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

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.

• Dont we need that $2^{2^n}+1$ is prime to apply the reciprocity law? Oct 17, 2019 at 4:11

$$\left({\frac {a}{p}}\right)={\begin{cases}1&{\text{if }}a{\text{ is a quadratic residue modulo }}p{\text{ and }}a\not \equiv 0{\pmod {p}},\\-1&{\text{if }}a{\text{ is a non-quadratic residue modulo }}p,\\0&{\text{if }}a\equiv 0{\pmod {p}}.\end{cases}} \tag{1}\label{eq1A}$$
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
$$\left(\frac{2^{2^n}+1}{5}\right) = -1 \tag{3}\label{eq3A}$$