The sequence $\{a_n\}$ is given recursively by $(1) \; a_n = 2a_{n-1} + a_{n-2}$, or explicitly as, $(2) \; {\textstyle a_n = \frac{\alpha^n - \beta^n}{2\sqrt{2}}}$. where $\alpha = 1 + \sqrt{2}$ and $\beta = 1 - \sqrt{2}$. Formula (2) can be proven as follows: $(3) \; a_{n+1}a_{n-1} - a_n^2 ={\textstyle \frac{\alpha^{n+1} - \beta^{n+1}}{2\sqrt{2}} \cdot \frac{\alpha^{n-1} - \beta^{n-1}}{2\sqrt{2}} - (\frac{\alpha^n - \beta^n}{2\sqrt{2}})^2 = \frac{\alpha^{2n} + \beta^{2n} - (\alpha \beta)^{n-1}(\alpha^2 + \beta^2)}{8} - \frac{\alpha^{2n} + \beta^{2n} - 2(\alpha \beta)^n}{8} = \frac{-(\alpha \beta)^{n-1}(\alpha - \beta)^2}{8}}$. Now $\alpha - \beta = (1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2}$ and $\alpha \beta = (1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1$, which inserted in (3) give us ${\textstyle a_{n+1}a_{n-1} - a_n^2 = \frac{-(-1)^{n-1}(2\sqrt{2})^2}{8} = (-1)^n}$, i.e. ${\textstyle a_{n+1} =\frac{a_n^2 + (-1)^n}{a_{n-1}}}$. Applying formula (2), we obtain ${\textstyle a_{m+1}^2 + a_m^2 = (\frac{\alpha^{m+1} - \beta^{m+1}}{2\sqrt{2}})^2 + (\frac{\alpha^m - \beta^m}{2\sqrt{2}})^2 = \frac{\alpha^{2m+2} + \beta^{2m+2} - 2(\alpha \beta)^{m+1}}{8} + \frac{\alpha^{2m} + \beta^{2m} - 2(\alpha \beta)^m}{8} = \frac{\alpha^{2m}(\alpha^2 + 1) + \beta^{2m}(\beta^2 + 1) - 2(\alpha \beta)^m(\alpha \beta +1)}{8}}$, which via the fact that $\alpha ^2 + 1 = (1 + \sqrt{2})^2 + 1 = 4 + 2\sqrt{2} = 2\sqrt{2}(1 + \sqrt{2}) = 2\sqrt{2} \, \alpha$, $\beta = (1 - \sqrt{2})^2 + 1 = 4 - 2\sqrt{2} = -2\sqrt{2}(1 - \sqrt{2}) = -2 \sqrt{2} \, \beta$ and $\alpha \beta = -1$ becomes ${\textstyle a_{m+1}^2 + a_m^2 = \frac{2\sqrt{2} \alpha^{2m+1} + 2\sqrt{2} \beta^{2m+1} - 2(-1)^m(-1 +1)}{8} = \frac{\alpha^{2m+1} - \beta^{2m+1}}{2\sqrt{2}}}$, which according to (2) means $a_{m+1}^2 + a_m^2 = a_{2m+1}$. $\;\;$ q.e.d.

The recurrence relation may be written as $a_{n+1}a_{n-1}={a_n}^2+(-1)^n$ Replacing $n$ by $n+1$ we get $a_{n+2}a_n={a_{n+1}}^2+(-1)^{n+1}$ Adding these relations we get $a_n(a_{n+2}-a_n)=a_{n+1}(a_{n+1}-a_{n-1})$ $\implies \frac{a_{n+2}-a_n}{a_{n+1}-a_{n-1}}=\frac{a_{n+1}}{a_n}$ $\implies \frac{a_{n+2}-a_n}{4}=\prod_{i=3}^{n+1}{\frac{a_{i+1}-a_{i-1}}{a_i-a_{i-2}}}=\prod_{i=3}^{n+1}{\frac{a_i}{a_{i-1}}={\frac{a_{n+1}}{2}}}$ $\implies a_{n+2}=2a_{n+1}+a_n$ The characteristic equation of the sequence is $x^2-2x-1=0$ with roots $\alpha=1+\sqrt{2},\beta=1-\sqrt{2}$.Testing the constants for the initial values we get $a_n=\frac{{\alpha}^n-{\beta}^n}{2\sqrt{2}}$. So ${a_{m}}^2+{a_{m+1}}^2$ $=\frac{1}{8}[({\alpha}^m-{\beta}^m)^2+({\alpha}^{m+1}-{\beta}^{m+1})^2]$ $=\frac{1}{8}[{\alpha}^{2m}(4+2\sqrt{2})+{\beta}^{2m}(4-2\sqrt{2})]$ $=\frac{1}{2\sqrt{2}}({\alpha}^{2m+1}-{\beta}^{2m+1})$ $=a_{2m+1}$