If this holds for all $n\gt3$ then we have another way of expressing Goldbach's conjecture as a product of two primes because the sum of the factors of the difference of two squares equals $2n$:
$n^2-m^2=(n+m)(n-m)$
$4^2 - 1^1 = 5\times3 = 15$
$2(4) = 8 = 5 + 3$
$11^2 - 6^2 = 17\times5 = 85$
$2(11) = 22 = 17 + 5$
$15^2 - 4^2 = 19\times11 = 209$
$2(15) = 30 = 19 + 11$
In this form Goldbach's conjecture can then be restated: For every $n\gt3$ there exists some $m$ where $0\lt m\lt n$ such that $n^2-m^2$ is a semiprime.