## Task 18/71

For which values of $t$ does the system of equations

$$
x^{2 n}+y^{2 n}=1000 \quad, \quad x^{n}+y^{n}=t
$$

with natural $n$ have positive real solutions?
It is

$$
x^{n}+y^{n}=t=\sqrt{x^{2 n}+y^{2 n}+2(x y)^{n}}=\sqrt{1000+2(x y)^{n}}
$$

If we denote by Min $t$ and Max $t$ the smallest and the largest value that $t$ can take, respectively, then

$$
\begin{gathered}
\operatorname{Min} t=\operatorname{Min} \sqrt{1000+2(x y)^{n}}=\sqrt{1000} \\
\operatorname{Max} t=\operatorname{Max} \sqrt{1000+2(x y)^{n}}=\sqrt{1000+2 \operatorname{Max}(x y)^{n}}
\end{gathered}
$$

It is well known that the arithmetic mean of non-negative numbers is never less than the geometric mean; from this it follows that

$$
1000=x^{2 n}+y^{2 n} \geq 2 \sqrt{x^{2 n} y^{2 n}}=2(x y)^{n}
$$

Thus, Max $t=\sqrt{1000+1000}=\sqrt{2000}$.

Therefore, the system of equations has positive real solutions for all $t$ with

$$
10 \sqrt{10}=\sqrt{1000}<t \leq \sqrt{2000}=20 \sqrt{5}
$$

The lower bound of the interval is open, as it is only attained by $t$ for $x=0$ or $y=0$; the upper bound is attained for $x=y$, which follows directly from the relationship between the arithmetic mean and the geometric mean.