Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$

Question

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?

Answer 1

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 $$(1-\sqrt{7})^3=22-10\sqrt{7}.$$

Answer 2

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:

$$\left(1-\sqrt 7\right)^3=22-10\sqrt 7.$$

Answer 3

Note that $2.5\lt\sqrt7\lt2.7$, so $-1.7\lt1-\sqrt7\lt-1.5$, while $-5\lt22-10\sqrt7\lt-3$. The only possible integer power of $n$ for which $(1-\sqrt7)^n=22-10\sqrt7$ is $n=3$. (The power cannot be even, since even powers are non-negative, and if $n\ge5$ then $(-1.5)^n\lt-5$.)

This doesn't prove that $n=3$ satisfies the equation, of course. It merely tells us what needs to be checked, namely

$$(1-\sqrt7)^3=1-3\sqrt7+3\cdot7-7\sqrt7=22-10\sqrt7$$