Question

Prove that there exist sequences of odd positive integers $\left(x_n\right)n\ge 3,\left(y_n\right)n\ge 3$ such that $7x_n^2+y_n^2=2^n$ for all $n\ge 3.$

Answer 1

For n = 3, define $x_3=y_3=1.$ Suppose that forn $n\ge 3$ there exisat odd integers $x_n$ and $y_n$ such that $7x_n^2+y_n^2=2n.$ Observe the integers $\frac{x_n+y_n}{2}$ and $\frac{x_n-y_n}{2}$ cannot be both even, since their sum is odd. If $\frac{x_n+y_n}{2}$ is odd, we define $x_{n+1=}\frac{x_n+y_n}{2},y_{n+1=}\frac{7x_n-y_n}{2}$ and the conclysion follows by noticing that $7x_{n+1}^2+y_{n+1}^2=\frac{1}{4}(7{(x_n+y_n)}^2+{(7x_n-y_n)}^2)=2(7x_n^2+y_n^2)=2^{n+1}$If $\frac{x_n-y_n}{2}$ is odd, we difine $x_{n+1}=\frac{x_n-y_n}{2},y_{n+1}=\frac{7x_n+y_n}{2}$ and a similar computation yields the result.