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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 $k$.

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    $\begingroup$ 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. $\endgroup$
    – Anurag A
    Jul 2, 2021 at 8:36

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Safwane
    Jan 4 at 10:03
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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))

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