SolFeb28

Claim: There is only one solution to $q^2 + q + 1 = p^n + p^{n-1} + \cdots + p + 1$ for $p, q$ primes, $n > 2$ even.

Proof: (Due to HiLander) Suppose $(p, q, n)$ satisfies the claim. Since $n$ is even, $n = 2 k$ for some $k$ , and since $n > 2$ , we have $k > 1$ and $p \ne q$ . Subtracting $1$ from both sides of the formual, we factor both sides:

$q(q+1)= p(p^{n-1} + p^{n-2} + \cdots + 1) = p\left(\frac{p^n - 1}{p-1}\right) = \left(\frac{p}{p-1}\right) (p^k-1)(p^k+1)$

Suppose first that $q \le p^k - 1$ . Then

$q(q+1) < (p^k-1)(p^k + 1) < \left(\frac{p}{p-1}\right)(p^k-1)(p^k+1),$

a contradiction.

Now suppose $q > p k + 1$ . Since $q$ is prime, and we know $q \nmid p$ and $q \nmid (p^k+1)$ , we must have $q \mid \frac{p^k-1}{p-1} < p^k$ , another contradiction.

Therefore $q \in \{p^k,p^k+1\}$ . If $q = p k$ , then $k = 1$ since both are prime, so instead we have $q = p k + 1$ . But that means

$q+1 = \frac{p}{p-1}(p^k-1) = (p^k-1) + \frac{p^k-1}{p-1} = (q - 2) + (1 + \cdots + p^{k-1}),$

and so

$3=1 + p + \cdots p^{k-1}.$

Since $p \ge 2$ and $k \ge 2$ , this implies both hold at equality, else the RHS of our last equation is too large. From this, we get $n = 2 k = 4$ , and $q = p k + 1 = 5$ , so $(2,5,4)$ is the only solution $(p, q, n)$ . $\square$

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