<s>[INST] Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$ Is there a natural number $n$ for which $$\sqrt[n]{22-10\sqrt7}=1-\sqrt7$$
My idea was to try to express $22-10\sqrt7$ as something to the power of $2$, but it didn't work $$22-10\sqrt7=22-2\times5\times\sqrt7$$ Since $5^2=25, \sqrt7^2=7$ and $25+7\ne22$. What else can we try?
 [/INST] To paraphrase the excellent answer given in the comments:
Because $\sqrt{7}$ is a pure quadratic surd, if there is such a natural number $n$, then also
$$\sqrt[n]{22+10\sqrt{7}}=1+\sqrt{7}.$$
As $(1+\sqrt{7})(1-\sqrt{7})=1^2-\sqrt{7}^2=-6$ it follows that
\begin{eqnarray*}
-6&=&(1+\sqrt{7})(1-\sqrt{7})\\
&=&\sqrt[n]{22+10\sqrt{7}}\sqrt[n]{22-10\sqrt{7}}\\
&=&\sqrt[n]{(22+10\sqrt{7})(22-10\sqrt{7})}\\
&=&\sqrt[n]{22^2-10^2\sqrt{7}^2}\\
&=&\sqrt[n]{-216}.
\end{eqnarray*}
It is then clear that the only remaining candidate is $n=3$, and a routine verification shows that
$$(1-\sqrt{7})^3=22-10\sqrt{7}.$$
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