<s>[INST] Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
 [/INST] for $n=3$, they are solutions ($x_1=1,y_1=1$).
If you have odd solutions $x_n$, $y_n$ for $2^n$, then
$$\left(\frac{7y_n-x_n}{2}\right)^2+7\left(\frac{x_n+y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n+1}$$
$$\left(\frac{7y_n+x_n}{2}\right)^2+7\left(\frac{x_n-y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n+1}$$
Hence $x_{n+1}=\frac{7y_n-x_n}{2}$ and $y_{n+1}=\frac{x_n+y_n}{2}$
or $x_{n+1}=\frac{7y_n+x_n}{2}$ and $y_{n+1}=\frac{x_n-y_n}{2}$
This is, in fact, the same answer as wendy.
Note that always one of the solution is odd and the other is even. Take the odd one !
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