Appendix on the abc- conjecture
I know many of you attended the full-house public lecture - sponsored by the Royal Irish Academy and the Irish Times - in Trinity College Dublin last October 14th, given by Andrew Wiles (who, as the saying goes, needs no introduction). In his talk, Wiles mentioned the abc-conjecture , and I thought there might be some interest because of that in this small final section.
A great new conjecture has been formulated in recent times (since the mid 1980s) by Oesterlé and Masser. There are many outstanding web references (I especially recommend Abderrahmane Nitaj's), and there is the highly motivating Nov 2002 AMS Notices article - It's as easy as abc - by Granville and Tucker.
Here, before stating the conjecture, my reader may be captivated by two well-known consequences of it:
Of course these consequences have already been independently established, but there are many, many remarkable consequences that are still open questions (Nitaj maintains a list of consequences).
The abc -conjecture . Let a and b be relatively prime natural numbers (i.e. they have no common prime divisor), let c be their sum: a + b = c , and let be any positive number; then there is some constant C (whose value depends only on ) such that
where are the distinct prime factors of a , b and c .
Granville and Tucker note it has been conjectured (by whom?) that the abc -conjecture holds good with = 1 and C = 1, which would give .
By way of illustrating the power of the latter form of the conjecture I will show that the only solutions of in natural numbers x , y and z would be (1, 1, 1) and (4, 2, 2).
First, though, my reader may wish to see some factored values of ( ), with x and y varying over a small range of values. Note the near miss at x = 8, y = 6, where ones sees that = *53.
Other similar near misses are *223, *8861, *1075159, *(an 18-digit prime), *(a 19-digit prime).
for x from 6 to 8 do
Applying the conjectured = 1, C = 1 form of the the abc -conjecture to hypothetical solutions of would give ('30' being 2*3*5), and so z could only be 1, 2, or 3. Then ( x , y , z ) = (1, 1, 1) and (4, 2, 2) would be the only solutions. (You probably want to ask me...)
Finally, my reader may be interested in a theorem of mine which looks as though it ought to be a consequence of the abc -conjecture, but isn't (I leave it as an exercise to see that my theorem isn't a consequence of abc ).
Theorem (motivated by reflecting on Euclid's proof of the infinitude of primes, MAA Monthly, Vol. 96, No. 4, April 1989). Let be the first n prime numbers, let be any one of those primes, and let be the product of those n primes, with omitted; then
is impossible for . (Note, for n = 3 and i = 1, that = 3*5+ 1 = 16 = .)
Also, is impossible for . (Note, for n = 3 and i = 2, that = 2*5 - 1 = 9 = .)
A numerical illustration for the case n = 15:
n := 15: N[n] := product(ithprime(m), m=1..n):
After August 31st 2007 please use the following Gmail address: jbcosgrave at gmail.com
This page was last updated 18 February 2005 15:10:01 -0000