A prime’s square dividing a sum of a geometric sequence

Question

For what $p$ primes does there exist two positive integers $(a,b)$, such that $$\dfrac{1-a^p}{1-a}=bp^2?$$

This is from a junior olympiad from Britian but I can’t solve it. Please help!

Answer 1

Consider instead $\frac{(a+1)^p-1}{a}-bp^2=a^{p-1}+\binom{p}{p-1}a^{p-2}+\cdots+\binom{p}{1}-bp^2$. Now apply Eisenstein's Criterion with prime $p=p$. $p$ doesn't divide $1$ infront of of $a^{p-1}$. Each binomial coefficient is divisible by $p$ (in particular since $p$ is prime). Finally $p-bp^2$ is divisible by $p$, but not by $p^2$. So the polynomial is irreducible over the rationals in $a$. There are no such $(a,b)$ for $p>2$.

Edit: Mike rightfully points out that the degree one case is special. In this case if $p=2$, then the polynomial becomes $a+2-4b=0$, which has solutions $a=2(2b-1)$, e.g. $a=2,b=1$, or in the original notation $(a=3,b=1 )$.