Let $f(n) = 10^n - (5 + \sqrt{17})^n - (5 - \sqrt{17})^n$ for all integers $n > 1$. Prove that $f(n)$ is always divisible by $2^{n+1}$.

Question

Let $f(n) = 10^n - (5 + \sqrt{17})^n - (5 - \sqrt{17})^n$ be a function which is valid for all integers $n > 1$. Prove that $f(n)$ is always perfectly divisible by $2^{n+1}$.

edit(12 JAN 2025)- when I posted my proof, I was particularly doubtful about my idea on which my proof was based, My primary aim is just get my idea verified by people at MSE. My proof: The given equation gave me the intuition of a third order recurrence relation solution (after trying for 55 minutes by induction :/ ), so from the given equation of $n$th term I created the characteristic equation of the recurrence relation which comes out to be $$ x^3 - 20x^2 + 108x - 80 = 0 $$

Now in recurrence relation form it can be rewritten as $$ a_{n+3} = 20a_{n+2} - 108a_{n+1} + 80a_n $$ Now to apply induction, where our inductive hypothesis is that every $a_n$ is divisible by $2^{n+1}$. We check the base cases, that is, $a_2=16$, $a_3=240$, and $a_4=3072$, which looks fine.

Then we assume its true for some $a_k$, $a_{k+1}$, and $a_{k+2}$, and then we get that $a_{k+3}$ is divisible by $2^{k+4}$ for some $k$ hence completing our induction. I have cut short the last part because from the equation only its quite visible.

Now is my proof correct? I don't have much experience in proof writing. Moreover, firstly I was trying to prove it using induction only but failed repetitively. As soon as this idea struck me, I published it here, so I apologise for all my errors.

If you have any helpful advice or suggestion please do drop down in the comments. Thanks!

Answer 1

For what it's worth, after I finished this answer, I noticed the comment by J.W.Tanner, which links to a method similar to mine. Since my answer is somewhat different, I am leaving it undeleted.

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Let $f(n) = 10^n - (5 + \sqrt{17})^n - (5 - \sqrt{17})^n$ be a function which is valid for all integers $n > 1$. Prove that $f(n)$ is always perfectly divisible by $2^{n+1}$.

Induction also can be made to work, although it is a little tricky.

Throughout this response, I will use $~M~$ to denote some (unspecified) odd positive integer, and $~E~$ to denote some (unspecified) even positive integer.

Let $~g(n)~$ denote $~\displaystyle (5 + \sqrt{17})^n - (5 - \sqrt{17})^n.$

$\underline{\text{Claim:}}$
$g(n) = 2^n \times M.$

Assuming that the claim is true, then you have that

$$10^n - g(n) = 2^n \times (5^n - M),$$

which is divisible by $2^{n+1}.$

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$\underline{\text{Proof of Claim:}}$

By inspection, $~f(2) = 2^2 \times 21, ~f(3) = 2^3 \times 95, ~f(4) = 2^4 \times 433, ~f(5) = 2^5 \times 1975.$

Assume that the claim is true for $~n = N,~$ and $~n = N-2.$

Then,

$$g(N) \times g(2) = M \times 2^{N+2}$$

$$= \left[ ~(5 + \sqrt{17})^N + (5 - \sqrt{17})^N ~\right] \times \left[ ~(5 + \sqrt{17})^2 + (5 - \sqrt{17})^2 ~\right]$$

$$=g(N+2)$$

$$ + \left\{ ~\left[ ~(5 + \sqrt{17})^N \times (5 - \sqrt{17})^2 ~\right] + \left[ ~(5 - \sqrt{17})^N \times (5 + \sqrt{17})^2 ~\right]~\right\}$$

$$= g(N+2) + \left\{ ~g(N-2) \times \left[ (5 + \sqrt{17})^2 \times (5 - \sqrt{17})^2 ~\right] ~\right\}$$

$$= g(N+2) + \left[ ~2^{N-2} \times M \times 2^6 ~\right]$$

$$= g(N+2) + \left[ ~M \times 2^{N+4} ~\right]$$

$$= g(N+2) + \left[ ~E \times 2^{N+2} ~\right].$$

Therefore,

$$g(N+2) = M \times 2^{N+2} - E \times 2^{N+2} = M \times 2^{N+2}.$$