Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers

Question

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 ??

Answer 1

The values of $2(\frac 12 +\frac 12 \sqrt{-7})^{n-2}$ give $x+y\sqrt{-7}$, which are a solution to the equation $x^2+7y^2 = 2^n$, where $x$ and $y$ are always odd.

These solutions are thence unique, because all representations of $x^2+7y^2$ reduce to a complex number $x+y\sqrt{-7}$, and that $\frac 12 \pm \frac 12 \sqrt{-7}$ is the unique primes of this form. $2$ is the product of these numbers, and since it requires both primes to give an even value of $x$ and $y$, powers of a single prime must lead always to odd numbers.

Answer 2

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.

Answer 3

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 !