Hint: We have ${{(1+\sqrt{2})}^{n}}=a+b\sqrt{2}=\sqrt{{{a}^{2}}}+\sqrt{2{{b}^{2}}}=\sqrt{m+1}+\sqrt{m}$, so we define ${{a}^{2}}=m+1,2{{b}^{2}}=m\Rightarrow {{a}^{2}}-2{{b}^{2}}=1$ and this is a Pell type equation, with an infinite many positive integer solution, and with $a,b$ we have the $m$

TuZo wrote: Hint: We have ${{(1+\sqrt{2})}^{n}}=a+b\sqrt{2}=\sqrt{{{a}^{2}}}+\sqrt{2{{b}^{2}}}=\sqrt{m+1}+\sqrt{m}$, so we define ${{a}^{2}}=m+1,2{{b}^{2}}=m\Rightarrow {{a}^{2}}-2{{b}^{2}}=1$ and this is a Pell type equation, with an infinite many positive integer solution, and with $a,b$ we have the $m$ The infinitude of the solutions isn't enough. You need to prove that solutions $(a,b)$ of the specified Pell equation suit the representation $(1+\sqrt 2)^n=a+b\sqrt 2$ for any $n$. Moreover, the equation should be changed for odd $n$s slightly.

Note that, $(\sqrt{2}+1)^n = a_n\sqrt{2}+b_n$ for some sequences $a_n,b_n$ of positive integers. Now, from $(a_n\sqrt{2}+b_n)(\sqrt{2}+1)=a_{n+1}\sqrt{2}+b_{n+1}$, we obtain that, $b_{n+1}=2a_n+b_n$, and $a_{n+1}=a_n+b_n$, with $a_1=b_1=1$. From here, we will prove via induction on $n$ that, $b_n^2-2a_n^2=\pm 1$ for every $n$, which will show that, $(\sqrt{2}+1)^n = \sqrt{2a_n^2}+\sqrt{b_n^2}$, which is going to yield the claim of the problem. Note that, this is clear for base cases. Assuming it holds for $n$, we have, $$ b_{n+1}^2-2a_{n+1}^2 = 4a_n^2+4a_nb_n +b_n^2 - 2a_n^2-4a_nb_n-2b_n^2, $$and therefore, $b_{n+1}^2-2a_{n+1}^2 = -(b_n^2-2a_n^2)$, and since by inductive hypothesis, $b_n^2-2a_n^2\in\{-1,1\}$, we conclude.

Bandera wrote: TuZo wrote: Hint: We have ${{(1+\sqrt{2})}^{n}}=a+b\sqrt{2}=\sqrt{{{a}^{2}}}+\sqrt{2{{b}^{2}}}=\sqrt{m+1}+\sqrt{m}$, so we define ${{a}^{2}}=m+1,2{{b}^{2}}=m\Rightarrow {{a}^{2}}-2{{b}^{2}}=1$ and this is a Pell type equation, with an infinite many positive integer solution, and with $a,b$ we have the $m$ The infinitude of the solutions isn't enough. You need to prove that solutions $(a,b)$ of the specified Pell equation suit the representation $(1+\sqrt 2)^n=a+b\sqrt 2$ for any $n$. Moreover, the equation should be changed for odd $n$s slightly. For every $n$ we have ${{\left( 1+\sqrt{2} \right)}^{n}}={{a}_{n}}+{{b}_{n}}\sqrt{2},{{\left( 1+\sqrt{2} \right)}^{n}}={{a}_{n}}-{{b}_{n}}\sqrt{2}\Rightarrow a_{n}^{2}-2b_{n}^{2}=\pm 1\Rightarrow m+1=a_{n}^{2},m=2b_{n}^{2}$ where ${{a}_{n}}=\frac{{{\left( 1+\sqrt{2} \right)}^{n}}+{{\left( 1-\sqrt{2} \right)}^{n}}}{2},{{b}_{n}}=\frac{{{\left( 1+\sqrt{2} \right)}^{n}}-{{\left( 1-\sqrt{2} \right)}^{n}}}{2\sqrt{2}}$, so the problem is completly solved! Done