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?

Alternate method : Binomial theorem
We have,
$$\begin{align}&\sqrt[n]{22-10\sqrt7}=1-\sqrt7\\
\implies &\left(1-\sqrt 7\right)^n=22-10\sqrt 7\end{align}$$
$22-10\sqrt 7<0$ tells us that, $n$ must be an odd integer. This implies $n≥3$ and $n=2k+1,\thinspace k\in\mathbb Z^{+}$.

Let, $N(n)$ be the integer part of $\left(1-\sqrt 7\right)^{n}$, such that if $\left(1-\sqrt 7\right)^n=a-\sqrt b$, then $N(n)=a$.

Since $\sqrt 7$ is irrational and  $\forall k\in\mathbb Z^{+}$ we have,
$$\begin{align}N(2k+1)&≥(2k+1)(\sqrt7)^{2k}+1\\
&=7^k(2k+1)+1\\
&≥22.\end{align}$$
Thus, we conclude that if $N(2k+1)=22$, then $k=1$.
This means, if equality $\left(1-\sqrt 7\right)^n=22-10\sqrt 7 $ is possible, then $n=3$.
Finally, we see that $n=3$ gives us the required equality:
$$\left(1-\sqrt 7\right)^3=22-10\sqrt 7.$$
