# Proving non-existence of solutions to $3^n-2^m=t$ without using congruences

I made a passing comment under Max Alekseyev's cute answer to this question and Pete Clark suggested I raise it explicitly as a different question. I cannot give any motivation for it however---it was just a passing thought. My only motivation is that it looks like fairly elementary number theory but I don't know the answer.

OK so one problem raised in the question linked to above was "prove there are no solutions to $3^n-2^m=41$ in non-negative integers" and Aleksevev's answer was "go mod 60". It was remarked afterwards that going mod 601 or 6553 would also nail it. For example, modulo 6553 (which is prime), 3 has order 39, 2 has order 117, but none of the 39 values of $3^n-41$ modulo 6553 are powers of 2 modulo 6553.

My question (really just a passing remark) is:

Is there an integer $t$ such that the equation $3^n-2^m=t$ has no solutions in non-negative integers $m$, $n$, but for which there are solutions modulo $N$ for all $N\geq1$? (By which of course I mean that for each $N\geq1$ the equation is satisfied mod $N$ for some integers $m,n\geq0$ depending on $N$; I am not suggesting that $m$ and $n$ be taken modulo $N$ or are independent of $N$).

This for me looks like a "Hasse principle" sort of thing---in general checking congruences doesn't give enough information about solvability of the polynomial in integers and there are many examples of such phenomena in mathematics. As exponential Diophantine equations are harder than normal ones I would similarly expect the Hasse Principle to fail here, but others seemed to be more optimistic.

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@Kevin: thanks for posting it. Yes, I too know nothing about exponential Diophantine equations, so I figure there is nowhere to go but up. (P.S.: I think a big part of the solution of H10 is that exponential DE's are not any harder than ordinary DE's. Indeed, in a certain precise sense, nothing is harder than ordinary DE's!) –  Pete L. Clark Jun 30 '10 at 10:52
@Pete: Not even partial DEs –  Victor Protsak Jun 30 '10 at 11:01
@Pete: on reflection perhaps the question above isn't the right one; perhaps the right one is something like "can someone give me an example of an exponential Diophantine equation which violates the Hasse principle over Z". Perhaps that is a more natural question, for this should be a standard counterexample in my arsenal. –  Kevin Buzzard Jun 30 '10 at 11:36
I have to say I assumed the OQ was homework. That was why I didn't do more than give hints. (I knew the case t = 1 as an easy thing - wasn't it the last step in one of the proofs of Hopf invariant 1?) Fixing t we get that log 3/log 2 is close to a rational number. With more technology, can we find some definite examples? Or hypotheses using the Artin conjecture on primitive roots? –  Charles Matthews Jun 30 '10 at 12:42
This is a just a heuristic. What is the probability that $3^n-2^m=t$ has an "accidental" solution mod $N$? For most $N$ it is large, but not all. Choose $N$ with $2,3$ both "small" multiplicative order, then it is lower. There are "many"(?) such N prime. The list where $o_2o_3 < p$ for is 683!, 6553, 34511, 599479!, 11321831, 32397061, 42521761!, 47763361, 48912491, 92896849 ... the exclams are $o_2o_3=p-1$, the others are smaller. For each the probability is "low" (away from 1) and independent(?) to get an accident. The $n,m$ probability space needs thinking too. –  Junkie Jun 30 '10 at 13:12

I believe this is closely related to the conjecture by Brenner and Foster here. They ask if an exponential Diophantine equation of the form $$\sum \epsilon_i p_i^{m_i}=t$$ where $\epsilon_i=\pm 1$ can be solved using modular arithmetic. I don't know if your special case is any easier but maybe that's a good place to start looking in case there is anything in literature.

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It's Problem 8.01 in the paper. My understanding is that they're suggesting that congruences should be enough in this special case, based presumably on the wealth of such equations that they solve in this way in the paper. –  Kevin Buzzard Jun 30 '10 at 13:21
I'm going to accept this answer because I'm not sure I'm going to get any better than this. I was wondering whether someone would try some sort of heuristic argument to justify that congruences should be enough, but perhaps this paper is evidence that the heuristic will work. –  Kevin Buzzard Jul 1 '10 at 6:32

Just in case people were not aware, in the case of the single exponential problem $a^{n}=t$, if there is a solution modulo all prime powers, then there is an integer solution. [A nice proof is given in Cojocaru and Murty's book "An introduciton to sieve methods and their applications".

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Does this proof involve sieve methods? I know how to prove it using Chebotarev's Density theorem. –  David Corwin Aug 22 '10 at 23:09

i prove that for a large $N$,such that $m,n>N$,the equation of $3^n-2^m=l$ has no solution

we rewrite above eqution to form:

$3^n=(2^{m/2}-il^{1/2})(2^{m/2}-il^{1/2})$

if $gcd(2^{m/2}-il^{1/2}),(2^{m/2}-il^{1/2})=d$

similar to $z[i]$,$NORM(2^{m/2}+\-il^{1/2})=2^m+l=3^n$ and $NORM(2^{{m/2}+1})=2^{m+2}$,then $d=1$

so $2^{m/2}+il^{1/2}=(a+ib)^n$ and $2^{m/2}-il^{1/2}=(a-ib)^n$,that $3=a^2+b^2$, we can write:

$2^{m/2}+il^{1/2}=(a+ib)^n=3^{n/2}(cos(nx)+isin(nx))$,and also

$2^{m/2}-il^{1/2}=(a+ib)^n=3^{n/2}(cos(nx)-isin(nx))$ and $tan(x)=b/a$

then $cos(nx)=2^{m/2}/3^{n/2}$ and $sin(nx)=l^{1/2}/3^{n/2}$

so $tan(nx)=l^{1/2}/2^{m/2}$by lagrange's theorem,there is a $c$,such that $c$ is between $a_1$ and $b_1$ and

$f'(c)=f(b_1)-f(a_1)/(b_1-a_1)$,then $(1+tan^{2}(c))=tan(nx)/nx$ ,or $tan(nx)>nx$

therfore $x<(l^{1/2})/(2^{m/2}n)$

for $m,n>N$, $x$ is almost 0,then$b$ is almost zero,since$2^{m/2}+il^{1/2}=(a+ib)^n$,$l$ should be almost zero and this is contradiction,so for large $N$ this equation has no solution for a fix$l$

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This is a special case of the (proven) Catalan Conjecture: en.wikipedia.org/wiki/Catalan's_conjecture. –  Pete L. Clark Aug 23 '10 at 1:41
catalan' conjecture is equation$x^m-y^n=1$,not this –  Hashem sazegar Aug 23 '10 at 5:32
Even if one would correct all the typos, your proof still would not make any sense. –  Franz Lemmermeyer Aug 23 '10 at 9:55