This took me a whole morning... We have $a_{m+1}a_{m-1}-a_m^2=-a_1^2=a_ma_{m-2}-a_{m-1}^2\Rightarrow a_{m-1}(a_{m+1}+a_{m-1})=a_m(a_m+a_{m-2})\Rightarrow \frac{a_{m+1}+a_{m-1}}{a_m}=\frac{a_{m}+a_{m-2}}{a_{m-1}}=\cdots=\frac{a_3+a_1}{a_2}=k$. $\Rightarrow a_{m+1}-ka_m+a_{m-1}=0$. We have $a_{m+1}^2+a_{m-1}^2=(a_{m+1}+a_{m-1})^2-2a_{m+1}a_{m-1}=k^2a_m^2-2(a_m^2-a_1^2)=(k^2-2)a_m^2+2a_1^2$ We will now prove inductively these two things: $a_{m+n}a_{m-n}=a_m^2-a_n^2$ and $a_{m+n}^2+a_{m-n}^2=\frac{k^2-4}{a_1^2}a_m^2a_n^2+2a_m^2+2a_n^2$ Assume for all $k\le n<m$, we have: $a_{m+k}a_{m-k}=a_m^2-a_k^2$ and $a_{m+k}^2+a_{m-k}^2=\frac{k^2-4}{a_1^2}a_m^2a_k^2+2a_m^2+2a_k^2$ We need to prove that $a_{m+n+1}a_{m-n-1}=a_m^2-a_{n+1}^2$ and $a_{m+n+1}^2+a_{m-n-1}^2=\frac{k^2-4}{a_1^2}a_m^2a_{n+1}^2+2a_m^2+2a_{n+1}^2$ $$a_{m+n+1}a_{m-n-1}=a_m^2-a_{n+1}^2$$$$\Leftrightarrow\frac{(a_{m+n}^2-a_1^2)(a_{m-n}^2-a_1^2)}{a_{m+n-1}a_{m-n+1}}=a_m^2-a_{n+1}^2$$$$\Leftrightarrow (a_{m+n}^2-a_1^2)(a_{m-n}^2-a_1^2)=(a_m^2-a_{n+1}^2)(a_m^2-a_{n-1}^2)$$$$\Leftrightarrow a_{m+n}^2a_{m-n}^2-a_1^2(a_{m+n}^2+a_{m-n}^2)+a_1^4=a_m^4-a_m^2(a_{n+1}^2+a_{n-1}^2)+a_{n+1}^2a_{n-1}^2$$$$\Leftrightarrow (a_m^2-a_n^2)^2-a_1^2(a_{m+n}^2+a_{m-n}^2)+a_1^4=a_m^4-a_m^2((k^2-2)a_n^2+2a_1^2)+(a_n^2-a_1^2)^2$$$$\Leftrightarrow a_1^2(a_{m+n}^2+a_{m-n}^2)=(k^2-4)a_m^2a_n^2+2a_m^2a_1^2+2a_n^2a_1^2$$which is true. We now need to prove that $a_{m+n+1}^2+a_{m-n-1}^2=\frac{k^2-4}{a_1^2}a_m^2a_{n+1}^2+2a_m^2+2a_{m+1}^2$. It is easy to verify by hand that $a_{m+2}^2+a_{m-2}^2=\frac{k^2-4}{a_1^2}a_m^2a_2^2+2a_m^2+2a_2^2$ We have $$a_{m+n+1}^2+a_{m-n-1}^2=\frac{k^2-4}{a_1^2}a_m^2a_{n+1}^2+2a_m^2+2a_{m+1}^2$$$$\Leftrightarrow (a_{m+n+1}^2+a_{m+n-1}^2)+(a_{m-n-1}^2+a_{m-n+1}^2)=\frac{k^2-4}{a_1^2}a_m^2(a_{n+1}^2+a_{n-1}^2)+4a_m^2+2(a_{n+1}^2+a_{n-1}^2)$$$$\Leftrightarrow (k^2-2)a_{m+n}^2+2a_1^2+(k^2-2)a_{m-n}^2+2a_1^2=\frac{k^2-4}{a_1^2}a_m^2((k^2-2)a_n^2+2a_1^2)+4a_m^2+2((k^2-2)a_n^2+2a_1^2)$$which is true.