Let $x$ and $y$ be integers such that $x ≡ 3$ (mod $9$) and y ≡$ 4$ (mod $9$). Is it possible that $ 20x + 3y^3 ≡ 6$ (mod $9$)

Question

I have problem answering this question. I know the answer is not possible but I by simply substitute the 3 and 4 but I have clue why so. Can anyone give me an explanation or a correct way to answer this question properly.

Any help is appreciated. Thanks :)

Answer 1

It's not possible, because if $x\equiv3\pmod9$ and $y\equiv4\pmod9$,

then $20x+3y^3\equiv20\times3+3\times4^3\equiv60+3\times1\equiv0\pmod9$.