5)
Let #d(n)# be the function which counts the number of digits of #n# then
5.1) #n ge 10^(d(n)) ge g(n)#
5.2) #(n-6)^2=g(n)->n = sqrt(g(n))+6# then
#g(n_k) = 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_1p_2 = m_1^2# and #4+p_1p_3=m_2^2#
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#
