Quadratic Modular Arithmetic

Question

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?

Answer 1

The whole point of modular arithmetic is reduction of the integers to finite sets. Thus you only have to account for a limited number of cases. In this problem, 0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 4, 4^2 = 1 in Z/5Z, as desired.

Answer 2

Hint:

Modulo $5$, $x\equiv0, \pm1, $ or $\pm 2$.

What is $x^2 $ in each case?

Answer 3

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.