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

## 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

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:

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

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.