Schatunowsky's Theorem

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Theorem

Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $\map w n$ denote the number of primes strictly less than $n$ which are not divisors of $n$.

Let $\map \phi n$ denote the Euler $\phi$ function of $n$.


Then $30$ is the largest integer $n$ such that:

$\map w n = \map \phi n - 1$


Proof

The above equation is equivalent to the property that all numbers greater than $1$ that are coprime to it but less are prime.

For an integer to have this property:

If it is greater than $p^2$ for some 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 $p_1 p_2 \cdots p_n \divides N$.

By Absolute Value of Integer is not less than Divisors, we have $p_1 p_2 \cdots p_n \le N$.


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 $n \ge 5$:

\(\ds N\) \(\ge\) \(\ds p_1 p_2 \cdots p_n\)
\(\ds \) \(=\) \(\ds 2 \times 3 \times 5 p_4 \cdots p_n\)
\(\ds \) \(>\) \(\ds 8 p_{n - 1} p_n\)
\(\ds \) \(>\) \(\ds 4 p_n^2\) Bertrand-Chebyshev Theorem
\(\ds \) \(>\) \(\ds p_{n + 1}^2\) Bertrand-Chebyshev Theorem
\(\ds \) \(\ge\) \(\ds N\) From assumption

This is a contradiction.


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

From the argument above we also have:

$2 \divides N$ for $4 < N \le 9$
$2, 3 \divides N$ for $9 < N \le 25$
$2, 3, 5 \divides N$ for $25 < N \le 49$
$2, 3, 5, 7 \divides N$ for $49 < N \le 121$

So we end up with the list $N = 1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.

This list is verified in Integers such that all Coprime and Less are Prime.

$\blacksquare$


Also see


Source of Name

This entry was named for Samuil Osipovich Shatunovsky.


Sources

  • May 2015: Victor PambuccianSchatunowsky's theorem, Bonse's inequality, and Chebyshev's theorem in weak fragments of Peano arithmetic (Math. Log. Quart Vol. 61, no. 3: pp. 230 – 235)