I am having troubles with this question:

Show that if p is prime and p >= 7, then there are always two consecutive residues of p (Hint: Show that at least one of 2,5,10 is a quadratic residue of p).

I was thinking of using the values of (2/p) and (5/p), but i am not sure where to go from there.

2. Here's how you show the hint : by Euler's criterion, $\displaystyle \left(\frac{a}{p}\right)=a^{(p-1)/2}$. This immediately implies that the Legendre symbol is multiplicative in the top argument; in perticular if neither $\displaystyle a,b$ are quadratic residues then $\displaystyle ab$ is a quadratic residue.

3. i feel silly, i was trying to use theorems from Gauss' lemma to show the hint

Okay now have i have if 2 is a quadratic residue of p, and 1 is a quadratic residue of every number, then we have 2 consecutive residues. However I'm having trouble with 5 and 10. Would it have to do with the fact that 5 and 10 are adjacent to 4 and 9 respectively?

4. Perhaps; I haven't really tried to think about the relevance of the hint to the original question.

Let's solve it in the most general manner possible. We will be working in $\displaystyle K=\mathbb{F}_p^\times$ and will not bother writing $\displaystyle \equiv \mod p$ but instead we'll just write $\displaystyle =$. We also take p sufficiently large (which is not quite large at all).

Lemma 1. For some $\displaystyle x \in K$, $\displaystyle x^2=y^2+z^2$ has a solution with $\displaystyle y,z \in K$.
Proof : find $\displaystyle m,n \in K$ with $\displaystyle (m/n)^2 \neq \pm 1$ (this is where we need $\displaystyle p$ large enough). Then $\displaystyle x=m^2+n^2,\: y=m^2-n^2,\: z=2mn$ is a solution.

Theorem. Every square in $\displaystyle K$ is the sum of two nonzero squares.
Proof : Suppose we want to write $\displaystyle a^2$ as a sum of two nonzero squares. We write $\displaystyle a^2=x^{2}b^2$ and then we have, using the lemma above, $\displaystyle a^2=(yb)^2+(zb)^2$.

In perticular, if we take $\displaystyle b=z^{-1}$ then we have $\displaystyle (xz^{-1})^2=(yz^{-1})^2+1$; i.e. there are two quadratic residues differing by 1.

Don't hesitate to ask if you're not convinced.
Sorry for not using the hint

5. See a more general statement here.

(Set $\displaystyle k=n=1$ for your particular case, you'd have to check the case p=7 separately though.)

EDIT: To add that in the case p=7, both 1 and 2 ( 3² = 9) are quadratic residues module 7.

6. I'm assuming that's field theory you did there bruno, and my knowledge of fields is very limited. It looks like a very rigourous proof but my number theory class doesn't use any results from abstract algebra. We're expected to prove this using properties of congruences and residues.