Find all integer solutions to $y^2=x^5+4.$ Is it true that the only solutions are $(x, y) = \{( 2,-6), ( 2,6),(0,2), (0,-2) \}.$ I think the following problem is difficult, any ideas for solution are welcome.
Find all integer solutions to $y^2=x^5+4.$
Is it true that the only solutions are $(x, y) = \{( 2,-6),  ( 2,6),(0,2), (0,-2) \}.$?
 A: Yes, these are the only examples.
Either $x$ is zero (your $(0, \pm 2)$ example) or not; assume the latter. Rewrite your equation as $(y - 2)(y + 2) = x^5$, and appeal to unique factorisation. Then $y \pm 2$ are either both fifth powers, or one is of the form $4a^5$ and the other is of the form $8b^5$. The former case is impossible, since no pair of fifth powers differ by 4.
Hence let $a, b$ be integers such that $a^5 \pm 1 = 2b^5$. Either $b$ is zero (your $(2, \pm 6)$ example) or not; assume the latter. Factorise the left-hand side over the integer ring of the cyclotomic field generated by a fifth root of unity, and again appeal to unique factorisation. You can show that of the five numbers $a \pm \omega^k$, where $\omega$ is a principal fifth root of unity, that four of them must be fifth powers and the other one (necessarily $a \pm 1$) is of the form $2c^5$. Either way, this involves finding a pair of fifth powers such that $d^5 - e^5 = \omega - \omega^4$, whence you can exhaustively determine there are no further solutions (again by unique factorisation).
