Quadratic Modular Arithmetic I want to prove that
$$ w^2 \equiv 2 \quad (\bmod{5})$$
has no solutions in integers.
What I tried:
$$ w^2 \equiv 2 \quad (\bmod{5})$$
$$ \Rightarrow w^2 =  2 + 5k,  \quad k \in \mathbb{Z} $$
Now, I don't know what to do, so I considered $x^2 = 2+ 5y$ over $\mathbb{R}$.
 We can solve this to see if the solutions are integers. 
If it has no solutions in integers, can I then say that this implies that $ w^2 \equiv 2 \; (\bmod{5})$ has no solutions?
Is there a different approach to this problem?

Without knowing how you finished the proof it can't be verified.  
This should prove it though:  
All multiples of $5$ end with the digit $5$ or $0$ in decimal base.
$$5+2=7$$
$$w^2\ne7+10n$$
$$w^2\ne2+10n$$
because no perfect squares end in $7$ or $2$ in decimal base.
