May 16, 2017

See below.

#### Explanation:

5)
Let $d \left(n\right)$ be the function which counts the number of digits of $n$ then

5.1) $n \ge {10}^{d \left(n\right)} \ge g \left(n\right)$
5.2) ${\left(n - 6\right)}^{2} = g \left(n\right) \to n = \sqrt{g \left(n\right)} + 6$ then
$g \left({n}_{k}\right) = {k}^{2} , k = 1 , 2 , 3 , \cdots$ so the equation to obey is

${n}_{k} = k + 6$ so the only number observing this relationship is

${n}_{1} = 7$

6)
Let $4 + {p}_{1} {p}_{2} = {m}_{1}^{2}$ and $4 + {p}_{1} {p}_{3} = {m}_{2}^{2}$
then

$\left\{\begin{matrix}{p}_{1} {p}_{2} = \left({m}_{1} - 2\right) \left({m}_{1} + 2\right) \\ {p}_{1} {p}_{3} = \left({m}_{2} - 2\right) \left({m}_{2} + 2\right)\end{matrix}\right.$

but ${p}_{1} , {p}_{2} , {p}_{3}$ being prime numbers then

$\left\{\begin{matrix}{p}_{1} = {m}_{1} - 2 \\ {p}_{2} = {m}_{1} + 2 \\ {p}_{1} = {m}_{2} + 2 \text{ otherwise } {m}_{1} = {m}_{2} \\ {p}_{3} = {m}_{2} - 2\end{matrix}\right.$

so

${m}_{1} = {m}_{2} + 4$ then

$\left\{\begin{matrix}{p}_{1} = {p}_{3} + 4 \\ {p}_{2} = {p}_{1} + 4\end{matrix}\right.$

and

${p}_{2} > {p}_{1} > {p}_{3}$

so the prime numbers are

${p}_{1} - 4 , {p}_{1} , {p}_{1} + 4$

or

${p}_{3} = 3 , {p}_{1} = 7 , {p}_{2} = 11$