Why does the equation $x^2\equiv2 \pmod 5$ have no solutions? Why does the equation $x^2\equiv2 \pmod 5$ have no solutions?
I did a remainders table and found that $$x^2\equiv0;1;4\pmod 5$$
But is there any way to justify this besides that?
The original equation was $2x^2\equiv9\pmod 5$ but I got it to the form above.

The best way is what you have done.


*

*First note that given any $x$, we have


$x\equiv 0\pmod5 \text{ or }x \equiv 1\pmod5 \text{ or }x \equiv 2\pmod5 \text{ or }x \equiv 3\pmod5 \text{ or }x \equiv 4\pmod5$


*Next note that if $x \equiv y\pmod{n}$, then $x^2 \equiv y^2\pmod{n}$. Hence, we have that


*

*If $x \equiv 0 \pmod5$, then $x^2 \equiv 0^2 \pmod5 \equiv 0 \pmod5$.

*If $x \equiv 1 \pmod5$, then $x^2 \equiv 1^2 \pmod5 \equiv 1 \pmod5$.

*If $x \equiv 2 \pmod5$, then $x^2 \equiv 2^2 \pmod5 \equiv 4 \pmod5$.

*If $x \equiv 3 \pmod5$, then $x^2 \equiv 3^2 \pmod5 \equiv 4 \pmod5$.

*If $x \equiv 4 \pmod5$, then $x^2 \equiv 4^2 \pmod5 \equiv 1 \pmod5$.



Hence, we can only have $x^2 \equiv 0,1,4\pmod5$.
