Problem 28423 from GMB 10/2022

            The problem is proposed for the 9th grade, on page 483.

In translation, thanks to Miss Google:

          "Show that the equation $x^{n+1}+y^n=z^n$ has an infinity of solutions in the set of nonzero natural numbers, for any natural number $n \geqslant 2$."

ANSWER CiP

$$x=\frac{(k^{n+1}+1)^n-1}{k^n},\;\;y=\frac{(k^{n+1}+1)^n-1}{k^{n+1}},\;\;z=\frac{(k^{n+1}+1)^{n+1}-k^{n+1}-1}{k^{n+1}}$$

represent solutions of the given equation, for $k \in \mathbb{N}\setminus \{0\}.$

                    Solution CiP

               Let us first show that the values given in the Answer are solutions of the equation.

          Let $l \underset{def}{=} k^{n+1}+1.$ Then $x=\frac{l^n-1}{k^n},\;y=\frac{l^n-1}{k^{n+1}},\;z=\frac{l^{n+1}-l}{k^{n+1}}$ and we have

$$x^{n+1}+y^n=\frac{(l^n-1)^{n+1}}{k^{n(n+1)}}+\frac{(l^n-1)^n}{k^{(n+1)n}}=$$  

$=\frac{(l^n-1)^n}{k^{(n+1)n}}\cdot [(l^n-1)+1]=\frac{(l^n-1)^n \cdot l^n}{k^{(n+1)n}}=\left ( \frac{l(l^n-1)}{k^{n+1}} \right )^n=\left ( \frac{l^{n+1}-l}{k^{n+1}} \right ) ^n\;=\;\;z^n.$

          We found the expressions in the answer by trying to find solutions of the given equation with $x=ky,\;k \in \mathbb{N} \setminus \{ 0 \}.$ We have

  $$x^{n+1}+y^n=k^{n+1} \cdot y^{n+1}+y^n=y^n(k^{n+1} \cdot y+1).$$  

For the above result to be a power of $n$ we must have $k^{n+1}y+1=l^n$ for some integer $l$, so $y=\frac{l^n-1}{k^{n+1}}.$ For $y$ to be an integer, it is enough to choose $l=k^{n+1}+1$ because according to the binomial theorem we have

$l^n-1=(k^{n+1}+1)^n-1=\sum_{i=0}^n \textrm{C}_n^i (k^{n+1})^{n-i}=$ 

$=[(k^{n+1})^n+\textrm{C}_n^1(k^{n+1})^{n-1}+ \cdots +\textrm{C}_n^{n-1}k^{n+1}+1]-1=k^{n+1} \cdot A$,

 for some integer $A$. That's how I found the answer.

$\blacksquare$

           REMARK CiP

          In particular cases, the solutions can be expressed differently. For $n=2$ the equation

$$x^3+y^2=z^2$$

$\Leftrightarrow \;x^3=(z-y)(z+y)$ can be solved by choosin

$$\begin{cases}z-y=x\\z+y=x^2\end{cases}$$

obtaining an infinite family of complete solutions

$$x=m,\;y=\frac{m^2-m}{2},\;z=\frac{m^2+m}{2},\;m\in \mathbb{Z}.$$

(I did not investigate if these represent the complete solutions of the equation.)

          The particular case has the following solutions

The general formula provides the solutions in the case of n=2