Notice that between $1,2,...,p$ we have at most $\frac{p+1}{2}$ primes. For all numbers $m=p_1^{e_1}...p_k^{e_k}$, define $O(m)=p_i$ if $p_i^{e_i}=\max(p_1^{e_1},...,p_k^{e_k})$. Since $\frac{p+3}{2} > \frac{p+1}{2}$, we have that $O(x+i)=O(x+j)=q$ for some $i < j$. Then if $q^a||x+i$ and $q^b||x+j$, we have that the minimum divides $j-i$. $$\frac{p+1}{2} \ge j-i> \min(q^a,q^b)>(x+i)^{\frac{1}{\frac{p+1}{2}}}$$

Gomes17 wrote: Notice that between $1,2,...,p$ we have at most $\frac{p+1}{2}$ primes. For all numbers $m=p_1^{e_1}...p_k^{e_k}$, define $O(m)=p_i$ if $p_i^{e_i}=\max(p_1^{e_1},...,p_k^{e_k})$. Since $\frac{p+3}{2} > \frac{p+1}{2}$, we have that $O(x+i)=O(x+j)=q$ for some $i < j$. Then if $q^a||x+i$ and $q^b||x+j$, we have that the minimum divides $j-i$. $$\frac{p+1}{2} \ge j-i> \min(q^a,q^b)>(x+i)^{\frac{1}{\frac{p+1}{2}}}$$ how can we get the last inequality