Elegance in Integers by Dorin Marghidanu

There's an obvious trivial solution, $x=y=z=t=u=0.$ Let's find a nontrivial solution. Since $4^{24}+4^{24}+4^{24}+4^{24}=4^{25},$ we conclude that

$4^{24}+(4^{12})^2+(4^{8})^3+(4^{6})^4=(4^{5})^5$

so that $(4^{24},4^{12},4^{8},4^{6},4^{5})$ is one set of solutions.

In general, if the quintet $(x_0,y_0,z_0,t_0,u_0)$ solves the given equation, i.e., satisfies $x_0+y_0^2+z_0^3+t_0^4=u_0^5,$ then multiplying by $k^{60},$ $k\in\mathbb{N},$ gives

$k^{60}\cdot x_0+(k^{30}\cdot y_0)^2+(k^{20}\cdot z_0)^3+(k^{15}\cdot t_0)^4=(k^{12}\cdot u_0)^5$

such that the quintet $(k^{60}x_0,\,k^{30}y_0,\,k^{20}z_0,\,k^{15}t_0,\,k^{12}u_0)$ is also a solution for any $k\in\mathbb{N}.$

Solution

A trivial solution $x_1=x_2=\ldots=x_n=y=0$ is obvious. For a nontrivial solution note that

$\underbrace{n^{n!}+n^{n!}+\ldots+n^{n!}}_{n\text{ terms}}=n\cdot n^{n!}$

which is equivalent to

$n^{n!}+(n^{\frac{n!}{2}})^2+(n^{\frac{n!}{3}})^3\ldots+(n^{\frac{n!}{n}})^n=n^{n!+1},$

making the $n-tuple$

$(x_1,x_2,x_3,\ldots,x_n,y)=(n^{n!},\,n^{\frac{n!}{2}},\,n^{\frac{n!}{3}},\ldots,n^{\frac{n!}{n}},\,n^{n!+1}),$

a solution. We get more solutions with any $k\in\mathbb{N}:$

$(k^{n!}x_1,\,k^{\frac{n!}{2}}x_2,\,k^{\frac{n!}{3}}x_3,\ldots,k^{\frac{n!}{n}}x_n,\,k^{n!}y).$

Due to the presence of the linear term on the left, both the problem and the generalization are in fact trivial. For the problem, $y,z,t$ could be chosen arbitrarily and $u$ sufficiently large to ensure $u^5-t^4-z^3-y^2\gt 0.$ The problem becomes less trivial when the linear term is left out:

Then the solution above can be adapted: since $3^{24}+3^{24}+3^{24}=3^{25},$ $(3^{12},3^{8},3^{6},3^5)$ serves a solution to the equation.

The problem admits generalizations different from the above. For example,

Since $3^{mnp}+3^{mnp}+3^{mnp}=3^{mnp+1},$ $(3^{np})^m+(3^{mp})^n+(3^{mn})^p=3^{mnp+1}$ and the problem is to find the factors of $mnp+1.$ For example, with $m=2,$ $n=4,$ and $p=6,$ $mnp+1=49,$ making the question of finding natural solutions to $y^2+z^4+t^6=u^7$ meaningful. $(3^{24},3^{12},3^{8},3^7)$ is one solution to the latest equation.

Dorin Marghidanu has kindly shared the above problem on the CutTheKnotMath facebook page, along with a solution of his and a generalization.