A new extreme abc-example
=========================

For coprime integers a, b, c such that a + b = c the well known
"abc-conjecture" says that 

   H(a,b,c) < c_epsilon . N(a,b,c)^(1+epsilon),
   
where H(a,b,c) = max(|a|,|b|,|c|) is the height of the triple,
and N(a,b,c) = product of primes dividing abc is the conductor.

We may define the 'quality' of an abc-example as the corresponding
value of 1 + epsilon when we force the constant to be 1, i.e.

   quality(a,b,c) = log H(a,b,c) / log N(a,b,c).
   
In 1985 I came up with the example

   11^2 + 3^2 5^6 7^3 = 2^21 23
   
with quality 1.625991, and a little later (when I remember well in 1987)
Eric Reyssat found

   2 + 3^10 109 = 23^5
   
with quality 1.629912. These two examples still are the best ones known.

For (nonzero) algebraic numbers a, b, c such that a + b = c the (uniform)
abc-conjecture can be formulated as follows:

   H(a/c,b/c,1) < c_epsilon^[Q(a/c):Q] (Disc(Q(a/c):Q) N(a/c,b/c))^(1+epsilon),
   
for appropriate height and conductor functions H and N. See Broberg's paper 
for details. This leads to the following quality function:

   quality(a,b,c) = log H(a/c,b/c,1) / (log Disc(Q(a/c):Q) + log N(a/c,b/c)).

Broberg searched for extreme examples for this quality function.
He found two examples with quality larger than that of Reyssat's example,
namely

   r^17 + (1-r)^5 (3-r) = (1+r)^5 (3+r), r = sqrt(2)
    
with quality = 1.768124, and

   r + ((1+r)/2)^13 = ((1-r)/2)^13, r = sqrt(-7),

with quality = 1.707221. His third example is also quite good:

   (1+r)^14 + 1 = (1+r)^7 r^3 13^2, r = sqrt(2),
   
with quality = 1.561437.

After a moment's thought I realised that Broberg's first two examples
both come from the largest solution of the Ramanujan-Nagell equation. 
The Ramanujan-Nagell equation is x^2 + 7 = 2^n. It was conjectured by
Ramanujan and proved by Nagell that x = 181, n = 15 is the largest
solution. Now we concentrate on the equality 181^2 + 7 = 2^15.

First we write it as 

   181^2 - 2 (2^7)^2 = -7,
   
and we factor over Q(sqrt(2)). Then we get (writing r = sqrt(2))

   181 + 2^7 r = (1+r)^5 (3+r).
   
The large number 181 will spoil the quality of an abc-example, as
it is not divisible by a large power. So we want to eliminate it. 
This can be done by subtracting the conjugate equation. Then
we get exactly Broberg's first example, the one with quality 1.768124.

Then we write the Ramanujan-Nagell largest solution equality as

   181^2 + 7 = 2^15,
   
and we factor over Q(sqrt(-7)). Then we get (writing r = sqrt(-7))

   (181+r)/2 = - ((1-r)/2)^13.
   
Again we subtract the conjugate equation to get rid of the annoying
181, and we arrive nicely at Broberg's second example, the one with
quality 1.707221.

Another well known diophantine equation with a nice large solution
is Ljunggren's equation x^2 - 2 y^4 = -1, with solution x = 239,
y = 13. Factoring over Q(sqrt(2)) and losing the number 239 we find
Broberg's third example. (Factoring 239^2 + 1 = 2 . 13^4 over
Q(sqrt(-1)) yields a less interesting example).

When I had seen all this, I wondered what more you could find along
these lines. Note that the quadratic equations above are directly
related to second order recurrence sequences in which something nice
happens (such as an extremely small element, or an extreme power as element).

The first example that thus, quite naturally I think, came to my mind 
was the extreme zero in the Berstel sequence. It turns out to be quite 
good from the abc-viewpoint too. Here it is in detail.

The Berstel sequence (u_n) is the recurrence sequence of order 3, 
defined by

   u_(n+1) = 2 u_n - 4 u_(n-1) + 4 u_(n-2), u_0 = u_1 = 0, u_2 = 1.
   
It is well known that u_n = 0 for n = 0, 1, 4, 6, 13 and 52. This sequence
is well known because it has so many zeroes, and also because there
is a zero at such a large index (52).

It is also well known that we may write for all n

   u(n) = ((gamma-beta) alpha^n + (alpha-gamma) beta^n + (beta-alpha) gamma^n) 
                             / (4 sqrt(-11)),
   
where alpha, beta, gamma are the roots of x^3 - 2 x^2 + 4 x - 4 = 0.
Thus the equation u_52 = 0, i.e. 

   (gamma-beta) alpha^52 + (alpha-gamma) beta^52 + (beta-alpha) gamma^52 = 0,
   
yields an abc-example with promisingly high exponents.

When I suggested this example to Niklas Broberg, he immediately computed the
quality to be

   quality = 1.920859.
   
Need I say more? 

Well, an interesting question is: do these good abc-examples over
extensions of Q lead to elliptic curves, defined over these extensions of Q,
of Frey-Hellegouarch type (or isogenous to such a curve), that have large 
Tate-Shafarevich groups (compare my paper "A + B = C and big Sha's",
Quarterly Journal of Mathematics (Oxford) (2) 49 [1998], 105 - 128)?
(I don't even know if this question makes sense.)


Benne de Weger
deweger@xs4all.nl

August 25, 1999

   
Reference
=========
Niklas Broberg, Some examples related to the abc-conjecture for algebraic 
number fields, Math. Comp., posted on May 17, 1999, PII: S 0025-5718(99)01153-9
(to appear in print).

Broberg's paper can be downloaded from Juliusz Brzezinski's homepage 
(www.math.chalmers.se/~jub/abc). On that homepage there's also a paper by 
Jerzy Browkin, with a slightly different notion of 'quality'. 

(Question to the experts: what is the 'natural' definition of quality
in the algebraic number field case? I will accept any answer that keeps 
the above example on the first place :-).)