$-3$ is a quadratic residue iff $p \equiv 1 \pmod 3$ So this is the question:
Let $p$ be an odd prime, prove that $-3$ is a quadratic residue modulo $p$ iff $p \equiv 1 \pmod 3$. 
My idea was:
$$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\cdot \left(\frac{3}{p}\right)$$
also: 
$$\left(\frac{-3}{p}\right)= (-3)^a$$
when $a$ stands for  $\frac{p-1}{2}$.
No idea how to continue... I would appreciate any help. Thanks!
 A: Your idea is right! But note that 

$$\left (\frac{-3}{p}\right)=1\iff p\equiv_61$$

is a stronger and better condition than $p\equiv_3 1$, why? Look at all the numbers $n\equiv_3 1$ (that is, $n=1+3k$): $$1,\color{red}{4},7,\color{red}{10},13,\color{red}{16},19,\color{red}{22},\dots$$ Cool, now what? Notice that you alternate between even and odd numbers, so you can just discard the even numbers, because none of them are prime. This happens because adding an odd number of $3$'s to $1$ gives an even number. But no even number on the form $1+3k$ can be prime! Therefore you need to add an even number of $3$'s, that is, adding $6$. Therefore you should look at $p\equiv_6 1$ instead: $$1,7,13,19,\dots$$ Now, back to business!
You can write $\left (\frac{-3}{p}\right)=\left (\frac{-1}{p}\right)\left (\frac{3}{p}\right)$, now we know that $\left (\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}$, so we only have to worry about what $\left (\frac{3}{p}\right)$ is. 
For this, you can use a theorem known as quadratic reciprocity:

For $p,q$ distinct odd primes$$\left (\frac{p}{q}\right)\left (\frac{q}{p}\right)=(-1)^{\frac{q-1}{2}\frac{p-1}{2}}$$

Then you have to check two cases:


*

*$\left (\frac{-1}{p}\right)=1$ and $\left (\frac{3}{p}\right)=1$

*$\left (\frac{-1}{p}\right)=-1$ and $\left (\frac{3}{p}\right)=-1$
Can you take it from here?

(Edit)
I'll help you with the first case. 
If we want $\left (\frac{-1}{p}\right )=1$, then $(-1)^{\frac{p-1}{2}}=1$, which means that $\frac{p-1}{2}$ must be even. This is the same as saying: $$\frac{p-1}{2}=2k\iff p=1+4k\iff p\equiv_4 1$$ From quadratic reciprocity we know that: $$\left( \frac{3}{p}\right) \left( \frac{p}{3}\right )=(-1)^{\frac{p-1}{2}}=1$$ Thus, $$\left( \frac{3}{p}\right)=\left( \frac{p}{3}\right)$$ (we can multiply and divide by $\left( \frac{3}{p}\right)$, since $\left( \frac{3}{p}\right)=\pm 1$)
Now we need to find out what $\left( \frac{p}{3}\right)$ is. Since we want $\left( \frac{3}{p}\right)=1$, we need to have $\left( \frac{p}{3}\right)=1$, that is, we need to find the primes $p$ such that $p$ is a quadratic residue modulo $3$. Let us look at the residues: $$\begin{align}1^2=1\\(-1)^2=1\end{align}$$ So the only quadratic residue modulo $3$ is $1$, therefore we want $p=1+3n\iff p\equiv_3 1$
We now have two conditions on $p$:


*

*$p\equiv_3 1$

*$p\equiv_4 1$
This means that $p\equiv_{12}1$. Now you only have to check the second case! Combining both cases you should get: $$p\equiv_6 1$$ which is stronger than $p\equiv_3 1$.
A: The idea to use $\left(\dfrac{-1}p\right)$ and $\left(\dfrac3p\right)$ to calculate $\left(\dfrac{-3}p\right)$ is a natural one...
But, IMHO it works BETTER if you go the opposite direction, and calculate $\left(\dfrac3p\right)$ using the other two!!
This is because $-3$ happens to be the discriminant of the polynomial
$$
\Phi_3(x)=x^2+x+1.
$$
And, coincidentally, the presence of zeros of $\Phi_3$ modulo a prime $p$ can be decided by other means. Namely, we have the factorization $\Phi_3(x)(x-1)=x^3-1$.
Implying that modulo $p>3$ the integer $a$ is a zero of $\Phi_3$ if and only if $a\not\equiv1\pmod p$ and $a^3\equiv1\pmod p$. In other words, $a$ is a zero if and only if it has order three in the group $\Bbb{Z}_p^*$. This group is cyclic of order $p-1$ (recall the existence of primitive roots modulo $p$), so, by Lagrange, such numbers $a$ exist if and only if $3\mid(p-1)$.
OTOH, by quadratic formula, $\Phi_3(x)$ has zeros modulo $p$ if and only if the discriminant $-3$ is a quadratic residue.

It follows that $\left(\dfrac{-3}p\right)=1$ if and only if $p\equiv1\pmod3$.

