Can we determine if $y(x)=\frac {c-x}{x+1}$ has any strictly positive integer solutions?

Question

The equation is as follows:

$$y=\frac {c-x}{x+1}$$

Where $c$ is a positive, even integer.

Is it possible to determine whether there are any positive integer solutions different than $x=0$?

Some examples:

$c=14$; $x=2$; $y=4$.

$c=10$; NO POSITIVE INTEGER SOLUTION

$c=20$; $x=2$; $y=6$.

$c=44$; $x=4$; $y=8$.

$c=76$; $x=6$; $y=10$.

Answer 1

Hint: You have

$$y = \frac{c - x}{x + 1} = \frac{c + 1 - 1 - x}{x + 1} = \frac{c + 1 - (1 + x)}{x + 1} = \frac{c + 1}{x + 1} - 1$$

Next, consider how whether or not $c + 1$ is prime, e.g., where $c = 10$, affects the existence of positive integer solutions for $x$ and $y$.

Answer 2

A slightly different wording than John Omielan's answer (but same idea) may make this more intuitively clearer.

$y = \frac {c-x}{x+1}$ will have solutions when $x+1|c-x$.

And $x+1|c-x \iff x+1|(c-x)+(x+1)= c+1$.

So for any $k|d=c+1$, we will have solutions $x = k-1$ (and $y = \frac {c-(k-1)}{(k-1)+1}=\frac {(c+1)-k}k = \frac dk -1$)

However if $d=c+1$ is prime we either $x= 0$ or $x = d-1=c$ (in which case $y=0$).

Answer 3

It was not perfectly clear to me whether this was a question about suitable $x$ (any solutions other than $x=0$?) or about suitable $c$ (listing various $c$ for which $x,y$ could be found). I provide an answer about $x$, and the implications for $c$.

$y$ is an integer when $c=k(x+1)+x$, viz: $y=\frac{k(x+1)+x-x}{x+1}=k$.

However, since $x$ and $x+1$ have different parities, $c$ cannot be even when $x$ is odd. When $x$ is even, $k=2n$ must be even for $c=2n(x+1)+x$ to be even. In that case, $y=\frac{2n(x+1)+x-x}{x+1}=2n$.

In sum, solutions exist for all $x$ and $y$ being any even integers; $c$ is limited to being of the form $c=2n(x+1)+x$ for any particular chosen $x$.