Interval containing Prime Number of forms 4n - 1, 4n + 1, 6n - 1, 6n + 1

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Theorem

Let $n \in \Z$ be an integer such that $n \ge 118$.


Then between $n$ and $\dfrac {4 n} 3$ there exists at least one prime number of each of the forms:

$4 m - 1, 4 m + 1, 6 m - 1, 6 m + 1$


Proof


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It is demonstrated that the result is true for $n = 118$:

$\dfrac {4 \times 118} 3 = 157 \cdotp \dot 3$

The primes between $118$ and $157$ are:

\(\ds 127\) \(=\) \(\ds 4 \times 32 - 1\)
\(\ds \) \(=\) \(\ds 6 \times 21 + 1\)
\(\ds 131\) \(=\) \(\ds 4 \times 33 - 1\)
\(\ds \) \(=\) \(\ds 6 \times 22 - 1\)
\(\ds 137\) \(=\) \(\ds 4 \times 34 + 1\)
\(\ds \) \(=\) \(\ds 6 \times 23 - 1\)
\(\ds 139\) \(=\) \(\ds 4 \times 35 - 1\)
\(\ds \) \(=\) \(\ds 6 \times 23 + 1\)
\(\ds 149\) \(=\) \(\ds 4 \times 37 + 1\)
\(\ds \) \(=\) \(\ds 6 \times 25 - 1\)
\(\ds 151\) \(=\) \(\ds 4 \times 38 - 1\)
\(\ds \) \(=\) \(\ds 6 \times 25 + 1\)
\(\ds 157\) \(=\) \(\ds 4 \times 39 + 1\)
\(\ds \) \(=\) \(\ds 6 \times 26 + 1\)


So it can be seen that:

the primes of the form $4 m - 1$ are:
$127, 131, 139, 151$
the primes of the form $4 m + 1$ are:
$137, 139, 157$
the primes of the form $6 m - 1$ are:
$131, 137, 149$
the primes of the form $6 m + 1$ are:
$127, 139, 151, 157$

and so the conditions of the result are fulfilled for $118$.


But then note we have:

$\dfrac {4 \times 117} 3 = 156$

and so the primes between $117$ and $157$ are the same ones as between $118$ and $157$, excluding $157$ itself.


Thus the conditions also hold for $117$.

It is puzzling as to why the condition $n \ge 118$ has been applied to the initial statement.


Sources

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