0

I don't know how to start thinking about this problem, I was going to try to prove it by induction, but I think I'm on the wrong path. Any hints?

Lowkey
  • 203
  • The order of $3$ in $\mathbb{Z}/(13\mathbb{Z})^$ is $3$ since $3^3=27\equiv 1\pmod{13}$. It follows that the remainder of $1+3^x+9^x\pmod{13}$ only depends on $x\pmod{3}$ and you just have to test $x\in{0,1,2}$ to get that $1+3^x+9^x\equiv 0\pmod{13}$ iff* $x\not\equiv 0\pmod{3}$. – Jack D'Aurizio Dec 07 '18 at 03:26
  • "It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact. – Lowkey Dec 07 '18 at 03:29
  • $3^{x+3k}\equiv 3^{x}\pmod{13}$, hence $3^{x}\pmod{13}$ is the same thing as $3^{x\pmod{3}}\pmod{13}$. – Jack D'Aurizio Dec 07 '18 at 03:33
  • Oh, of course, thank you. – Lowkey Dec 07 '18 at 03:34

3 Answers3

1

Note that $3^{3n}\equiv 1 \bmod 13$, and that $9^{3n}\equiv 1 \bmod 13$.

Edit: Typo!

RandomMathGuy
  • 129
1

Below put $\,x = 3,\,\ \{A,B,C\} = \{2n,n,0\}\ $ for any $\,n\not\equiv 0\pmod{\!3}$

Lemma $\ x^{2}\!+\!x\!+\!1\mid x^A\! +\! x^B\! +\! x^C\ $ if $ \ \{A,B,C\}\equiv \{2,1,0\}\pmod{\!3}.\ $

Proof $ $ Special case of a simple proof in this answer.

Bill Dubuque
  • 282,220
0

Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$

If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity

Consider $n=3m+1,3m+2$

lab bhattacharjee
  • 279,016
  • I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at? – RandomMathGuy Dec 07 '18 at 02:24
  • @RandomMathGuy, Yes we have derived a generalization. Set $y=3$ – lab bhattacharjee Dec 07 '18 at 02:29
  • @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity. – RandomMathGuy Dec 07 '18 at 02:33
  • @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$ – lab bhattacharjee Dec 07 '18 at 02:36
  • @lab_bhattacharjee And that is still not a complete proof of your proposition. How do you intend to generalize that away from cube roots of unity to modular arithmetic? – RandomMathGuy Dec 07 '18 at 02:38
  • 1
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $\bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want. – RandomMathGuy Dec 07 '18 at 03:00
  • @RandomMathGuy It's a special case of the result I cite in my answer. For many examples see this linked questions list – Bill Dubuque Dec 07 '18 at 03:40
  • @BillDubuque I had actually never seen this before, and I love it! Thanks for posting your answer and a link to that post. – RandomMathGuy Dec 07 '18 at 03:42
  • @RandomMathGuy Yes, it isn't as well know as it should be. But it is quite natural (and obvious) once one masters modular arithmetic. – Bill Dubuque Dec 07 '18 at 03:44