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Between two triangular numbers there is at least one prime number. Is there a mathematical proof for this statement?

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I don't think there is a proof yet. – Shahab Mar 2 '14 at 3:39
No............................ – Will Jagy Mar 2 '14 at 4:03
Can you prove it, Will Jagy? – Ruben Gil Mar 2 '14 at 4:06
Given the fact that all triangular numbers are of the form $\dfrac{n^2+n}2$ , this seems related to Legendre's conjecture, for which no proof is currently known. – Lucian Mar 2 '14 at 4:46
Ruben: Notice that you asked "Is there a mathematical proof...?" so Will's answer means "No, there is not (yet) a proof." That is a statement about the mathematical literature, not about mathematics. – Pete L. Clark Mar 2 '14 at 4:52

Standard answer: there are proofs of statements of this type: for large enough real numbers $x,$ there is a prime between $x$ and $x + x^{21/40};$ the important part is that $21/40$ is bigger than $1/2.$ Here, as in all such results, nobody knows how big "large enough" needs to be. Such results are collectively called "ineffective," as they cannot be used to prove anything about small and medium numbers. See

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I think for this particular problem, we don't have a proof even for "large enough" numbers. – ShreevatsaR Mar 2 '14 at 6:17
@ShreevatsaR, we do with exponent 21/40, not with exponent 1/2. – Will Jagy Mar 2 '14 at 6:19
Right, that was my point. – ShreevatsaR Mar 2 '14 at 6:20

Note that if $m \geq 2$ and $T_{n-1} < m^2 $ then

$$(n-1)n < 2 m^2 $$

From here is easy to deduce that $n < m$.

Also note that $T_{n+1}-T_n=n+1$.

Therefore, for each positive integer $m \geq 2$ if $n$ is the largest integer so that $T_{n-1} <m^2$, then we have by the above we have

$$T_{n+1} = T_n+n+1 < T_{n-1}+2n+1< m^2+2m+1$$

This shows that

$$m^2 \leq T_n <T_{n+1}< (m+1)^2$$

This shows that the statement you ask for, if true, would imply the Legendre conjecture.

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With gaps between consecutive triangular numbers, for $x \to \infty$ it's equivalent to gaps between consecutive squares, which means a prime between $x$ and $x + 2x^{1/2}$. When $x$ gets too large, the number two is smaller than $x^{1/40}$. So then the proof for $x + x^{21/40}$ would not apply to gaps between consecutive triangular numbers. This is why it's "ineffective", right?

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