The elements of the sequence $\mathrm{Az}\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,

$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$

Determine the elements of the sequence.
The elements of the sequence $\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,

$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$

Determine the elements of the sequence.

Solution. We will prove by induction that $x_{n}=n$ for every positive integer $n$. For $n=1$, the equation is:

$$
2 x_{1}^{4}=x_{1}^{5}+x_{1}^{7}
$$

Rearranging and factoring:

$$
x_{1}^{4}\left(x_{1}^{2}+x_{1}+2\right)\left(x_{1}-1\right)=0
$$

The only positive real root of this equation is 1.

Assume that $k>1$, and that we have repeated this assertion for $n=1,2, \ldots,(k-1)$, i.e., that

$$
x_{1}=1, x_{2}=2, \ldots, x_{k-1}=k-1
$$

is the unique positive real solution of the defining equation of the sequence when $n=k-1$. We will show that for $n=k$, (1) has a unique solution: $x_{k}=k$.

Write down the $k$-th and $(k-1)$-th equations, and subtract the latter, using the identity $1+2+\ldots+(k-1)=\frac{k(k-1)}{2}$:

$$
\begin{gathered}
2\left(1+2+\ldots+(k-1)+x_{k}\right)^{4}=\left(1^{5}+2^{5}+\ldots+(k-1)^{5}+x_{k}^{5}\right)+\left(1^{7}+2^{7}+\ldots+(k-1)^{7}+x_{k}^{7}\right) \\
2(1+2+\ldots+(k-1))^{4}=\left(1^{5}+2^{5}+\ldots+(k-1)^{5}\right)+\left(1^{7}+2^{7}+\ldots+(k-1)^{7}\right) \\
2\left(\frac{k(k-1)}{2}+x_{k}\right)^{4}-2\left(\frac{k(k-1)}{2}\right)^{4}=x_{k}^{5}+x_{k}^{7}
\end{gathered}
$$

Perform the exponentiations and rearrange the equation:

$$
x_{k}^{7}+x_{k}^{5}-2 x_{k}^{4}-4 k(k-1) x_{k}^{3}-3 k^{2}(k-1)^{2} x_{k}^{2}-k^{3}(k-1)^{3} x_{k}=0
$$

We can divide by $x_{k}$, since it is positive; rearranging the parentheses:

$$
x_{k}^{6}+x_{k}^{4}-2 x_{k}^{3}-\left(4 k^{2}-4 k\right) x_{k}^{2}-\left(3 k^{4}-6 k^{3}+3 k^{2}\right) x_{k}-\left(k^{6}-3 k^{5}+3 k^{4}-k^{3}\right)=0
$$

It can be verified by substitution that $x_{k}=k$ is a solution. Factor out the $\left(x_{k}-k\right)$ root:

$$
\left(x_{k}-k\right)\left(x_{k}^{5}+k x_{k}^{4}+\left(k^{2}+1\right) x_{k}^{3}+\left(k^{3}+k-2\right) x_{k}^{2}+\left(k^{4}-3 k^{2}+2 k\right) x_{k}+\left(k^{5}-3 k^{4}+3 k^{3}-k^{2}\right)\right)=0
$$

We claim that every coefficient in the second factor is positive. It is sufficient to prove this from the fourth term onwards. Indeed,

$$
\begin{aligned}
& k^{3}+k-2 \geq k^{3}+2-2>0 \\
& k^{4}-3 k^{2}+2 k>k^{4}-4 k^{2}+4=\left(k^{2}-2\right)^{2} \geq 0 \text { and } \\
& k^{5}-3 k^{4}-3 k^{3}-k^{2}=k^{2}(k-1)^{3}>0
\end{aligned}
$$

Therefore, since $x_{k}>0$, the second factor is always positive, and the only positive root of the equation is: $x_{k}=k$.

This completes the proof of our statement.