Explanation

We need natural q>p with q^2<5p+4. The smallest admissible q is p+1, so (p+1)^2<5p+4, i.e. p^2-3p-3<0, giving p<(3+sqrt(21))/2 which is about 3.79. Thus p<=3, and p=3 works (q=4: 16<19). The largest such p is 3. Note p is bounded because q must exceed p while q^2 stays below 5p+4.