Calculating Legendre symbol : $\left(\frac{3}{2^{2^n}+1}\right)$ , n being positive.

Question

I want to find the Legendre symbol : $\left(\frac{3}{2^{2^n}+1}\right)$ for any positive n.

Also I want to find the Legendre symbol for: $\left(\frac{5}{2^{2^n}+1}\right)$, when $n>1$

Solution:

Using Gaussian reciprocity law: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$, we get $\left(\frac{3}{2^{2^n}+1}\right)$=$\left(\frac{2^{2^n}+1}{3}\right)$.Similarly, $\left(\frac{5}{2^{2^n}+1}\right)$=$\left(\frac{2^{2^n}+1}{5}\right)$.

But I am unable to proceed any further. Thanks in advance.

Answer 1

Wikipedia's Legendre symbol page says the definition is

$$\left({\frac {a}{p}}\right)={\begin{cases}1&{\text{if }}a{\text{ is a quadratic residue modulo }}p{\text{ and }}a\not \equiv 0{\pmod {p}},\\-1&{\text{if }}a{\text{ is a non-quadratic residue modulo }}p,\\0&{\text{if }}a\equiv 0{\pmod {p}}.\end{cases}} \tag{1}\label{eq1A}$$

Since $2^2 = 4 \equiv 1 \pmod 3$ and $2^{2^n} = 4^{2^{n-1}}$, then $2^{2^n} + 1 \equiv 1 + 1 \equiv 2 \pmod{3}$, with $2$ being a non-quadratic residue (as only $1$ is a quadratic residue), you have that

$$\left(\frac{2^{2^n}+1}{3}\right) = -1 \tag{2}\label{eq2A}$$

Similarly, since $2^{4} = 16 \equiv 1 \pmod 5$ and $2^{2^n} = 16^{2^{n-2}}$ for $n \gt 1$, you have that $2^{2^n} + 1 \equiv 1 + 1 \equiv 2 \pmod{5}$, with $2$ being a non-quadratic residue (as only $1$ and $4$ are quadratic residues), you get

$$\left(\frac{2^{2^n}+1}{5}\right) = -1 \tag{3}\label{eq3A}$$