From: k.buzzard@ic.ac.uk (Kevin Buzzard)
Subject: A fairly-innocent looking equation
Date: 2 Feb 99 18:29:40 GMT
Newsgroups: sci.math.numberthy
Keywords: x^2 + 7 nearly a power?

My boss (a group theorist) has run into an equation that he'd like
to "solve": it's

x^2+7=8p^n

where x is an integer, n>=0 is an integer, and p>=2 is prime.

My subject area isn't really equations of this sort, but I had a bash
anyway because it looked like fun. Initially he wanted to know the set
of solutions. But even for n=1 the question basically reduces
to something like "which primes are of the form f(m)" where f is a polynomial
of degree 2. So I guess one could happily conjecture that there
were infinitely many solutions and also be pretty sure that it was
not currently possible to prove this.

Having been presented with this obstruction, my boss said that
it would suffice for him if it were possible to show that
there was an integer N such that there were no solutions
to the equation with n>N. I don't even know whether it is
reasonable to conjecture that such an N exists, let alone
to prove that one does. But it struck me that there could
be people reading this mailing list that might know about this kind of
thing, or who would know something in the literature...

OK, that was the question. Now here's what I know.

For n=0 we can solve by hand. For n=1 the question is to find the
primes of the form (x^2+7)/8 and there are almost certainly infinitely
many.

For n=2 it gets a little more interesting. The equation becomes
one of "Pell type", and if we forget for a minute that p has to be
prime then we can solve for x and p in integers using standard techniques
and the question has become "here's a degree 2 recurrence relation,
how many primes occur in it?". Well I don't even know whether it's
conjectured that there's infinitely many primes in, say, the
Fibonacci series, so here I'm a bit stuck. I computed the first
few thousand terms of the (two) series in question and found some
_very_ impressive-looking solutions where p had about 600
digits (and I guess also was only technically "probably prime").
So maybe one could conjecture that there were infinitely many
solutions with n=2. But this doesn't bother us because we're
only looking for an N.

For n=2t, t>1, we can use the same analysis above and the question
becomes "here's a series generated by a recurrence relation, how
many t-th powers of primes are in it?". I guess there's only finitely many
squares in the Fibonacci sequences, but there are other sequences
(e.g. 1,2,3,4,5,...) defined by a second-order recurrence relation
that have infinitely many squares of primes in. For higher powers I've no idea.

For n=3 the equation is an elliptic curve and can be handled using
current techniques---in fact one can find all solutions with p prime
and n any multiple of 3 this way. In fact the only one is p=2,n=12,x=181.
This is in fact the largest n for which I know a solution. So
are there _any_ solutions with n>12?

A brute force search found p=2,n=4,x=11 as another solution, and this
is the only other solution I know with n>2. A rather fast loop
in pari convinced me that other than those 2 solutions mentioned
above with p=2 and n>2, there were no other solutions with 3<=n<=20
and p<10^7. I've searched much further with n=5,7 and found nothing.

The only method I knew which would work for general n is the
generalisation of the method (due to Nagell in some form
but which I read about in Cassels' book on local fields)
used to prove the result of Nagell that the only
solutions to x^2+7=2^m in integers x,m>=0 have m<=15.
The equation we're interested in clearly has this as a
special case. Having stared at the method of proof, I convinced
myself that in general one would be able to prove that
for any fixed prime p, there are only finitely many n>=0 for which
x^2+7=8p^n has a solution. But this doesn't seem to be strong
enough to give me a bound at all, let alone one which is uniform in p.

One last observation is of course that for n>2 fixed there will
only be finitely many solutions (by Faltings if you like).

And that's where I stand at the minute. Given that Nagell's result
is in the literature, does anyone know of anything else in the
literature that may help?

Kevin Buzzard
==============================================================================

From: k.buzzard@ic.ac.uk (Kevin Buzzard)
Subject: Re: A fairly-innocent looking equation
Date: 5 Feb 99 21:02:54 GMT
Newsgroups: sci.math.numberthy

Hello to all. I recieved several replies about the equation

x^2+7=8p^n

and all were very gratefully recieved. Apologies to those
of you to whom I didn't reply individually. Because I'm not 100%
sure who replied to the mailing list and who just mailed
me, I thought I'd summarise some of the results I learnt
over the last day or two about this equation. Just as a
passing remark: it's incredible how the internet can enable
someone (e.g. me!) to go from seeing an equation like this to
basically knowing the state of the art regarding its solution,
in just a few days.

A reminder: the "real" question was: give a bound N such that
there are no solutions to the equation above, in integers,
with p prime and n>N. In my initial posting I made the remark
that I wasn't 100% sure whether such a bound should exist,
but, as was pointed out to me by several people, the existence
of a bound is a consequence of the A-B-C conjecture. So at
least one can conjecture that N exists, to begin with.

The replies I got were from both algebraists and transcendence
theorists (and I had never met most of the transcendence theorists
so I'm extra-grateful that they replied!). I guess nowadays
there's a very healthy rivalry between these two groups when
it comes to Diophantine equations :) I'll summarise
what some of the algebraists said first, then some of
the transcendence guys afterwards.

Jordan Ellenberg clearly had a lot of fun with the equation: he pointed
out that if A^2+7=8p^n is a solution,
then one can consider the elliptic curve E:y^2=x^3+(3^3*7)x-(2*3^3*7)A.
(Other people remarked that the theory of elliptic curves and modular
forms might get somewhere, as it did for FLT etc).
The point is that this curve has good reduction away from 2,3,7 and p.
Furthermore, the reduction is semi-stable at p. Now one can start considering
the representation on the q-torsion of E for various primes q
and try to get somewhere using the theory of modular forms, for
example one could try and prove that only a finite set of primes q
could divide n. Unfortunately Jordan couldn't prove existence of
N unconditionally.

Frits Beukers reminded me of his 1980 paper where he proves
that a large class of sequences obtained by recurrence relations of order 2
can only take the value +-1 at most 3 times [See Cassels' exposition
of Nagell's proof for p=2 to see why recurrence relations have anything
to do with this question]. In fact, for the case in hand, Frits thinks
that his method seems to show that, for a _fixed_ p>2, there is
at most one possible value of n>0 for which x^2+7=8p^n is solvable
with x an integer. I thought this result was fascinating but
unfortunately it doesn't answer the question either.

Some people gave me some useful buzz-words: "Exponential Diophantine
Equation" and "generalized Ramanujan-Nagell equation" were among
them.

Finally, what can one say about the question using transcendence
techniques? Well, actually, a whole lot more! Several people pointed out
to me that, in fact, the existence of N _is_ known, effectively
and unconditionally, from a theorem of Schinzel and Tijdeman. A reference
for all this is Theorem 10.2 of "Exponential Diophantine
Equations" by Shorey and Tijdeman. Interestingly enough, the
elliptic curve algebric approach might well use the fact that
p is prime, but the transcendence approach shows

Theorem: Say f is a non-zero poly in Z[X] with at least 2 distinct roots, b
is a non-zero integer, m>=0 and y>1 are integers, x is an integer
and f(x)=by^m. Then m is less than some effective constant
depending only on f,b.

This shows the existence of N without even assuming p is prime. One
wonders whether one could do any better using transcendence methods
and this assumption, but this is certainly something I know very
little about.

Anyway, the big question left is "What is N"? According to a couple
of people, nowadays N is probably about 10^15 although it might
be reducible to 1000 or so, using various clever techniques.

Anyway, this turns out to be no use for my boss, who
has to do calculations in GL_N(Z/pZ) where N is the bound.
But basically I guess he _really_ wanted to know whether one could
attempt to prove that N=12 was the correct bound nowadays;
unfortunately I think that we still can't. Still, at least I had a whole
lot of fun! (And learnt a lot too).

Thanks a lot to everyone,

Kevin Buzzard