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)