Mathematical Anecdotes
"Bombieri's Napkin Problem" :
A Botched Challenge (1979)
During the 1979 Queen's Number Theory Conference
in Kingston (Ontario),
Enrico Bombieri (Fields Medalist in 1974)
offered in jest a challenge analoguous to
Fermat's Last Theorem
to some colleagues he was having diner with, including
Roger Apéry and Michel MendèsFrance (who
reported
the anecdote).
Prove that there are no nontrivial solutions,
in positive integers, to the following equation (involving
choice numbers):
C(x,n) + C(y,n) = C(z,n)
The next morning, Apéry offered a solution:
n = 3, x = 10, y = 16, z = 17.
Bombieri just replied, with a straight face: "I said nontrivial."
The puzzle is much older than this anecdote:
It is casually mentioned in the popular book
Tomorrow's Math: Unsolved Problems for the Amateur (1972 edition, at least)
by Dr.
Charles Stanley Ogilvy
(19132000)
who echoes a misleading presentation that could easily have fooled the likes of Bombieri
or MendèsFrance in an era when computer access wasn't easy...
20060130: Some
Solutions to C(x,n) + C(y,n) = C(z,n)
n  x  y  z 

n 
2n1  2n1  2n 

1 
x  y  x+y 

2 
3m+3  4m+3  5m+4 

3m+4  4m+6  5m+7 
5m+5  12m+10  13m+11 
5m+6  12m+15  13m+16 
15m+10  8m+5  17m+11 
15m+21  8m+12  17m+24 
x  ½ x(x1)  ½ x(x1) + 1 
3 
10  16  17 

22  56  57 
32  57  60 
41  72  76 
63  104  111 
86  92  112 
36  120  121 
4 
132  190  200 

6 
14  15  16 

19  19  21 
35 
118  118  120 

40 
103  104  105 

With y=x+i and z=x+j (i<j) we may factor out C(x,nj) to obtain an equation
of degree j in x and n.
The second line of the above table (n=1) corresponds to j=1.
For j=2, we obtain two quadratic
diophantine equations (i=0 or i=1)
respectively yielding the following two infinite families of explicit solutions:
1)
When 8n^{2}+1
is a perfect square (q^{2 })
a solution is: x = ½ (4n3+q),
y = x, z = x+2
Such values of n (and those of q) obey the recurrence:
a_{i+2} = 6 a_{i+1}  a_{i}
n 
1  6  35  204  1189  6930 
40391  235416  7997214 

x 
2  19  118  695  4058  23659 
137902  803759  27304195 

y 
2  19  118  695  4058  23659 
137902  803759  27304195 

z 
4  21  120  697  4060  23661 
137904  803761  27304197 

2)
If 5n^{2}2n+1
is a square (q^{2 })_{ } then a solution is:
x = ½ (3n3+q),
y = x+1, z = x+2
Such values of n obey the recurrence:
a_{i+2} = 7 a_{i+1}  a_{i}  1
(Every fourth Fibonacci number is a value of q : 2, 13, 89, 610, 4181, 28657...)
n 
1  6  40  273  1870  12816 
87841  602070  4126648 

x 
1  14  103  713  4894  33551 
229969  1576238  10803703 

y 
2  15  104  714  4895  33552 
229970  1576239  10803704 

z 
3  16  105  715  4896  33553 
229971  1576240  10803705 

Bombieri's Napkin Problem is now discussed on the "Math Overflow" forum :)
For undisclosed reasons,
Wadim Zudilin states for the record that he "personally doesn't like" the above presentation :(
