Since p is prime we can clearly see that p=3 pr p ≡ 1(mod3) In a similar way we can see that p=2 or p ≡ 1,3(mod8) Thus we’ll get that p=3 or p ≡1,19(mod24) since p<100 there for p=3,43,67,73,97 Which all works with (x,y)=(3,1),(7,1),(11,3),(17,6),(11,2) as solution respectively I still havent find a effective method to find p in a wider range tho