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If $x_1$ and $x_2$ are the roots of the polynomial $x^2-6x+1$ then , for every non-negative integer, prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ .

My trying:

$ x_1 = 3+2\sqrt{2}$ and $ x_1 = 3-2\sqrt{2}$

So $ x_1^n +x_2^n = (3+2\sqrt{2})^n + (3-2\sqrt{2})^n$ = $ \dfrac{(3+2\sqrt{2})^{2n}+1}{(3+2\sqrt{2})^n}$

Then what should I do to solve this problem ?

Tim Ratigan
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Way to infinity
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  • how about proving this with induction??? Though i guess this should not be done like this... –  Dec 02 '13 at 06:16
  • for $n=1$ you would end up with $3+2\sqrt{2}+3-2\sqrt{2}=6$ not divisible by $5$... did you tired in this way? –  Dec 02 '13 at 06:18
  • Is there any other way to prove this except induction ? I want to prove this by arithmatically – Way to infinity Dec 02 '13 at 06:23
  • i am not very familiar with your usage arithmetically... what do you want to use in proving this –  Dec 02 '13 at 06:25
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    You can, and it can be done by expanding both binomials and cancel out opposite terms and use mod5. – DeepSea Dec 02 '13 at 06:25
  • @DeepSea: I´m trying to develop you proposal using binomial expansion, see the linked post. If you have some idea on how I can proceed, please let me know. – bluemaster Nov 06 '18 at 15:47

1 Answers1

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We have $x_1 + x_2 = \text{ Sum of roots }=6$. From the equation, we have $$x_1^2 -6x_1 + 1 = 0 \text{ and }x_2^2 -6x_2 + 1 = 0$$ Adding both we get $$x_1^2 + x_2^2 = 6\underbrace{(x_1+x_2)}_{\text{Integer}} - 2 = \text{Integer}$$ Now use strong induction and make use of the fact that $$x_1^{n+2} -6x_1^{n+1} + x_1^n = 0 \text{ and }x_2^{n+2} -6x_2^{n+1} + x_2^n = 0$$ i.e., $$x_1^{n+2} + x_2^{n+2} = 6(x_1^{n+1}+x_2^{n+1}) - (x_1^n + x_2^n)$$

Use the same idea to show that $$x_1^n + x_2^n \equiv \begin{cases} 1 \pmod5 & n\equiv 1 \pmod{4}\\ 4 \pmod5 & n\equiv 0, 2 \pmod{4}\\ 3 \pmod5 & n\equiv 3 \pmod{4}\\\end{cases}$$

  • the last equation doesn't seem to give the desired conclusion because of its logic. If a, and b are both not divisible by 5, it could be that 6a - b still divisible by 5. Take a = 7, b = 12, then 6a - b = 30 divisible by 5 while you want it not divisible by 5. – DeepSea Dec 02 '13 at 06:23
  • @MatikKen The first half is used to show that it is an integer; the later half must be used in addition to show that it is not divisible by $5$. –  Dec 02 '13 at 06:24