In an answer to my own question, I showed that if $p$ is an odd prime and $p=a^m \pm b^n$ with positive integers $a,b$ relatively prime to $p$, then $p$ does not simultaneously divide the Fermat quotients $\frac{a^{p-1}-1}{p}$ and $\frac{b^{p-1}-1}{p}$.

Once upon a time, this result would have been of some interest because it implies the first case of Fermat's Last Theorem (FLT) for certain prime exponents $p$. More precisely, Wieferich, Mirimanoff, and others showed that if the first case of FLT fails for the prime $p$, then $\frac{a^{p-1}-1}{p}$ will be divisible by $p$ for every prime number less than or equal to 89.

In particular, the above two paragraphs show that the first case of FLT holds if $p = a^m \pm b^n$ for some primes $a, b \leq 89$.

Although FLT was proved by independent methods, results like the one in my other question still have some interest. For instance, as far as I know, it is still unknown whether there are nontrivial solutions $x,y,z$ in the cyclotomic field $\mathbb{Q} (\zeta_p)$ (whether they are all relatively prime to $p$ or not) that satsify

$$ x^p + y^p = z^p$$

In ``On the first case of the fermat theorem for cyclotomic fields", Kolyvagin proves that there are no such solutions relatively prime to $p$ if each of the numbers $\frac{a^{p-1}-1}{p}$ with $a$ prime and $\leq 89$ is relatively prime to $p$. In particular, this is true when $p$ has the form $p=a^m \pm b^n$ with $a, b$ primes $\leq 89$.

I would presume that the set of such primes is quite scarce (although the possibility $p = a^m-b^n$ throws me a bit), so my question is:

Are there known heuristics or results for the set of primes $p=a^m \pm b^n$ with $a,b$ primes less than $89$ (one of them, of course, being 2)?


2 Answers 2


Non-expert disclaimer here.

Applying some heuristics, you get that there are $O(\log N)$ primes of the form $a^m-b^n$ for fixed $a$ and $b$.

To see this, let's count the values taken by $a^m-b^n$ in the range $[0,N]$. I claim there are $O(\log^2N)$ such terms. If $m$ and $n$ are in the range $[0,2\log N]$, this accounts for $O(\log^2N)$ pairs, a good fraction of which are in the given range.

If $m > 2\log N$, then there is at most one $n$ such that $a^m-b^n$ is in the range. To complete the bound, we need to give an upper bound on the $m$'s for which there can exist $n$'s such that $a^m-b^n$ is in the range.

Consider $m > 2\log N$ and assume that $|a^m-b^n| < N$ so that $n \gtrsim \log N$ also. Notice that $|a^m-b^n|=a^m|1-b^n/a^m|\approx a^m\log(b^n/a^m)=a^m(n\log b-m\log a)$. By Baker et al's results on logarithmic forms, $|n\log b-m\log a|>m^{-k}$ for a constant $k$ that depends on $a$ and $b$, so that $|a^m-b^n|\gtrsim a^m/m^k$. In particular, if $a^m > N(\log N)^k$, you can't be in range. This shows that there are at most $\log^2N$ terms of the form $a^m-b^n$ in range. Of those, how many are prime?

The heuristic would be that the "density" of difference powers is $\log N/N$ (the derivative of $\log^2N$). The density of primes is $1/\log N$ by the PNT. So since they are `obviously' independent, the density of primes of this form is $1/N$. So... there should be infinitely many according to this heuristic, but there should be $O(\log N)$ such primes up to $N$.

  • $\begingroup$ You might be able to simplify this analysis by noting that one of a or b must be 2. Gerhard "Ask Me About Even Primes" Paseman, 2012.06.07 $\endgroup$ Jun 8, 2012 at 0:46
  • $\begingroup$ I'm guessing it wouldn't change much... $\endgroup$ Jun 8, 2012 at 0:59

Hi there, If you're only interested in heuristics, then if you replace $89$ by $x$, fix $m$ and $n$, fix $\pm$ to be $+$ or $-$, fix one of $a,b$ to be $2$, and let the other variable range over primes less than $x$, then it is in the right form for the Bateman-Horn conjecture to be applied to. See for example,

P. T. Bateman, R. A. Horn, A heuristic asymptotic formula concerning the distribution of prime numbers, Mathematics of Computation 16 (1962), 363--367.

(Say $a=2$, $b$ ranges over primes less than $x$, $m$ and $n$ are fixed, and $\pm$ is $+$. Then one can use the case of Bateman-Horn for two polynomials simultaneously, one polynomial being $z$, the other one being $z^n+2^m$.) Then depending on which other variables you want to vary, you can apply Bateman-Horn to each case like this.

Don't know how accurate it would be for $x=89$ ! But it should be more accurate for $x \rightarrow \infty$. The usual caveat about this still being a conjecture applies.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.