I was going back through my past exams in a Discrete Mathematics Course and came across this problem which I failed to solve- Does there exist $x$ and $y$ s.t $x^3+y^3 \equiv 3\pmod 7$? Give a convincing proof of your assertion.
I went through some examples and couldn't find any such $x$ and $y$. My attempt was that of considering parity. For an integer which is $3\pmod 7$, it can be either even or odd. If it is even then either $x$ and $y$ must be even or $x$ and $y$ must be odd. Considering the even case, we get that $8a^3+8b^3 \equiv 3\pmod 7$. But This expression is always even and I wasn't really sure where to go from here. I feel like considering parity would be the right approach but I did similar things considering odds and other cases and couldn't gain any traction.
Any help would be appreciated.
Best Answer
Hint: Use Fermat little theorem:
If $7\nmid x$ then $$ x^6\equiv 1 \pmod 7 $$ and so $$ x^3\equiv \pm 1 \pmod 7 $$