Consider this Diophantine equation:
3
a + 4
b + 5
c = 6
d, where each of a, b, c, and d is a positive integer.
The famous ancient Greek philosopher and mathematician Plato
was aware that a=b=c=d=3 is a solution.
Does there exist any further solution?
Authenticate your answer with valid arguments.
The comfortable limit for type "double" is d=17, found via logs below. Extra answer is provided by allowing zero exponent, though disallowed in the puzzle statement.
The only set other than (3, 3, 3, 3) that included only positive integers is (3, 1, 1, 2) for 3^3 + 4^1 + 5^1 = 6^2.
Allowing zero in the set yields (0, 1, 0, 1), or 3^0 + 4^1 + 5^0 = 6^1.
3^3 + 4^1 + 5^1 = 6^2
clc, clearvars
for d=0:floor(log(10^14)/log(6))
remain1=6^d;
for c=0:floor(log(remain1)/log(5))
remain2=remain1-5^c;
for b=0:floor(log(remain2)/log(4))
remain3=remain2-4^b;
for a=0:floor(log(remain3)/log(3))
remain4=remain3-3^a;
if remain4==0
disp([a b c d])
end
end
end
end
end
having yielded
a b c d
0 1 0 1
3 1 1 2
3 3 3 3
|
|
Posted by Charlie
on 2022-06-09 10:08:23 |
testing only to the limit of double precision floating point
