Suppose that $a$, $b$, and $c$ are any three consecutive primes other than the triple $2$, $3$, and $5$. Do they satisfy the triangle inequalities: $a + b > c$; $b + c > a$; $c + a > b$? In other words, can we always form a triangle with sides being the $3$ successive prime numbers? Is this a well-known result? Where can I read about its proof or refutation? Thanks in advance.
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3Have you tried any small examples, say starting at $2$? – Erick Wong Jun 06 '13 at 17:08
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1Let me clarify my question. I know that the first 3 successive primes, 2, 3, and 5 do not satisfy the triangular inequalities. If we discard them, does the above condition hold true for the rest of the consecutive primes? – user67724 Jun 06 '13 at 17:08
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Hint: Use a stronger form of Bertrand's postulate, which states that $p_ {n+1} < 1.1 \times p_{n}$ for large enough $n$.
As such, $p_{n-1} + p_{n} > p_{n+1}$ satisfies the triangle inequality.
This means that we only need to check finitely many small cases, which is easy to do.
N. S.
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Calvin Lin
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It suffices to use $p_{n+1} < 1.6 p_n$, which is true for $n\ge 3$ and covers all cases. – Erick Wong Jun 06 '13 at 17:17
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@Calvin. Thanks, but I don't understand your hint. Could you please elaborate? Are you saying that the conjecture is true? Is this a well-known result? – user67724 Jun 06 '13 at 17:20
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@user67724 It follows immediately from the Prime Number Theorem, but it is probably easy to prove directly by the Erdos method too. I don't know if it easy to estimate what "for large $n$" means, which is important because one need to manually check those. – N. S. Jun 06 '13 at 17:22
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Let $p_n,p_{n+1}, p_{n+2}$ be three consecutive primes.
You need to show that
$$p_{n+2}< p_{n}+p_{n+1}$$
This follows immediately from the following stronger version of the Betrand Postulate: For $n \geq 7$ there are two primes between $n$ and $2n$.
The case $2,3,5$ is obviously a counterexample, an easy check up to 7 shows there is no other.
P.S. Does anyone have a good reference to this stronger version of the BP? It follows from last statement of this Paper, but I remember seeing once that statement written explicitely..
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I just love answering old questions :) ... basically:
- Bertrand's Postulate $p_{n+1} < 2 \cdot p_n$
- From (Limit inferior of the quotient of two consecutive primes) $$\lim_{n \to \infty } \frac{p_{n+1}}{p_{n}}=1 \Rightarrow \lim_{n \to \infty } \frac{p_{n+1}}{p_{n-1}}=\lim_{n \to \infty } \frac{p_{n+1}}{p_{n}} \cdot \frac{p_{n}}{p_{n-1}}=1$$ or $$p_{n+1} < 2 \cdot p_{n-1}$$ from some $n$.
As a result:
$$p_{n-1}+p_{n}\geq 2 \cdot \sqrt{p_{n-1} \cdot p_{n}}= \sqrt{2 \cdot p_{n-1} \cdot 2 \cdot p_{n}} > p_{n+1}$$
