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 $9$) [duplicate]

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 :)

• Welcome to mathematics Stack Exchange. Do you mean $3y^3$? – J. W. Tanner Mar 23 at 0:52
• Yeah...just corrected it – Qi Yuan Mar 23 at 0:52
• Such substitution is valid by the Polynomial Congruence Rule, an inductive extension of the Sum and Product Rules. – Bill Dubuque Mar 23 at 1:10
• Yes, substitution works. Prove that if $a\equiv a' \pmod N$ and $b \equiv b' \pmod N$ then $a+b \equiv a' + b'\pmod N$ and $ab \equiv a'b' \pmod N$ and the $a^k \equiv a'^k \pmod N$. Those rules imply if $P(x,y)$ is a polynomial with two variable then $P(a,b) \equiv P(a',b')$. ... to prove those rules you have $a = jN +a'$ and $b= kN+b'$ and however you manipulate those only manipulations directly relating to $a'$ and $b'$ won't be multiplied byt a multiple of $N$.... – fleablood Mar 23 at 1:27
• It's good practice to work it out once for yourself. If $x = 9*j + 3$ and $y = 9k+4$ then $20x + 3y^3 = 20(9j + 3) + 3(9k + 4) = 20*9j + 2([9k]^3 + 3*4*[9k]^2 + 3*4^2*9k) + 20*3 + 3*4^3$ and the only things that aren't a multiple of $9$ are the $20*3 + 3*4^3$. I.e. a substitution. – fleablood Mar 23 at 1:35

It's not possible, because if $$x\equiv3\pmod9$$ and $$y\equiv4\pmod9$$,
then $$20x+3y^3\equiv20\times3+3\times4^3\equiv60+3\times1\equiv0\pmod9$$.