<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] The result is true for $n=3$, since $2^3=x^2+7y^2$ with $x=y=1$. Moreover, $x\equiv y\pmod{4}$.
We show that if there exist $x$ and $y$, odd, with $x\equiv y\pmod{4}$, such that $x^2+7y^2=1$, then there exist $x_1,y_1$, odd, with $x_1\equiv y_1\pmod{4}$,  such that $x_1^2+7y_1^2=2^{n+1}$. Let
$$x_1=\frac{x+7y}{2}\quad\text{and}\qquad y_1=\frac{x+y}{2}.$$
First note that $\frac{x+y}{2}$ and $\frac{x-7y}{2}$ are odd. For $\frac{x+y}{2}$, we use the fact that $x$ and $y$ are odd and $x\equiv y\pmod{4}$ to conclude $\frac{x+y}{2}$ is odd. The argument for $\frac{x-7y}{2}$ is similar. It is easy to verify that $x_1\equiv y_1\pmod{4}$. 
To show that $x_1,y_1$ work, we compute
$$\frac{(x-7y)^2}{4}+7\frac{(x+y)^2}{4}.$$
This simplifies to $2x^2+14y^2$, which is $2^{n+1}$. (We used a version of the Brahmagupta Identity.)
Remark: It is an artefact of the proof that the solutions involve negative integers. We can get positive solutions by taking the absolute value. 
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