There can't be
another Fibonacci number like 21.
I'll prove the
following slightly more general statement:
The Fibonacci numbers 5, 8, and 21 are the only ones which can
be written as the sum of exactly three positive, non-composite
Fibonacci numbers with distinct indexes.
The claimed sums
5 = 3+1+1;
8 = 5+2+1;
21 = 13+5+3.
As the number 1 is no more
considered a prime, only last sum involves three primes.
Note that 1 can be used
twice without contradicting my statement because it appears
twice in the Fibonacci sequence, at indexes 1 and 2.
For the proof, I will use
two known properties of Fibonacci numbers.
Recurrence relation: F(n)
= F(n-1) + F(n-2).
Strong divisibility: gcd(F(m),F(n))
A Fibonacci number F(n) with n >= 5 can be written in only one
way as the sum of exactly three positive Fibonacci
numbers with distinct indexes:
F(n) = F(n-1) + F(n-3) + F(n-4).
Proof, using the recurrence
F(n) = F(n-1) + F(n-2) < F(n-1) +
F(n-2) + F(1), so the sum can't use both F(n-1) and F(n-2).
F(n) = F(n-1) + (F(n-3)+F(n-4)) >
F(n-2) + (F(n-3)+F(n-4)), so the sum must use F(n-1), and it
can't use F(n-2).
F(n) = F(n-1) + F(n-3) + F(n-4) > F(n-1) + F(n-4) + F(n-5), so
the sum must use F(n-3) too.
Then last term must be F(n) - F(n-1) - F(n-3) = F(n-4).
The sequence R(n) = F(n) mod 4 is periodic, with period 6.
R(0) = F(0) = 0 and R(1) = F(1) =
1; by applying the recurrence relation:
Let us check the six possible
triplets F(n-1), F(n-3), F(n-4) mod 4.
The first column lists n mod 6.
| 0 || 1 | 2 | 1 |
| 1 || 0 | 3 | 2 |
| 2 || 1 | 1 | 3 |
| 3 || 1 | 0 | 1 |
| 4 || 2 | 1 | 0 |
| 5 || 3 | 1 | 1 |
If n is congruent
to 1, 3 or 4 mod 6, one term of the sum is zero or a composite
multiple of 4.
If n is congruent to 0 mod
6, then F(n-3) is an odd multiple of 2 and is composite, unless F(n-3)
= 2; equality holds only for n = 6.
If n is congruent to 2 mod 6, then
n = 6*k+8; by the strong divisibility property, the number
F(n-4) = F(6*k+4) is divisible by the smaller number F(3*k+2)
and is composite, unless F(3*k+2) = 1; equality holds only for k
= 0, so n = 8.
If n is congruent to 5 mod 6, then
n = 6*k+5; again, the number F(n-1) = F(6*k+4) is composite,
unless k = 0; so n = 5.
Therefore, the only candidates are
F(5) = 5, F(6) = 8, and F(8) = 21.
For each candidate, the only
possible sum is the one listed above, which actually involves no
composites, as claimed.
There are no other such sums !
Indeed, let F(n) denote the nth Fibonacci number.
Suppose we have 3 < a < b < c and F(a) + F(b) + F(c) = F(d).
Then we must have c = d - 1.
Indeed, c <= d - 2 would imply :
F(a) + F(b) + F(c) <= F(c-2) + F(c-1) + F(c)
F(d) <= F(d-4) + F(d-3) + F(d-2)
<= F(d-4) + F(d-1) < F(d-2) +
F(d-1) = F(d),
Thus, F(a) + F(b) + F(d-1) = F(d)
which implies that F(a) + F(b) must be equal to F(d-2),
which is only possible when a = d-4 and b = d-3.
The only cases in which two
consecutive Fibonacci number are both prime are :
2 and 3 (but this is not
giving a solution)
3 and 4 which give the
solution 3 + 5 + 13 = 21.