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Working out some similar questions with my systematic approach as done for instance in this older question (and in some other questions) I could not yet find the equivalent proof of nonexistence of a solution in $a,b \gt 0$ $$ 5^{3+a} = 11^{2+b} + 4 \tag 1$$ which I usually rewrite as $$ {5^a -1 \over 11^2} = {11^b - 1\over 5^3 } \tag 2$$

Possibly there is a simpler method than mine by some cleverer modular considerations.

Well, I didn't check whether we can deduce from the fact that $4=2^2$ being a square number there is only a finite (already known) list of possible solutions for $x^a-y^b = z^2$ for $a,b>2$ which would solve this immediately. However, I'd prefer the other way:

  • I'd like to see a proof using modularity to understand, why or where my own method fails or why it might need only more/excessive effort.
Gottfried Helms
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    According to the generalized Catalan-conjecture (See here : https://en.wikipedia.org/wiki/Catalan%27s_conjecture ) , the only powers $n$ , for which $n+4$ is a powe as well, are $4,32,121$. This would imply the claim. If a pair with positive $b$ is a solution, $b$ must be greater then $10^5$. – Peter May 10 '17 at 12:49
  • A modular-equation proof cannot work because $5^3=11^2+4$ is a solution. But methods like the infinite descent might work. – Peter May 10 '17 at 12:51
  • @Peter: thanks for that link, I didn't remember the keyword/name, I had something named after S.Pillai in mind and think Mike Bennet has done canonical work with this but had not the papers. At your second comment: please see my worked answer. However, I'd like to see infinite-descent at work here. Do you have an idea how to start with this? – Gottfried Helms May 10 '17 at 13:08
  • No, I just thought of a method that could do the job. With brute force, I showed that $b$ must be very large (larger than $10^5$) and intuitively, such large numbers of the form $11^{b+2}+4$ will not be a power of $5$, but I have no idea for a general proof. – Peter May 10 '17 at 13:10
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    My analysis $5^a=\dfrac{11^{2+b}+4}{125} $ $=\dfrac{(5^3-4)*11^b+4}{125}$ $=11^b+\dfrac{4(1-(10+1)^b)}{125}=11^b+4\dfrac{-10^b-bC_110^{b-1}-・・・-100\frac{b(b-1)}{2}-10b}{125}$ $=5^3A+11^b-\dfrac{2b(b-1)}{5}-\dfrac{8b}{25}$

    Then $b$ must be multiple of $5^2$. Let $b=25kb'$

    $11^{25kb'}-10kb'(25b'-1)-8kb'$

    $k=2,7,・・・$

     – Takahiro Waki May 10 '17 at 13:19
  • @Peter: at your 2nd comment: for a similar problem there exist -as I see it- elegant and concise modular considerations, see my link in the question and also the link to an answer in AoPS / math-olympiad http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=418604 They are using squares and cubes of primes as moduli to arrive at the contradictions - much more concise than my ansatz, only not so general and systematic. – Gottfried Helms May 10 '17 at 13:22
  • @TakahiroWaki : perhaps you better form a full answer? I'd like to see your approach fully expanded. – Gottfried Helms May 10 '17 at 13:23
  • @GottfriedHelms What I meant, we cannot disprove the equation by taking it modulo some number $m$. Additional tricks, like quadratic residues, or that a number of the form $6k+5$ cannot have only prime factors of the form $6k+1$ , can be successful. Maybe, the use of the ring of Gaussian integers does the job. – Peter May 10 '17 at 13:34
  • @GottfriedHelms And proofs using such special tricks are hard to generalize. And additionally, they are usually very difficult to find. – Peter May 10 '17 at 13:36
  • @Peter: "they are very difficult to find" - true, that was the reason that I tried to find some approach which is general enough. The ansatz which I layed out at the link in my question is such a general attempt and solved *blindly* many questions of this type (provisorically canned in a Pari/GP procedure) up to some highest exponents and bases. (Only I had a glitch in my current problem-configuration, thus my surprised current question) – Gottfried Helms May 10 '17 at 13:43
  • What I do not quite understand. How does your ansatz COMPLETELY solve the problem ? We can rule out many exponents and get larger and larger lower bounds, but how do we know that NO positive $b$ solves the equation ? – Peter May 10 '17 at 13:46
  • @Peter: in our example, when in the rhs we need some exponent such that the primefactor $5$ occurs too often - then there cannot be equality with the lhs. The comparision of the primefactorizations and the (iterative) mutual adaptions imply always more factors in the exponents - and at one step a (large) primefactor is required in the rhs whose order to base $11$ contains $5^3$ as a factor. Then we're done... Perhaps look also at the answer of mine to which I've linked in my Q or look at http://go.helms-net.de/math/expdioph/CyclicSubgroups_work.pdf (pg 18 ff) where I explain it in much detail – Gottfried Helms May 10 '17 at 13:55
  • $5^3≒11^2$ shows $5^{3m}≒11^{2n}$ but since $5^{3m}$ is far from $11^{2n}+4$, $5^{3m}=11^{2n}+4$ has no solution on $(m,n)>(1,1)$. Therefore it is precisely assumed that there is no solution on this equation. – Takahiro Waki May 10 '17 at 16:16
  • @TakahiroWaki : hmm, I don't think that the consideration of $3m$ and $2n$ is enough. Continued fractions give us other combinations with higher values whose relative approximation improves - what would help us here is that the absolute approximation has a pattern which in general/in asymptotics should be useable ... but perhaps I do not express this correctly here... – Gottfried Helms May 11 '17 at 02:13

1 Answers1

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I solved the problem myself - I'd just overlooked one test. Here it is, how it went.

The ansatz is to determine the required structures in the exponents $a$ and $b$ to allow compatible sets of primefactors in the lhs and rhs: $$ {5^a-1 \over 11^2 } = {11^b-1 \over 5^3 } \tag 1$$

Initialization

To have exactly $11^2$ as factor in the numerator of the lhs we need that $a=5 \cdot 11 \cdot a_2$ : $$ \begin{array}{} \text{lhs :} & \large {5^a-1 \over 11^2 } &=\large {5^{55 \cdot a_2}-1 \over 11^2 } & \underset{a_2=1}= & 2^2.71.103511.\text{<big>} \end{array}$$

To have exactly $5^3$ as factor in the numerator of the rhs we need that $b=5^2 \cdot b_2$ $$ \begin{array}{} \text{rhs :} & \large {11^b-1 \over 5^3 } &=\large {11^{25 \cdot b_2}-1 \over 5^3 } & \underset{b_2=1}= & 2.3001.3221.\text{<big>} \end{array}$$

Step 1 : primefactor $2$

Next, we adapt the exponents so that lhs and rhs have the primefactor $2$ to the same exponent. We need $a_2$ and $b_2$ being even, so we define $a_2=2 a_3$ and $b_2 = 2 b_3$ finding

$$ \begin{array}{} \text{lhs :} & \large {5^{55\cdot 2 \cdot a_3}-1 \over 11^2 } & \underset{a_3=1}= & 2^3.3.23.67.71.521.5281.\text{<big>} \\ \text{rhs :} & \large {11^{25 \cdot 2 \cdot b_3}-1 \over 5^3 } & \underset{b_3=1}= & 2^3.3.3001.3221.13421.\text{<big>} \end{array}$$

Step 2 : primefactor $3001$

Next, we adapt the exponent in the lhs that it has primefactor $3001$. We need the exponent containing $250$ so $a_3$ needs to provide the factor $5^2$ so we define $a_3= 5^2 a_4$ finding

$$ \begin{array}{} \text{lhs :} & \large {5^{55 \cdot 2 \cdot 5^2 \cdot a_4}-1 \over 11^2 } & \underset{a_4=1}= & 2^3.3.23.67.71.101.251.401.521.1901.3001.\text{<big>} \end{array}$$

Step 3 : primefactor $251$

Finally, we adapt the exponent in the rhs that it has primefactor $251$. We need thus the exponent containing $250$ as on the lhs, so $b_3$ needs to provide the factor $5$ so we define $b_3= 5 b_4$ finding

$$ \begin{array}{} \text{rhs :} & \large {11^{25 \cdot 2 \cdot 5 \cdot b_4}-1 \over 5^3 } & \underset{b_4=1}= & 2^3.3.5.251.3001.3221.13421.\text{<big>} \end{array}$$

Proof by contradiction on primefactor $5$

Here in the primefactorization of the rhs we find an additional factor $5$ which cannot be compensated by varying the exponent in the lhs. We needed only partial adaptions of exponents (but of course all steps are mandatory), and still arrived at the expected contradiction:

We cannot have exponents $a,b>0$ providing equality in (1) and thus the only solution is the initial one $ 5^3 = 11^2 + 4$

What we have explicitely used were the primefactors $2,3001,251$, and likely direct considerations on modularities (as usual with such problems) should use that moduluses (possibly one does not need the primefactor $2$). However this does not exclude that my basic approach could uncover more/alternative solutions for the disproof.

Appendix: a provisorical Pari/GP routine for solving such problems systematically can be obtained if desired
For an extended explanation look at this older MSE-answer or this

Gottfried Helms
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