XXXVIII OM - II - Problem 5

Determine all prime numbers $ p $ and natural numbers $ x, y $, for which $ p^x - y^3 = 1 $.
Suppose that the numbers $ p $, $ x $, $ y $ satisfy the given conditions. Then






Let's consider two cases:
1. $ y = 1 $. Then (1) reduces to the equation $ p^x = 2 $, from which $ p = 2 $, $ x = 1 $.
2. $ y > 1 $. Then both factors of the product on the right side of (1) are numbers greater than $ 1 $ and as divisors of the power of the prime number $ p $, they must themselves be positive powers of this prime number:






Now notice that






Thus, $ 3 $ is divisible by $ p $, which means that $ p = 3 $. Equation (3) takes the form






which is






The right side of the last equation is not divisible by $ 3 $. Hence $ l - 1 = 0 $ and $ 2k-1=k $, so $ k = l = 1 $ and by the relation (2) $ x = 2 $, $ y = 2 $.
In summary: the solutions $ (p, x, y) $ of the given equation can only be the triples $ (2, 1, 1) $ and $ (3, 2, 2) $. Verifying that these are indeed solutions is immediate.