Is there a prime between every pair of consecutive triangular numbers? Between two triangular numbers there is at least one prime number. Is there a mathematical proof for this statement?
 A: 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 http://en.wikipedia.org/wiki/Prime_gap#Upper_bounds
A: 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.
A: 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?
