Odd Integers whose Smaller Odd Coprimes are Prime

Theorem
$105$ is the largest integer such that all smaller odd integers greater than $1$ which are coprime to it are prime.

The full list of odd integers is:
 * $1, 3, 5, 7, 9, 15, 21, 45, 105$

Proof
First it is demonstrated that $105$ itself satisfies this property.

Let $d \in \Z_{> 1}$ be odd and coprime to $105$.

Then $d$ does not have $3$, $5$ or $7$ as a prime factor.

Thus $d$ must have at least one odd prime as a divisor which is $11$ or greater.

The smallest such composite number is $11^2$.

But $11^2 = 121 > 105$.

Thus $d$ must be prime.

Thus it has been demonstrated that all odd integers greater than $1$ and smaller than $105$ which are coprime to $105$ are prime.

Using an argument similar to the above, we see that for an integer to have this property,

if it is greater than $p^2$ for some odd prime $p$, then it must be divisible by $p$.

If not, it will be coprime to $p^2$, a composite number.

Let $p_n$ denote the $n$th prime.

Suppose $N$ has this property.

By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_2 p_3 \cdots p_n \divides N$.

By Absolute Value of Integer is not less than Divisors, we have $p_2 p_3 \cdots p_n \le N$.

Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$.

Thus we have $2 p_n > p_{n + 1}$.

Hence for $n \ge 5$:

This is a contradiction.

Hence we must have $N \le p_5^2 = 121$.

From the argument above we also have:


 * $3 \divides N$ for $9 < N \le 25$
 * $3, 5 \divides N$ for $25 < N \le 49$
 * $3, 5, 7 \divides N$ for $49 < N \le 121$

So we end up with the list $N = 1, 3, 5, 7, 9, 15, 21, 45, 105$.