# At least one prime between N and N-(sqrtN)

I don't know if this is an already existing conjecture, or has been proven: There is at least one prime number between $N$ and $N-\sqrt{N})$.

Some examples: $N=100$

$\sqrt{N}=10$ Between and 90 and 100, there is a prime: 97

$N=36$

$\sqrt{N}=6$ Between 30 and 36, there is a prime: 31

$N=64$

$\sqrt{N}=8$ Between 56 and 64, there are the primes: 59 and 61

N=12

$\sqrt{N}=3.46..$ Between 8.54 and 12, there is a prime: 11

If this hasn't been brought up before, I'm calling this the Dwyer Conjecture.

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No prime lies in $[125-\sqrt{125},125]$, and ditto for $126$. –  Ian Coley Feb 8 at 23:31
Take $11$. $\sqrt{11} = 3.3166$. $11 - \sqrt{11} = 7.6834$. There is no prime from $7.6834$ to $11$ unless you include the upper limit as well. –  hjpotter92 Feb 9 at 14:30

the usual form of such things is to say that something is true for sufficiently large numbers. That is likely to be the case, but actual proofs, including the "sufficiently large" condition, have still not reached what you need. The best results are something like $x + x^{0.525};$ for large enough positive real number $x,$ there is guaranteed to be a prime between $x$ and $x + x^{0.525}.$
So, while everyone believes there is a prime between $x$ and $x + \sqrt x$ for sufficiently large $x,$ we may never be sure. Some very believable conjectures suggest that "sufficiently large" is $x \geq 5504.$ Does not make it true, just reasonable. I can tell you that the $x + \sqrt x$ thing is true for $5504 \leq x \leq 4 \cdot 10^{18}.$
I'm afraid this is false whether we consider the strong form in which endpoints are allowed, or not. $\sqrt{126}<12$ and $113$ is the next prime below $126$.