Prove that $(\sqrt2 − 1)^n, \forall n \in \mathbb{Z^+}$ can be represented as $\sqrt{m} − \sqrt{m−1}$ for some $m \in \mathbb{Z^+}$ (no induction).

Step 1

Let us define $$ \psi(p,q) = p \sqrt{2} + q. $$ Let us define $$ \Psi = \big\{ \psi(p,q) | p, q \in \mathbb{Z} \big\}. $$ We have $$ \big(p \sqrt{2} + q \big) \big(r \sqrt{2} + s \big) = \big( p s + q r \big) \sqrt{2} + \big( 2 p r + q s \big). $$ Therefore

$$ \forall \psi_1, \psi_2 \in \Psi : \psi_1 \psi_2 \in \Psi. $$

But also

$$ \forall \psi \in \Psi, \forall n \in \mathbb{N} : \psi^n \in \Psi. $$

Whence for $\psi = \sqrt{2} + 1 \in \Psi$, we get $$ \forall n \in \mathbb{N} : \big( \sqrt{2} + 1 \big)^n \in \Psi\\ \Downarrow $$

$$ \exists p, q \mathbb{Z} : \big( \sqrt{2} + 1 \big)^n = p \sqrt{2} + q. $$

Step 2

Let us define $p_n \in \mathbb{Z}$ and $q_n \in \mathbb{Z}$ such that $$ \big( \sqrt{2} + 1 \big)^n = p_n \sqrt{2} + q_n. $$ Thus $$ p_{n+1} \sqrt{2} + q_{n+1} = \big( \sqrt{2} + 1 \big) \big( p_n \sqrt{2} + q_n \big) = \big( p_n + q_n \big) \sqrt{2} + \big( 2 p_n + q_n \big). $$ So we obtain the recursion relation

$$ \left[ \begin{array}{rcl} p_0 &=& 0\\\\ q_0 &=& 1\\\\ p_{n+1} &=& p_n + q_n\\\\ q_{n+1} &=& 2 p_n + q_n \end{array} \right. $$

Step 3

Whence $$ p_1 = p_0 + q_0 = 1 $$ and $$ p_{n+2} = p_{n+1} + q_{n+1} = p_{n+1} + 2 p_n + q_n = p_{n+1} + 2 p_n + p_{n+1} - p_{n} = 2 p_{n+1} + p_{n}. $$ Therefore

$$ \left[ \begin{array}{rcl} p_0 &=& 0\\\\ p_1 &=& 1\\\\ p_{n+2} &=& 2 p_{n+1} + p_{n}\\\\ \hline\\ q_{n} &=& p_{n+1} - p_{n} \end{array} \right. $$

Step 4

The recursion $$ p_{n+2} = 2 p_{n+1} + p_{n} $$ is a brother of Fibonacci, as Fibonacci is given by $F_{n+2} = F_{n+1} + F_{n}$.

We can write $$ p_{n+2} = 2 p_{n+1} + p_{n}\\ \Downarrow\\ p_{n+2} + \big( \phi - 2 \big) p_{n+1} = \phi p_{n+1} + p_{n}\\ \Downarrow\\ \phi p_{n+2} + \big( \phi^2 - 2 \phi \big) p_{n+1} = \phi \big( \phi p_{n+1} + p_{n} \big). $$ The case $$ \phi^2 - 2 \phi = 1 $$ yields $$ \phi p_{n+2} + p_{n+1} = \phi \big( \phi p_{n+1} + p_{n} \big). $$ So $$ \phi p_{n+1} + p_{n} = \phi^n \big( \phi p_{1} + p_{0} \big). $$

As $$ \phi^2 - 2 \phi = 1 \Rightarrow \phi_\pm = 1 \pm \sqrt{2} $$ we obtain $$ \begin{array}{rclc} \phi_+ \phi_- p_{n+1} + \phi_+ p_{n} &=& \phi_+ \phi_-^n \big( \phi_- p_{1} + p_{0} \big)\\ \phi_+ \phi_- p_{n+1} + \phi_- p_{n} &=& \phi_- \phi_+^n \big( \phi_+ p_{1} + p_{0} \big)\\ &&&-\\ \hline\\ \big( \phi_+ - \phi_- \big) p_{n} &=& \phi_+ \phi_-^n \big( \phi_- p_{1} + p_{0} \big) - \phi_- \phi_+^n \big( \phi_+ p_{1} + p_{0} \big) \end{array} $$ Whence $$ p_{n} = - \phi_+ \phi_- \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } p_1 + \phi_+ \phi_- \frac{ \phi_+^{n-1} - \phi_-^{n-1} }{ \phi_+ - \phi_- } p_0. $$ As $p_0=0$, $p_1=1$ and $\phi_+ \phi_- = -1$, we obtain

$$ p_{n} = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \in \mathbb{Z}. $$

As $$ q_n = p_{n+1} - p_n, $$ we get $$ q_n = \frac{ \phi_+^{n+1} - \phi_-^{n+1} }{ \phi_+ - \phi_- } - \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } = \frac{ [\phi_+ - 1 ]\phi_+^n - [ \phi_- - 1] \phi_-^n }{ \phi_+ - \phi_- } = \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2}, $$ so we obtain

$$ q_{n} = \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2} \in \mathbb{Z}. $$

Step 5

Eventually we obtain $$ ( \sqrt{2} + 1 )^n = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2} + \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \sqrt{2}. $$

So $$ ( \sqrt{2} + 1 )^n = \sqrt{ 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 } + \sqrt{ 2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2 } $$

Note that $$ p_{n} = \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \in \mathbb{Z} \Rightarrow 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 \in \mathbb{Z}, $$ Now comes the fun part: $$ 2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2 = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 + 8 \frac{ \big( \phi_+ \phi_- \big)^n }{ \big( \phi_+ - \phi_- \big)^2 }, $$ and as $\phi_+ \phi_- = -1$ and $\phi_+ - \phi_- = 2 \sqrt{2}$, we get $$ 2 \left( \frac{ \phi_+^n + \phi_-^n }{ \phi_+ - \phi_- } \right)^2 = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 + (-1)^n. $$ Let $$ m = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 + \frac{1 + (-1)^n}{2}. $$ Whence

$$ ( \sqrt{2} + 1 )^n = \sqrt{m} + \sqrt{m-1}. $$

Step 6

Note that $$ ( \sqrt{2} + 1 ) ( \sqrt{2} - 1 ) = 1 $$ and $$ ( \sqrt{m} + \sqrt{m-1} ) ( \sqrt{m} - \sqrt{m-1} ) = 1, $$ then $$ ( \sqrt{2} - 1 )^n = \frac{1}{ ( \sqrt{2} + 1 )^n } = \frac{1}{ \sqrt{m} + \sqrt{m-1} } = \sqrt{m} - \sqrt{m-1}. $$ Whence

$$ ( \sqrt{2} - 1 )^n = \sqrt{m} - \sqrt{m-1}. $$

Conclusion

$$ ( \sqrt{2} \pm 1 )^n = \sqrt{m} \pm \sqrt{m-1}, $$ where $$ m = 2 \left( \frac{ \phi_+^n - \phi_-^n }{ \phi_+ - \phi_- } \right)^2 + \frac{1 + (-1)^n}{2}, $$ and $$ \phi_\pm = 1 \pm \sqrt{2}. $$

If we look it as general such as:for every $n,m\in \mathbb{N}$ have $\quad \exists k\in \mathbb{N}\quad \\ \\ $

$$ \left( \sqrt { m } -\sqrt { m-1 } \right) ^{ n }=\sqrt { k } +\sqrt { k-1 } $$ using binomial formula,we will get

$$\left( \sqrt { m } \pm \sqrt { m-1 } \right) ^{ n }=\sum _{ i=0 }^{ n }{ { C }_{ n }^{ i }\left( \sqrt { m } \right) ^{ n-i }\left( \pm \sqrt { m-1 } \right) ^{ i } } \\ $$

in case $n=2j\left( j\in \mathbb{N} \right)$ it we will get: $$\left( \sqrt { m } \pm \sqrt { m-1 } \right) ^{ n }=\sum _{ i=0 }^{ j }{ { C }_{ n }^{ 2i }\left( \sqrt { m } \right) ^{ 2j-2i }\left( \sqrt { m-1 } \right) ^{ 2i } } \pm $$ $$\pm \sum _{ i=1 }^{ j }{ { C }_{ n }^{ 2i-1 }\left( \sqrt { m } \right) ^{ 2j-2i+1 }\left( \sqrt { m-1 } \right) ^{ 2i-1 }= } $$ $$ \\ \\ =\sum _{ i=0 }^{ j }{ { C }_{ n }^{ 2i }{ m }^{ j-i }\left( m-1 \right) ^{ i }\pm \sqrt { m\left( m-1 \right) } \sum _{ i=1 }^{ j }{ { C }_{ n }^{ 2i-1 } } } { m }^{ j-i }\left( m-1 \right) ^{ i-1 }=a\pm b\sqrt { m\left( m-1 \right) } \\ $$

where $a,b\in \mathbb{Z}^{ + }$ and in case $n=2j-1\left( j\in \mathbb{N} \right) $ we will get :

$$\left( \sqrt { m } \pm \sqrt { m-1 } \right) ^{ n }=\sum _{ i=0 }^{ j-1 }{ { C }_{ n }^{ 2i }\left( \sqrt { m } \right) ^{ 2j-1-2i }\left( \sqrt { m-1 } \right) ^{ 2i } } \pm $$ $$ \pm \sum _{ i=1 }^{ j }{ { C }_{ n }^{ 2i-1 }\left( \sqrt { m } \right) ^{ 2j-2i }\left( \sqrt { m-1 } \right) ^{ 2i-1 }= } $$ $$\\ \\ =\sqrt { m } \sum _{ i=0 }^{ j-1 }{ { C }_{ n }^{ 2j }{ m }^{ j-i-1 }\left( m-1 \right) ^{ i }\pm \sqrt { m-1 } \sum _{ i=1 }^{ j }{ { C }_{ n }^{ 2i-1 } } } { m }^{ j-i }\left( m-1 \right) ^{ i-1 }=c\sqrt { m } \pm d\sqrt { m-1 } $$ where $c,d\in \mathbb{Z }^{ + }$ in both case we have equitions $$\left( \sqrt { m } \pm \sqrt { m-1 } \right) ^{ n }=\sqrt { k } \pm \sqrt { l } $$ for $k,l\in \mathbb{Z}^{ + }$ and $$k-l=\left( \sqrt { k } +\sqrt { l } \right) \left( \sqrt { k } -\sqrt { l } \right) =\left( \sqrt { m } +\sqrt { m-1 } \right) ^{ n }\left( \sqrt { m } -\sqrt { m-1 } \right) ^{ n }=\\ =\left( \left( \sqrt { m } \right) ^{ 2 }-\left( \sqrt { m-1 } \right) ^{ 2 } \right) ^{ n }=1\\ $$

hence $$l=k-1$$ and $$\\ \left( \sqrt { m } +\sqrt { m-1 } \right) ^{ n }=\sqrt { k } +\sqrt { k-1 } $$ as you can see your problem part of it (in case m=2) i hope you will understand,i tried to write few words,because of my poor english