5)
Let d(n)d(n) be the function which counts the number of digits of nn then
5.1) n ge 10^(d(n)) ge g(n)n≥10d(n)≥g(n)
5.2) (n-6)^2=g(n)->n = sqrt(g(n))+6(n−6)2=g(n)→n=√g(n)+6 then
g(n_k) = k^2, k = 1,2,3,cdotsg(nk)=k2,k=1,2,3,⋯ so the equation to obey is
n_k = k+6nk=k+6 so the only number observing this relationship is
n_1=7n1=7
6)
Let 4+p_1p_2 = m_1^24+p1p2=m21 and 4+p_1p_3=m_2^24+p1p3=m22
then
{(p_1p_2=(m_1-2)(m_1+2)),(p_1p_3=(m_2-2)(m_2+2)):}
but p_1,p_2,p_3 being prime numbers then
{(p_1=m_1-2),(p_2=m_1+2),(p_1=m_2+2 " otherwise " m_1=m_2),(p_3=m_2-2):}
so
m_1 = m_2+4 then
{(p_1=p_3+4),(p_2=p_1+4):}
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
