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Background

I was playing around with the function $y=x+\frac{1}{x}$ and noticed the following pattern:

$x^2+\frac{1}{x^2}=y^2-2$

$x^3+\frac{1}{x^3}=y^3-3y$

The issue

I am trying to prove $x^n+\frac{1}{x^n}=y^n-n\times y^{n-2}$ for natural numbers $n$.

Showing this is true for $n=1$ is easy, so I decided proof by induction proof may be the way to go.

Showing $n=k+1$ is implied by $n=k$ is giving me trouble and I suspect either my idea is wrong or there is an algebraic trick that I have missed.

Expressing $x^{k+1}+\frac{1}{x^{k+1}}$ in terms of $x^k+\frac{1}{x^k}$ was my first idea which leads to

$(x+\frac{1}{x})(x^k+\frac{1}{x^k})=x^{k+1}+\frac{1}{x^{k+1}}+x^{k-1}+\frac{1}{x^{k-1}}$

but that doesn't seem to be going anywhere.

Any and all ideas are welcome.

Edit (23/May) Thanks for all the comments - the counter-example(s) have verified my suspicions that my formula was wrong. Much appreciated - thanks to all.

Red Five
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  • $n=1,4,5…$ are counter-examples – Soham Saha May 23 '24 at 08:16
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    Your expression fails for $x=1$ on $n=4$. Left side is always 2 at $x=1$, the right side is $2^4-4\cdot 2^2 = 0$. – Randy Marsh May 23 '24 at 08:21
  • Are you familiar with the binomial theorem? Pascal's triangle? Both say that $(A+B)^4 = A^4\ +\ 4A^3B\ +6A^2B^2 + 4AB^3 + B^4$. The first and the last terms will match the pattern, $4y^2= 4(x + \frac1x)^2$, matches the middle terms. However, at first glance the pattern for the middle term breaks down at $6x^3 (1/x)^3)=6 \neq 8=4 * 2$ – nickalh May 23 '24 at 08:21
  • Thanks @RandyMarsh - I had a suspicion it wasn't going to work. Thanks heaps for verifying. If you want to add this as an answer I can accept it. – Red Five May 23 '24 at 09:39
  • Thanks @nickalh - yes I was thinking about the binomial theorem a lot but couldn't find a way to work it into the proof. As you and others have now (thankfully) shown me, it is because the rule simply does not work. Much appreciated. – Red Five May 23 '24 at 09:47
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    An old answer of mine contains a correct recurrence formula. Take a look! It is based on your last equation, but it is a two-step recurrence. – Jyrki Lahtonen May 23 '24 at 10:33
  • Thanks @JyrkiLahtonen - I am very interested in this so will do. – Red Five May 23 '24 at 10:35
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    Oh, I should have started with the proper buzzword. You want to learn about Dickson polynomials. Particularly with the parameter $\alpha=1$. That WP page (and the links in there) give you a better thought out account of them than my old answer (tailored for a particular need). – Jyrki Lahtonen May 24 '24 at 09:22

1 Answers1

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The proposed equality does not hold for all $x$. For $x=1$, $y$ evaluates to $2$. Therefore, at $x=1$ the proposed equality is $2 = 2^{n-2}(4-n)$. Clearly for $n\geq 4$ the right-hand-side is less than or equal to $0$.

Randy Marsh
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