We'll say that positive integers $x$ and $y$ are connected and write $x$~$y$ if there exists a positive integer $t$ and a sequence $w_1, w_2, ..., w_t$ consisting of positive integers, such that $w_1 = x,w_t = y$, and $(w_i + w_{i+1}) | w_iw_{i+1}$ for all $i = 1,2,..., t - 1$. Obviously the relation ~ is symmetric and transitive. It has to be proven that every positive integers $x$ and $y$ greater than 2 are connected. Lemma 1: if $q|p$, then $p$~$p(q-1)$. Proof: Let's show that $(p+p(q-1))|p^2(q-1)$. This is the same as $pq|p^2(q-1)$, which is true because $pq|p^2$ and $p^2|p^2(q-1)$. Lemma 2: $p$~$p!$ for every positive integer $p$. Proof: It is enough to prove that $p$~$p(p-1)...(p-r)$ for every non-negative integer $r$ which is not greater than $p-1$. This can be proven via induction: for $r = 0$ we get that $p$~$p$, which is true. Let's assume $p$~$p(p-1)...p(p-r)$ for some non-negative integer $r$ which is not greater than $p-2$. since $(p-r)|p(p-1)...(p-r)$, by Lemma 1 $p$~$p(p-1)...(p-r)$~$p(p-1)...(p-r-1)$. The proof via induction is done. Now for $r=p-1$ we get that $p$~$p!$. Lemma 3: If p is a non-prime integer greater than 4, then $(p-1)!$~$(p-2)!$ Proof: It is enough to prove that $((p-1)!+(p-2)!)|(p-1)!(p-2)!$, which is the same as $(p-2)!p|(p-1)!(p-2)!$ or $p|(p-1)!$ and this is true since p is not a prime. Lemma 3': If $p$ is a prime number greater than 4, then $(p-1)!$~$(p-2)!$ Proof: By Lemma 2 $(p-1)!$~$(p-1)$~$(p-1)(p-2)$ and $(p-2)!$~$(p-2)$~$(p-2)(p-3)$. Let's prove that $((p-1)(p-2)+(p-2)(p-3))|(p-1)(p-2)^2(p-3)$, which is the same as $2(p-2)^2|(p-1)(p-2)^2(p-3)$ or $2|(p-1)(p-3)$. Since p is a prime number greater than 2, p is odd and p-1 is even, so 2|(p-1)(p-3). Now by using Lemma 3, Lemma 3' and induction we can get that $x!$ ~ $y!$ for every positive integers $x$ and $y$ greater than 2, which is the same as $x$~$y$ since $x$~$x!$ and $y$~$y!$ (Lemma 2 and transitivity) Q.E.D.