you different values for $$a,b,c$$ such that the equation has infinitely many solutions.

• Now, by using your method, if $(a,b,c)=(rt,rt,2rt)$ where $r$ is a constant and $t$ is a integer variable then the equation has finite number of solutions. Jan 4 at 10:03

Diophantine equation can never have infinitely many positive integer solution if a>0 and b>0; Define two numbers l,m , where l=ceil(c/a), m=ceil(c/b); Taking x>l will make the first term more which cannot be decreased by second term and same argument can be said for y>m So the number of positive integer solutions is upper bounded by max(ceil(c/a), ceil(c/b))