Solutions to Some Advanced Methods in Solving Diophantine Equations

Abstract

1.Solve the equation

$$x^2 + 4 = y^n,$$

where n is an integer greater than 1. Solution. For n = 2, the only solutions are (0, 2) and (0, –2). For n = 3, we have seen in Example 4 that the solutions are (2, 2), (–2, 2), (11, 5), and (–11,5). Lef now n ≥ 4. Clearly, for n even, the equation is not solvable, since no other squares differ by 4. For n odd, we may assume without loss of generality that n is a prime p ≥ 5. Indeed, if n = q ^k, where q is an odd prime, we obtain an equation of the same type: x ² + 4 = (y ^k)^q.