Conjecture: There are infinitely many $N \in \Bbb{N}$ such that $p$ a prime $p \leq \sqrt{N+1} \implies p \mid N$?

Question

Conjecture. There are infinitely many $N$ such that if $p$ is a prime $\leq \sqrt{N+1}$ then $p \mid N$.

Is this another hard to prove number theory conjecture, or do you have some idea of how to solve it?

Answer 1

If I'm interpreting your conjecture correctly, it is false. Let's say a number $N$ is an enjoyable number if, for all primes $p\le \sqrt{N+1}$, $p$ divides $N$. So for example, $30$ is an enjoyable number because $\sqrt{31} \approx 5.6$, and all primes $\le 5.6$ divide $30$.

Why can there not be infinitely many enjoyable numbers? Suppose $N$ is an enjoyable number. Then $N$ is divisible by every prime smaller than or equal to $\sqrt{N+1}$. In particular, $N$ is divisible by the product of all such primes. The prime counting function has the lower bound $$ \pi(x)> \frac{x}{\log x} $$ The product of all primes less than or equal to $\sqrt{N+1}$ is bounded below by $$2^{\pi(\sqrt{N+1})}>2^{\frac{\sqrt{N+1}}{\log(\sqrt{N+1})}} = 4^{\frac{\sqrt{N+1}}{\log(N+1)}}$$ which is clearly asymptotically bigger than $N$, and computationally is bigger than $N$ for all $N\ge 1473$. Hence all enjoyable numbers must be smaller than $1473$.

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Update: I did a computer of all integers up to $1473$. The complete set of enjoyable numbers is $\{1,2,4,6,12,18,30\}$. My Haskell code below:

-- Integer square root 
isqrt :: (Integral a, Enum a, Ord a) => a -> a 
isqrt n = pred $ head $ filter (\k -> k^2 > n) [1..]

-- checks if the first argument is divisible by the second 
divis :: Integral a => a -> a -> Bool
divis n = (0 == ) . (rem n)

-- checks if the first argument is not divisible by the second 
sivid :: Integral a => a -> a -> Bool
sivid n = (0 /= ) . (rem n)

-- list of all primes 
primes :: (Integral a, Enum a) => [a]
primes = 2:(filter (\ k -> and $ map (sivid k) $ takeWhile (not . ( > k) . (^2)) primes) [3..])

-- multiplies all the primes up to n 
pp :: (Integral a, Enum a, Ord a) => a -> a
pp n = product $ takeWhile ( not . (> n) ) primes

-- checks if a number is enjoyable 
is_enjoyable :: (Integral a, Enum a, Ord a) => a -> Bool
is_enjoyable n = divis n $ pp $ isqrt $ succ n

-- set of all enjoyable numbers 
enjoyables :: [Integer]
enjoyables = filter is_enjoyable [1..1473]

Answer 2

If $N$ is divisible by every prime $p\le\sqrt{N+1}$, then $N$ is divisible by the product of all those primes. But the product of all the primes up to $x$ is known to be asymptotic to $e^x$, and $e^{\sqrt{N+1}}$ grows much much faster than $N$. So there are only finitely many such $N$.