Q:

Integer solutions to $\left(p + \frac{r}{a}\right)\left(p -1 + \frac{r}{a}\right) - \left(m + \frac{1}{a}\right)\left(m - 1+ \frac{1}{a}\right) = 0$

My question concerns the expression $$\left(p + \frac{r}{a}\right)\left(p -1 + \frac{r}{a}\right) - \left(m + \frac{1}{a}\right)\left(m - 1+ \frac{1}{a}\right) = 0.$$
Here $p,r,a,m\in \mathbb{Z}^+$ and $r < a$. I am trying to show that the above equation is satisfied if and only if $r = 1$ (which implies that $p = m$). The forward direction is trivial. I am struggling with the reverse direction. I don't really know where to start, or what tools to use. How might one prove that the above expression implies $r = 1$ (equivalently $p = m$)?
I tried multiplying out the expressions only to (unsurprisingly) gain a relation that is similar to the one above, except with more terms and less "symmetry" between the two products. I also tried by assuming $p \neq m$ and writing $p = m + s$ for some integer $s$ with the hope of concluding $s = 0$, but the same issue as in my first attempt emerged.

A:

Your expression is
$$\left(p + \frac{r}{a}\right)\left(p -1 + \frac{r}{a}\right) - \left(m + \frac{1}{a}\right)\left(m - 1+ \frac{1}{a}\right) = 0 \tag{1}\label{eq1}$$
If you let
$$b = p + \frac{r}{a} \tag{2}\label{eq2}$$
$$c = m + \frac{1}{a} \tag{3}\label{eq3}$$
and moving the second term to the right, \eqref{eq1} then becomes
$$b(b - 1) = c(c - 1) \tag{4}\label{eq4}$$
In general, either $b = c$ or they are $2$ distinct values of $x$ where $y = f(x) = x(x - 1)$ have the same value of $y$. Note that $f(x)$ is a concave up parabola with a minimum & line of horizontal symmetry at $x = \frac{1}{2}$. As such, for $b \neq c$, this means there is $t \neq 0$ such that $b = \frac{1}{2} + t$ and $c = \frac{1}{2} - t$, which from \eqref{eq2} and \eqref{eq3} gives 
$$\frac{1}{2} + t = p + \frac{r}{a} \tag{5}\label{eq5}$$
$$\frac{1}{2} - t = m + \frac{1}{a} \tag{6}\label{eq6}$$
Adding \eqref{eq5} and \eqref{eq6} gives
$$1 = p + m + \frac{r + 1}{a} \tag{7}\label{eq7}$$
However, since $p,r,a,m\in \mathbb{Z}^+$, this equation is not possible as the right side is $> 2$. Thus, the only possibility is that $b = c$, i.e.,
$$p + \frac{r}{a} = m  + \frac{1}{a} \; \implies \; \frac{r - 1}{a} = m - p \tag{8}\label{eq8}$$
The right side is an integer and since $1 \le r \lt a$, the left side can only also be an integer if $r = 1$, which then also means that $p = m$. This proves that the only solution is the one you stated.

