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We have $a(a^{p-1} - 1) = pb^2$. An example is $a=9$, $p=2$, $b=6$. Since the factors on the left are coprime, it must be the case that $\frac{a}{p}$ of $a$ is a perfect square. In the first case $a$ must be a multiple of $p$ and so checking $a=1,2,\ldots,8$ (with the corresponding possible $p$) gives no solution. When $a$ is a perfect square, it remains to rule out only $a=4$. If $p=2$, then the power of $2$ on the left is $2$ but on the right it is odd, contradiction. So $p\geq 3$, $b=2t$ and $4^{p-1} - 1 = pt^2$. Factor to get $(2^{p-1} - 1)(2^{p-1} + 1) = pt^2$, with the factors on the left being coprime. If $2^{p-1} \pm 1 = t^2$, then there is no solution (factor up to $(t - 2^k)(t + 2^k) = \pm 1$ where $k = \frac{p-1}{2}).