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Looking at prime multiples of $P=[1,1]$ on the curve $y^2=x^3+x-1$ the size of the denominator grows quite rapidly. So $5P=[\frac{685}{11^2},\frac{-18157}{11^3}],7P=[\frac{[154513}{443^2},\frac{-45623219}{443^3}]$

Looking purely at the x coordinate we have

k       sqrt(denominator(k*P))
3       1      
5       11      
7       443     
11      656243  
13      1048516201 
17      4451192987631289   
19      429330677071423079   
23      77700648618312971843469771251 
29      7517777790273327975717181745948599636777678033 
31      30820930821843565941522654527823658160891296876267921  
37      587308602722146793206557071567309610618121704728654635117962467133627539771     
41      22916712844996907328126003624983339852284728578820686400041407536873657173496053280584200979

I know this involves the Neron-Tate height somehow and is part of the reason finding a factor-base for calculating discrete logs on elliptic curves over $F_p$ seems impossible.

I was wondering if there is a choice of point or curve or prime multiple for which the size of the denominator of large prime multiples of that point would be small, or at least a partial factorization could be achieved.

Is this possible?

It would be very interesting to find a large composite number $n$ which could be shown to be a factor of the denominator of a a roughly $n^\frac13$-sized prime multiple of a point.

user72700
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    Don't you have more important things to do Mr. President? – Baby Dragon Apr 15 '13 at 15:44
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    More important than pure mathematics? – Abel Apr 15 '13 at 15:58
  • That's what I also thought at first reading, @Abel, but then I realized that not much mathematics, pure or whatever, can be done if one has to broom the floor after a north korean nuke lands on one's office... – DonAntonio Apr 15 '13 at 16:10
  • If $P$ is a torsion point, then the denominators will stay bounded. So you obviously are making an implicit assumption that $P$ is not torsion. – Álvaro Lozano-Robledo Apr 15 '13 at 16:23
  • @ÁlvaroLozano-Robledo Interesting.. If P is a torsion point how big can the denominators get? Can you give me an example of large ones? – user72700 Apr 15 '13 at 17:27
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    @BarackObama, if your curve is defined over $\mathbb{Q}$, and your model has integral coefficients, then the denominator of the $x$-coordinate of a torsion point is at most $4$. But if you allow your model to have $\mathbb{Q}$-coefficients, then the torsion points may also have large denominators. – Álvaro Lozano-Robledo Apr 15 '13 at 17:33
  • For instance, the curve $y^2+xy=x^3+4x+1$ has a $2$-torsion point $(-1/4,1/8)$. – Álvaro Lozano-Robledo Apr 15 '13 at 17:35

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The strong form of Siegel's theorem says that the numerator and denominator are about the same size for large multiples. Here's the statement. Let $P\in E(\mathbb Q)$ have infinite order. Write $x(nP)=A_n/B_n$ in lowest terms. Then $$ \lim_{n\to\infty} \frac{\log|A_n|}{\log|B_n|} = 1. $$ Since we also know that $$ \lim_{n\to\infty} \frac{\log\max\{|A_n|,|B_n|\}}{n^2} = \hat h(P) > 0, $$ This shows that both $\log|A_n|$ and $\log|B_n|$ grow rapidly.

Joe Silverman
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