Library / /18 of 69

, 2011

Number theory Difficulty 4.7 AIME Prove it South Africa

Prove that the equation $x^5 + y^3 = z^2$ has infinitely many solutions in the natural numbers.

Solution

Put $x = k^6$ , $y = 2k^{10}$ and $z = 3k^{15}$ for some natural number $k$ . Then
$x^5 + y^3 = (k^6)^5 + (2k^{10})^3 = k^{30} + 8k^{30} = 9k^{30} = (3k^{15})^2 = z^2$
as required. Since there are infinitely many possible values for $k$ , there are infinitely many triples $(x, y, z)$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.