Noam Elkies is one of

those persons I seem to bump into (figuratively speaking) wherever my

interests take me. At the moment I’m reading (long overdue, I

know, I know) the excellent book Notes on

Fermat’s Last Theorem by Alf Van der

Poorten. On page 48, Elkies figures as an innocent bystander in the

1994 April fools joke e-perpetrated by

Henri Darmon in the midst of all confusion about ‘the

gap’ in Wiles’ proof.

There has

been a really amazing development today on Fermat’s Last Theorem.

Noam Elkies has announced a counterexample, so that FLT is not true

after all! He spoke about this at the institute today. The solution to

Fermat that he constructs involves an incredibly large prime exponent

(larger than $10^{20}$), but it is constructive. The main idea seems to

be a kind of Heegner-point construction, combined with a really

ingenious descent for passing from the modular curves to the Fermat

curve. The really difficult part of the argument seems to be to show

that the field of definition of the solution (which, a priori, is some

ring class field of an imaginary quadratic field) actually descends to

$\\mathbb{Q}$. I wasn’t able to get all the details, which were

quite intricate…

Elkies is also an

excellent composer of chess problems. The next two problems he composed

as New Year’s greetings. The problem is : “How many shortest

sequences exists (with only white playing) to reach the given

position?”

$\\begin{position}

\\White(Kb5,Qd1,Rb1,Rh1,Nc3,Ne5,Bc1,Bf1,a2,b2,c4,d2,e2,f3,g3,h2)

\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard

}xc $

Here’s Elkies’ solution

:

There are 2004 sequences of the minimal length 12.

Each consists of the sin- gle move g3, the 3-move sequence

c4,Nc3,Rb1, and one of the three 8-move sequences

Nf3,Ne5,f3,Kf2,Ke3,Kd3(d4),Kc4(c5),Kb5. The move g3 may be played at

any point, and so contributes a factor of 12. If the King goes

through c5 then the 3- and 8-move sequences are independent, and can

be played in $\\binom{11}{3}$ orders. If the King goes through c4 then

the entire 8-move sequence must be played before the 3-move sequence

begins, so there are only two possibilities, depending on the choice

of Kd3 or Kd4. Hence the total count is $12(\\binom{11}{3}+2)=2004$ as

claimed.

A year later he composed the

problem

$\\begin{position}

\\White(Kh3,Qe4,Rc2,Rh1,Na4,Ng1,Bc1,Bf1,a2,b2,c3,d3,e2,f4,g2,h2)

\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard

}xd $

of which Elkies’ solution is

:

There are 2005 sequences of the minimal length 14.

This uses the happy coincidence $\\binom{14}{4}=1001$. Here White

plays the 4-move sequence f4,Kf2,Kg3,Kh3 and one of the five

sequences Nc3,Na4,c3,Qc2,Qe4,d3,Bd2(e3,f4,g5,h6),Rc1,Rc2,Bc1 of

length 10. If the Bishop goes to d2 or e3, the sequences are

independent, and can be played in $\\binom{14}{4}$ orders. Otherwise

the Bishop must return to c1 before White plays f4, so the entire

10-move sequence must be played before the 4-move sequence begins. Hence

the total count is $2 \\binom{14}{4}+3 =

2005$.

With just a few weeks remaining, anyone in for

a 2006 puzzle?

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