For hundreds of years Fermat's Last Theorem, which stated
simply that for n > 2 there exist no integers a, b, c> 1 such that a=
b
+ c
,
has remained elusively unproven. (A recent proof is believed to be correct,
though it is still undergoing scrutiny.)
It is possible, however, to find integers greater than
1 that satisfy the "perfect cube" equation a=
b
+ c
+
d
(e.g. a quick calculation will
show that the equation 12
= 6
+ 8
+ 10
is indeed true).
This problem requires that you write a program to find
all sets of numbers {a,b,c,d} which satisfy this equation for a 100.
The output should be listed as shown below, one perfect
cube per line, in non-decreasing order of a (i.e. the lines should be sorted
by their a values).
The values of b, c, and d should also be listed in non-decreasing
order on the line itself.
There do exist several values of a which can be produced
from multiple distinct sets of b, c, and d triples.
In these cases, the triples with the smaller b values
should be listed first.
Note that the programmer will need to be concerned with
an efficient implementation.
The official time limit for this problem is 2 minutes,
and it is indeed possible to write a solution to this problem which executes
in under 2 minutes on a 33 MHz 80386 machine.
Due to the distributed nature of the contest in this
region, judges have been instructed to make the official time limit at
their site the greater of 2 minutes or twice the time taken by the judge's
solution on the machine being used to judge this problem.
The first part of the output is shown here:
Cube = 6, Triple = (3,4,5)
Cube = 12, Triple = (6,8,10)
Cube = 18, Triple = (2,12,16)
Cube = 18, Triple = (9,12,15)
Cube = 19, Triple = (3,10,18)
Cube = 20, Triple = (7,14,17)
Cube = 24, Triple = (12,16,20)