Find $X^2 \equiv 9 \pmod {13}$ without consulting table?
The answer is given as $3, 10$, but without consulting tables it would mean :
$X^2 - 9 \equiv 0 \pmod {13}$
So, either $(X-3)$ or $(X+3) $ is divisible by $13$, as difference between the two is $6$, and $(6,13)=1$.
$$ \begin{align} (X-3) \equiv 0\pmod{13} & \ ======(X+3) \equiv 0\pmod{13}\\ X \equiv 3\pmod{13} & \ ====== X \equiv 10\pmod{13}\\ \end{align} $$
Issue is answer does not say that either $3$ or $10$ is a root, and I have stated only one is a root.
Addendum Have second question and am unable to be solve (as per the answer given) by working on the lines of Q.1.
- Find $X^2 \equiv 1 \pmod {8}$ without consulting table?
The answer is given as : $1, 3 , 5,7$.
But, am able to get only the two values: $1, 7$ by following the approach followed in Q.1.
It is possible to get these 4 values by verification/substitution of each of the 8 values to be substituted for $X$, but what about by formula?
Not much else to do (in fact, you are at the point of citing seemingly irrelevant facts, such as $6$ being coprime with $13$), except remembering the definition of prime element of a ring (in this case, the ring would be $\Bbb Z$). You know that $13\mid x^2-9=(x-3)(x+3)$. Since $13$ is prime in $\Bbb Z$, and it divides $(x-3)(x+3)$, it must divide either $x-3$ or $x+3$. Equivalently, either $x\equiv 3\pmod {13}$ or $x\equiv -3\pmod {13}$. Perhaps, if you so desire, you may observe that $-3\equiv 10\pmod{13}$.