Find all triples such that , where is a prime and are natural numbers.

**Hint 1:** p cannot be 2. For if p is 2 then is even which implies is even or y is even. Suppose y = 2y’. Then . Now x cannot be 1 or 2 as must be greater than 4. Hence . Let x = x’ + 3. Thus

which is impossible.

**Hint 2:** Suppose p is odd. Then . Hence there can be two cases:

**Case 1:** (as ). Solving the second equation for y gives us solution y = 1 , x= 1 and p =5.

**Case 2:** where x = m +n and m > n . Subtracting the second equation from the first we get . This implies p divides y as p does not divide 4 (as p is not equal to two). But considering the equation since p divides left side, p must divide right side and hence p must divided . If p divides y then p will also divide 4 which cannot be.

Hence there are no other solution. Only solution is (5, 1, 1)