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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 :)

J. W. Tanner
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Qi Yuan
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  • Welcome to mathematics Stack Exchange. Do you mean $3y^3$? – J. W. Tanner Mar 23 '20 at 00:52
  • Yeah...just corrected it – Qi Yuan Mar 23 '20 at 00:52
  • Such substitution is valid by the Polynomial Congruence Rule, an inductive extension of the Sum and Product Rules. – Bill Dubuque Mar 23 '20 at 01:10
  • 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 '20 at 01:27
  • Example $a=jN + a'$ and $b = kN + b'$ means $a+b = (j+k)N + (a'+b')$ and $ab = N(jkN + j+ k) + (a'b')$ and $(jN + a')^k =$... a bunch of terms involving $N$ to powers .... $+ a'^k$. – fleablood Mar 23 '20 at 01:29
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    It's good practice to work it out once for yourself. If $x = 9j + 3$ and $y = 9k+4$ then $20x + 3y^3 = 20(9j + 3) + 3(9k + 4) = 209j + 2([9k]^3 + 34[9k]^2 + 34^29k) + 203 + 34^3$ and the only things that aren't a multiple of $9$ are the $203 + 34^3$. I.e. a substitution. – fleablood Mar 23 '20 at 01:35

1 Answers1

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

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

J. W. Tanner
  • 60,406