Finding the perfect square of a given operation

Question

Given $a, b, c \in \mathbb{N}$ such that $a+b=b*(a-c)$ and $c+1=p^2$, such as $p$ is prime, prove that $a+b$ or $ab$ is a perfect square.

So, from the given information I already know that $ab=a+bp^2$ and that $p^2=\frac{ab-a-b}{b}+1$. However, everytime I try to replace these "things", I always get to obvious results, such as $0=0$ or $a=a$ or $b=b$, and never get to the solution. How do I find a number $r$ such as $r^2 = a+b$ or a number $s$ such as $s^2=ab$ (i.e., how do I find a perfect square in this case)?

I'd like a hint... Thanks in advance!

Answer 1

Hint: You're given that

$$a + b = b(a - c) \; \; \to \; \; a = b(a - c - 1) \tag{1}\label{eq1A}$$

$$c + 1 = p^2 \tag{2}\label{eq2A}$$

Substituting \eqref{eq2A} into \eqref{eq1A} and simplifying gives, as you've determined, that

$$ab = a + bp^2 \; \; \to \; \; a(b - 1) = bp^2 \tag{3}\label{eq3A}$$

Substituting the right side of \eqref{eq1A} into the right side of \eqref{eq3A} gives

$$b(a - c - 1)(b - 1) = bp^2 \; \; \to \; \; (a - c - 1)(b - 1) = p^2 \tag{4}\label{eq4A}$$

Remember that $p$ is prime, so that limits what $a - c - 1$ may be. As you asked for just a hint, the rest of the answer is in the spoiler below.

Since $p$ is prime, this means with $$a - c - 1 = k \tag{5}\label{eq5A}$$ that, since $k$ is positive as \eqref{eq1A} has $a$ and $b$ both being positive, then $k$ is $1$, $p$ or $p^2$. Using \eqref{eq1A}, the first case gives that $a = b$ so $ab = b^2$ and the third case gives $ab = (bp^2)b = (bp)^2$, i.e., $ab$ is a square in either situation. For the second case, i.e., $k = p$, then $$a = pb \tag{7}\label{eq7A}$$ so using this in the right side of \eqref{eq3A} gives $$bp(b - 1) = bp^2 \; \; \to \; \; b - 1 = p \; \; \to \; \; b = p + 1 \tag{8}\label{eq8A}$$ Using \eqref{eq7A} and \eqref{eq8A} then gives that $$\begin{equation}\begin{aligned} a + b & = bp + b \\ & = b(p + 1) \\ & = (p + 1)(p + 1) \\ & = (p + 1)^2 \end{aligned}\end{equation}\tag{9}\label{eq9A}$$ This confirms that either $a + b$ or $ab$ are a perfect square.