Find all \(x, y \in \Z_+\) and primes \(p\) such that
\[p^x + 1 = y^2\]
Solution
We begin by rearranging and transforming our equation:
\[ \begin{aligned} p^x + 1 &= y^2 \\ p^x &= y^2 - 1 \\ p^x &= (y+1)(y-1) \end{aligned} \]
Therefore, we have
\[ \left.\begin{aligned} y+1&=p^a, a \in \Z_+ \\ y+1&=p^b, b \in \Z_+, a > b \end{aligned}\right\} \Longrightarrow (y+1)-(y-1) = p^a - p^b \Longrightarrow 2 = p^b(p^{a-b} - 1) \]
This leaves us with two cases:
- Case 1: \(p^b = 2; p^{a-b} - 1 = 1\). This gives us \(p=2,a=2,b=1\).
- Case 2: \(p^b = 1; p^{a-b} - 1 = 2\). This gives us \(p=3,a=1,b=0\).
And therefore the solutions are:
\[\begin{aligned} \textbf{p=2, x=3, y=3} \\ \textbf{p=3, x=1, y=2} \end{aligned}\]
And we are done.