My equation is:

$2^{(p-1)/n} \equiv 1 \pmod p$

And this is always true: $1 \equiv p \pmod n$

I want to know which form $p$ must have to make the first equation true.

I know for $n=3$ (cubic reciprocity) the form is $p=x^2+27y^2$.

For $n=4$ (biquadratic reciprocity) the form is $p=x^2+64y^2$

I know the solution has to deal with Gaussian integers and Artin's reciprocity theorem.

Can someone please show me how to exactly do this maybe with the example $n=5$.

If $p \equiv 1 \pmod n$, what additional conditions are needed to ensure that $$2^{(p-1)/n} \equiv 1 \pmod p?$$

I know:

• For $n=3$ (cubic reciprocity) the form is $p=x^2+27y^2$.
• For $n=4$ (biquadratic reciprocity) the form is $p=x^2+64y^2$.
• The solution has to deal with Gaussian integers and Artin's reciprocity theorem.

Can someone please show me how exactly to do this, maybe with the example $n=5$.

My equation is:

$2^{(p-1)/n} \equiv 1 \pmod p$

And this is always true: $1 \equiv p \pmod n$

I want to know which form $p$ must have to make equation 1 true.

I know for $n=3$ (cubic reciprocity) the form is $p=x^2+27y^2$.

For $n=4$ (biquadratic reciprocity) the form is $p=x^2+64y^2$

I know the solution has to deal with Gaussian integers and Artin's reciprocity theorem.

Can someone please show me how to exactly do this maybe with the example $n=5$.

My equation is:

$2^{(p-1)/n} \equiv 1 \pmod p$

And this is always true: $1 \equiv p \pmod n$

I want to know which form $p$ must have to make the first equation true.

I know for $n=3$ (cubic reciprocity) the form is $p=x^2+27y^2$.

For $n=4$ (biquadratic reciprocity) the form is $p=x^2+64y^2$

I know the solution has to deal with Gaussian integers and Artin's reciprocity theorem.

Can someone please show me how to exactly do this maybe with the example $n=5$.