Exercise 5.28

Proof. If $p\equiv 1[4]$ and if there exists $x\in \mathbb{Z}$ such that $x^4\equiv 2[p]$, then

From Ex. 5.27, where $p=a^2+b^2,a$ odd, we know that

Since $f^2\equiv -1[p]$, the order of $f$ modulo $p$ is 4, thus $4\mid \frac{\mathit{ab}}{2}$, so $8\mid \mathit{ab}$.

As $a$ is odd, $8|b$, then $p=A^2+64B^2$ (with $A=a,B=b∕8$).

Conversely, if $p=A^2+64B^2$, then $p\equiv A^2\equiv 1[4]$.

Let $a=A,b=8B$. Then

As $2^{\frac{p-1}{4}}\equiv 1[p]$, $x^4\equiv 2[p]$ has a solution in $\mathbb{Z}$ (Prop. 4.2.1), i.e. $2$ is a biquadratic residue modulo $p$.

Conclusion :

Note : the equation $x^4\equiv 2[p]$ has also solutions if $p\equiv -1[8]$.

Indeed, the equation $x^4\equiv 2[p]$ has a solution in $\mathbb{Z}$ iff $2^{\frac{p-1}{d}}=1$, where $d=4\wedge (p-1)=2$, thus iff $2^{\frac{p-1}{2}}\equiv 1[p]$, which is true since $\left(\frac{2}{p}\right)=1$.

For instance, $8^4\equiv 2(\mathit{mod}23)$, with $23\equiv -1(\mathit{mod}8)$. □