n unique sums of squares for 2p^(2n) for prime p (1 mod 4) and n natural number >= 1?

Question

After looking at many sums of squares, this is something I am hypothesizing. I'd love to know if there is any work related to this, if it is easy to solve or even if there exist counterexamples.

Conjecture: Let $p$ be some prime number where $p \equiv 1 \bmod 4$. Then, for $n \in \mathbb{N}$ and $n \geq 1$, $2p^{2n}$ will give $n$ unique pairs $a,b \in \mathbb{N}$ such that:

\begin{align*} a^2 + b^2 &= 2p^{2n} \end{align*}

In the image below I have provided 3 different examples along with the $a,b$ sums of squares for more details.

Ex 1: $p=617$ and $n=1$.

Ex 2: $p = 13$ and $n=3$.

Ex 3: $p = 5$ and $n=7$.

link to image with examples described above

Answer 1

There’s an exact formula for the number of ways to express given integer as a sum of two squares(by the theory of modular forms). You can use this to prove your statement.

Answer 2

None of your resulting triples will be primitive because that requires that the RHS be an odd number. To find them we use Euclid's formula for $\quad A^2+B^2=C^2\quad$ which is $ \quad A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2\quad$ We solve for $k$ and test a defined range of values to see which, if any yield integers.

\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\quad\text{for}\quad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation}

The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

We will use $C$ instead of $C^2:\quad$ $\quad C^2=2p^{2n}\implies C=2p^n$ $$C=2\cdot 617^1=1234 \implies \bigg\lfloor\frac{ 1+\sqrt{1234-1}}{2}\bigg\rfloor = 25 \le m \le \big\lfloor\sqrt{1234-1}\space\big\rfloor=35\\ \quad\text{and we find} \quad m\in\{35\}\implies k\in\{3\}\\ F(35,3)=(1216,210,1234)\quad \\$$

$$C=2\cdot 13^3=4394 \implies \bigg\lfloor\frac{ 1+\sqrt{4394-1}}{2}\bigg\rfloor = 47 \le m \le \big\lfloor\sqrt{4394-1}\space\big\rfloor=66\\ \quad\text{and we find} \qquad m\in\{55,65\}\implies k\in\{37,13\}\\ F(55,37)=(1656,4070,4394)\qquad F(65,13)=(4056,1690,4394)\quad \\$$

$$C=2\cdot 5^7=156250 \implies \bigg\lfloor\frac{ 1+\sqrt{156250-1}}{2}\bigg\rfloor = 280 \le m \le \big\lfloor\sqrt{156250-1}\space\big\rfloor = 395\\ \quad\text{and we find} \quad m\in\{307,325,375,395\}\implies k\in\{249,225,125,15\}\\ f(307,249)=(32248,152886,156250)\\ f(325,225)=(55000,146250,156250)\\ f(375,125)=(125000,93750,156250)\\ f(395,15)=(155800,11850,156250)$$