How can I prove that these three sets have no common values:
- A: {prime numbers}
- B: {Fibonacci numbers}
- C: {8|n+1}
- C: for example 15: 15+1 = 16 => 8|16 <= 16/8 = 2
- C: for example 23: 23+1 = 24 => 8|24 <= 24/8 = 3
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What you mean is that there isn't a single value which is in all three sets? – Mark Bennet Oct 22 '17 at 19:34
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I need to find out if there is or isn't a value which all three contain. It would be ideal if there wasn't any number that all three contained because it would solve my problem. – Kr3Ep Oct 22 '17 at 20:12
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The claim is that Fibonacci primes $F_n$ (which implies that $n$ is prime, except for $n=4$) do not satisfy $F_p\equiv 7\bmod 8$. Let $a(n)$ denote the sequence of Fibonacci primes. Then they satisfy $a(n)\equiv 1,5\bmod 8$, see OEIS, because of $a(n) \equiv 1 \bmod 4$ (Cadwell) for $n>2$.
Dietrich Burde
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So does this means that there supposed to be a number that all three sets contain? – Kr3Ep Oct 22 '17 at 19:56
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1If there were such a number, we would have a Fibonacci prime of the form $8k+7$, a contradiction. So, no. – Dietrich Burde Oct 23 '17 at 07:38
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Consider the Fibonacci numbers mod $8$:
$$1,1,2,3,5,0,5,5,2,7,1,0,\ldots$$
We see that $F_m\equiv7$ mod $8$ if and only if $m\equiv10$ mod $12$, which implies $m$ is even (and greater than $4$). But $F_n\mid F_{2n}$ for all $n$, so $F_m$ cannot be a prime if $8\mid F_m+1$.
Remark: The theorem $F_n\mid F_{2n}$ for all $n$ (and, more generally, $F_n\mid F_{kn}$ for all $k$ and $n$) boils down to the banal observation that $F_n\equiv0$ mod $F_n$. This means that the sequence of Fibonacci numbers mod $F_n$ begins
$$1,1,2,3,\ldots,F_{n-1},0,F_{n-1},F_{n-1},2F_{n-1},3F_{n-1},\ldots,F_{n-1}^2,0,\ldots$$
That is, once you hit the first $0$, the sequence repeats itself, multiplied by (powers of) $F_{n-1}$.
Barry Cipra
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