4
$\begingroup$

I found the following theorem in a paper by Yann Bugeaud (page 12) ,

enter image description here

the theorem was not written in detail, to be specific,following two lines on page 13 were not understandable-

enter image description here

I think this theorem may exist in other books, paper, it seems general result (I could be wrong though). Where can I get the detail version of the proof ? or can anyone prove (providing the detail) above two lines in detail? or is there an error /typo? Because a member, Gerhard Paseman, suggested in his answer that-

As of this date, as written, I still think the question is not good for this forum. However, it raises one which I think is a good one for this forum. Here's the good question: Does (16) have a typo?

PS:

One of the member tried to prove it but failed, his answer is given as below-

If we divide both sides of equation (15) of page 12 by $\frac{\gamma}{\gamma-\delta} \gamma^{sd}$, we get- $\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}{\left(\frac{\gamma^s}{\alpha^r}\right)}^{-d}-1=\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}$. Then, we have $\left|\frac{\alpha(\gamma-\delta)}{\gamma(\alpha-\beta)}{\left(\frac{\gamma^s}{\alpha^r}\right)}^{-d}-1\right|=\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}\right|$ Since $\frac{\delta}{\gamma}\approx 0$ (i.e. $0<\frac{\delta}{\gamma}<1$ ) from $\gamma\ge |\delta|^{1+\eta}\gg|\delta|$, we have $\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}-\left(\frac{\delta}{\gamma}\right)^{1+sd}\right|\approx \left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|$. Note that as the values of $\gamma$ become larger and larger, $\frac{\delta}{\gamma}$ become smaller and smaller and become close to $0$, so, we ignored $\frac{\delta}{\gamma}$ by writing $\frac{\delta}{\gamma}=0.$ Similarly, $\frac{\delta}{\gamma}\approx 0$ and $\frac{\beta}{\alpha}\approx 0$ from $\alpha\ge |\beta|^{1+\eta}\gg |\beta|$, we have, $\frac{\gamma-\delta}{\alpha-\beta}=\frac{1-\frac{\delta}{\gamma}}{1-\frac{\beta}{\alpha}}\cdot \frac{\gamma}{\alpha}\approx\frac{1-0}{1-0}\cdot\frac{\gamma}{\alpha}=\frac{\gamma}{\alpha}$ and so $\left|\frac{\gamma-\delta}{\alpha-\beta}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|\approx \left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|$. Since $\alpha^r\approx \gamma^s$ (how?), we have $\left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma^{1+sd}}\right|\approx \left|\frac{\gamma}{\alpha}\cdot \frac{\beta^{1+rd}}{\gamma\cdot \alpha^{rd}}\right|=\frac{\gamma}{\alpha}\cdot \frac{|\beta|^{1+rd}}{\gamma\cdot \alpha^{rd}}=\frac{|\beta|^{1+rd}}{\alpha^{1+rd}}$. By inspection, $\alpha\ll \alpha^{1+(1-\eta)}$ and from $\alpha\ge |\beta|^{1+\eta}\gg |\beta| \implies \alpha^{rd}\gg |\beta|^{rd} $ so we get, $\frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{rd}}{\alpha^{1+rd}}$ $\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$(how?) $ \implies \frac{|\beta|^{1+rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1+(1-\eta)rd}}{\alpha^{1+rd}}=\alpha^{-\eta rd}$.

This answer is taken from another member which is not complete.

$\endgroup$
  • 2
    $\begingroup$ The fact that this comes from the equality of two terms of Lucas sequences (which is an expansion of (15)) is crucial, and should have been included in this post. Think about what (15) means, and then you should come to see why alpha to the r is close to gamma to the s. It should then reveal more to you, including how to understand the proof offered by the other user. Gerhard "Understanding Basics Gives Basic Understanding" Paseman, 2019.07.25. $\endgroup$ – Gerhard Paseman Jul 25 '19 at 18:16
  • $\begingroup$ @GerhardPaseman there is another "how" :) .... go for the detail answer and take 100 points!!! $\endgroup$ – Mike SQ Jul 26 '19 at 4:03
  • $\begingroup$ I would prefer you learn. (16) is an algebraic consequence, with the complication that alpha to the power -eta rd is roughly a constant off from ( beta to the power rd)/(alpha to the power rd). Then carefully check ( this is the work you need to do) that Corollary 1 is applicable to (16 ) to get the first of the chain of approximate inequalities, with the other two following from the assertions of the end of that sentence. Then check that this chain yields that d is bounded. Gerhard "My Reputation Is Quite Different" Paseman, 2019.07.25. $\endgroup$ – Gerhard Paseman Jul 26 '19 at 5:18
  • 1
    $\begingroup$ @GerhardPaseman dude! $\frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{rd}}{\alpha^{1+rd}}$ $\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$**(how?)** is not a simple "algebraic consequence", the member who provided it deleted his answer because he could not prove it (he programmed and found positive result but could not prove it)... we both tried , probably there is a 'general theorem' involved that goes without mentioning in the number math.. $\endgroup$ – Mike SQ Jul 27 '19 at 12:19
  • 1
    $\begingroup$ I'm afraid I can't help with your question, but since @GerhardPaseman has made some detailed comments/suggestions I suggest engaging with them (and trying to remember that not everyone is a fan of leetspeak) $\endgroup$ – Yemon Choi Aug 1 '19 at 5:35
1
+100
$\begingroup$

Update 2019.08.01 GRP: As of this date, as written, I still think the question is not good for this forum. However, it raises one which I think is a good one for this forum. Here's the good question: Does (16) have a typo?

I suspect (16) as written in the paper holds (but I am going to use Roman instead of Greek letters), but I do not see an easy justification, and the idea that B is like A^(1-e) appeals to me, although really B is more like A^(1/(1+e)), and I am not fully comfortable with the switch.

If we rewrite (16) so that instead the right hand side bound is now B^(-erd), the game changes. This is an obvious and derivable bound from (15z), and I can see log A $\lt \lt$ log B holding in this context. I would then rewrite the last line using beta instead of alpha, check carefully that using beta does not muck up the application of corollary 1, and come to the same conclusion for the theorem.

I encourage the original poster to amend the question by putting this new question at the top, and then leaving the rest as is if needed (although including the middle section of the quoted Theorem would be better still). An even better route is to have the original poster contact the author and put forth this question using a more polite tone (e.g. are you sure it is alpha and not beta on RHS of (16)?). In any case, there is an issue here to resolve, and hopefully someone on MathOverflow has enough familiarity with the paper to convince us that applying Corollary 1 is still justified.

End Update 2019.08.01 GRP

This is not a good question for this forum. Most of the manipulations are straightforward, with the crucial one being an application of the mentioned corollary. Also the cited paper makes some simplifications for ease of exposition, as well as cites another reference where a similar result is derived.

As I commented above, some material is left out of the question post which would make most of the derivation above understandable. Here are some missing pieces. To save on my typing, I change some of the notation.

(15) is a reference to an equation hypothesizing that a number is common to two distinct Lucas sequences. Before (15) is used, some remarks are made that certain values are allowed to be sufficiently large, that the critical parameter $\eta$ (which I will call $e$) is assumed less than 1/2, and that the constants implied by use of $\lt \lt$ are effectively computable. I will use $A,B,C,D,e$ in place of the Greek letters used above, and recommend reading the paper to review the material on Lucas sequences which I do not include here. I also leave out absolute value signs and assume positivity as needed.

The Roman version of (15) is $$ (A^n - B^n)/(A-B) = (C^m - D^m)/(C-D). $$ This is rewritten in an equation just before (16), using n=1+rd and m=1+sd, into $$ (A/(A-B))A^{rd} - (C/(C-D))C^{sd} = (B/(A-B))B^{rd} - (D/(C-D))D^{sd}.$$ I call this last equation (15z).

Now to address $A^r$ is close to $C^s$. By assumption, $A$ is larger than both $C$ and a significant power of $B$. This means $A/(A-B)$ is closer to $1$ than $2$ in value, because $B$ is sufficiently large. So $B^{rd}$ is way way smaller than $A^{rd}$. This means the right hand side of (15z) is like an integer with r bits, while the left hand side is the difference of two s bit integers and s is like 2*r. Not really, but if the most significant bits of $A^r$ match the most significant bits of $C^s$, those two quantities have a ratio pretty close to 1. Since this is about linear forms in logarithms, one expects these two quantities to be close in this sense. In particular, $r \log A$ and $s \log C$ are very close.

Now getting to (16) from (15z) is straightforward. The left hand side of (15z) is rearranged into a form for application of Corollary (1), which is on the left of $\lt \lt$ of (16). The right hand side of (15z) is divided by a quantity close to $A^{rd}$, and the puzzle is to figure out how $(B/A)^{rd}$ is like $A^{-erd}$ on the right of (16), since we are hiding constant multiplicative factors with $\lt \lt$. Here is where we use that $B$ is like $A^{1-e}$. The right hand side of (15z) is much smaller than $B^{rd}$, and for the step after (16) we can afford the approximation given by (16).

Applying corollary 1 is the only hard work, and requires detailed checking that it is applicable, which I do not do here, as more basic issues seemed to be asked to be resolved. This gives the leftmost relation in the last line, and the other two are straightforward, as is concluding the theorem from the last line of relations.

Again, the question is not a good one for this forum, and the main answer that might be appropriate for this forum amounts to doing work that is not explicitly requested, namely understanding the technicalities in the application of Corollary 1. Indeed, leaving out the middle section of the proof from the question is bad form. If it had been left in, this probably could have been answered elsewhere, and the real issue of using Corollary 1 brought forth.

Gerhard "Not In It For Bounty" Paseman, 2019.07.31.

$\endgroup$
  • $\begingroup$ I may have switched m and n above, or perhaps r and s. It is challenging doing this on a cell phone, so instead of editing and checking I will just apologize. Gerhard "It's Past My Bed Time" Paseman, 2019.07.31. $\endgroup$ – Gerhard Paseman Jul 31 '19 at 7:39
  • $\begingroup$ thanks for yor time but at least provide rigorous proof of, $\implies \frac{|\beta|^{1} \beta|^{rd}}{\alpha^{1+rd}}\ll \frac{\alpha^{1}|\beta|^{(1-\eta)rd}}{\alpha^{1+rd}}$**(how it is possible ?$\eta<1/2$ makes things spooky!)** and $\alpha^r\approx \gamma^s$ (how?),that is what we are looking to know + learn (forget the application of corollary 1). If possible do that before 9 hours. $\endgroup$ – Mike SQ Aug 1 '19 at 3:03
  • 1
    $\begingroup$ The derivation in your comment above (between asterisks) does not make sense to me, nor do I think it is needed, and I won't try for a rigorous proof of it. I spent several words trying to avoid that derivation and convince you why the desired relation holds between alpha^r and gamma^s. If you still don't get that, I don't know how to convince you. If you don't see how a trillion is close to (a trillion plus a million), you are definitely at the wrong forum. Gerhard "You're Welcome For The Time" Paseman, 2019.07.31. $\endgroup$ – Gerhard Paseman Aug 1 '19 at 4:45
  • $\begingroup$ not much help and disagree hugely about your comment about right or wrong... anyway bounty awarded to you for your time.. $\endgroup$ – Mike SQ Aug 1 '19 at 7:42
  • $\begingroup$ I have updated and change the question, can you suggest any academic to whom I can mail for solution? $\endgroup$ – Mike SQ Aug 5 '19 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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