Compositeness test for repunits Is this proof acceptable ?
Definition
Let $R_p=\frac{10^p-1}{9} $ with $p$ prime be a repunit number .
Theorem
If $R_p$ is prime then $7^{\frac{R_p-1}{2}} \equiv -1 \pmod {R_p}$
Proof
Let $R_p$ be a prime , then by Euler's criterion :
$7^{\frac{R_p-1}{2}} \equiv \left(\frac{7}{R_p}\right) \pmod {R_p}$ , where $\left(\frac{7}{R_p}\right)$ denotes Legendre symbol .
If $R_p$ is prime then $R_p \equiv 1,2,4 \pmod 7$ and therefore : $\left(\frac{R_p}{7}\right)=1$
Since $R_p \equiv 3 \pmod 4$ according to the law of quadratic reciprocity it follows that : $\left(\frac{7}{R_p}\right)=-1$ .
Hence ,  $7^{\frac{R_p-1}{2}} \equiv -1 \pmod {R_p}$

Looks all good but as Crostul I feel that it is good to add why $R_p \equiv 1,2,4 \pmod 7$. It is easy to prove that $R_p \equiv 1,2 \pmod 7$ for odd $p$:
As $16 \cdot 63 = 1008$, we have $10^3 \equiv -8 \pmod {63}$ and thus $10^6 \equiv 1 \pmod {63}$. This implies $$10^{1+6n} \equiv 10 \pmod {63}$$ and $$10^{5+6n} \equiv 10^5 \pmod {63} \equiv -800 \pmod {63} \equiv 19 \pmod {63}$$ and thus (any odd $p$ is $\equiv 1,5 \pmod 6$):
either $R_p \equiv 1 \pmod 7$ or $R_p \equiv 2 \pmod 7$.
And of course, when $p=2$, then $R_p = 11 \equiv 4 \pmod 7$
