Is $p_1p_2...p_k - 1$ ever a perfect power?

Question

Is $p_1p_2...p_k - 1$ ever a perfect power, where $p_1 < p_2 < ... < p_k$ are the smallest primes?

The case for squares is simple, but I am struggling to generalise to all powers.

Answer 1

If you exclude the case $2-1=1^2$ then the answer is NO.

Suppose you have $$p_1p_2...p_n=x^2+1$$

Then you would have $3|x^2+1$ and this is a contradiction.

You can similarly exclude all even powers.

As far as generalisation goes, I'll think about it a bit more and post if I make some progress.

Answer 2

I think I have worked it out myself:

Let $p_1p_2...p_k = x^n + 1$. As pointed out by asdf, $n$ cannot be even. We can reduce our search to the case when $n$ is prime, since any power where $n$ is composite can be reduced to the power of a prime.

Since all the primes up to $p_k$ do not divide $x$, we have that $x > p_k$. This in turn gives us that $n < k$. But, since $k < p_k$, $n$ must be a prime contained in the product $p_1p_2...p_k$. Hence, let $n=P$, where $p_1 < P \le p_k$.

By Fermat's Little Theorem, $x^P \equiv x\mod P$. From the condition in the question, we also have that $x^P \equiv -1 \mod P$. Therefore, $P \mid x+1$.

Hence, let $x=tP-1$, where $t>0$. We therefore have that

$$p_1p_2...p_k = (tP-1)^P+1 = (tP)^P+...+\binom{P}{1}(tP)$$

However, since $P^2$ divides the RHS but not the LHS, we have a contradiction for $k>1$.