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Question :

If n is a natural number and k is an odd number

$k\in \mathbb{Z}$ then show that $2^{n+2}$ is a divisor of

the number $k^{2^{n}}-1$

My try as following :

I'm think use induction :

$n=1$ then $2^{3}=8$ since $k$ is odd so : $k^{2}=1\mod 8$ We find : $2^{3}$ divisor of $k^{2}-1$ (correct )

Now for $n+1$ how I prove $2^{n+3}\mid (k^{2^{n+1}}-1)$

And if can prove it without induction ??

Trevor Gunn
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Ellen Ellen
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3 Answers3

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$$\frac{k^{2^{n+1}}-1}{2^{n+3}}=\frac{k^{2^{n}}-1}{2^{n+2}}\times \frac{k^{2^n}+1}{2}$$ The first fraction on RHS is an integer by induction hypothesis and second one is integer because $k$ is odd, hence LHS is integer and you are done!

To prove without induction Observe that $$k^{2^n}-1=(k^2-1)\prod_{r=1}^{n-1}(k^{2^r}+1)$$ $$\Longrightarrow \frac{k^{2^n}-1}{2^{n+2}}=\frac{k^2-1}{2^3}\times \prod_{r=1}^{n-1}\frac{k^{2^r}+1}{2}$$ which is an integer because $k^2\equiv 1\pmod 8$ and all powers of $k$ are odd.

Martund
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A simple solution for non-advanced people like me:

By using the factorization of the difference of squares: $$k^{2^{n}}-1=(k^{2^{n-1}}+1)(k^{2^{n-2}}+1)(k^{2^{n-3}}+1)\ldots(k+1)(k-1)$$ This obviously has $n+1$ terms (as an example factor $x^2-1$ or $x^8-1$). And if $k$ is odd all of the terms in the product is even and thus $k^{2^{n}}-1$ is divisible by $2^{n+1}$.

Of course, the question was testing divisibility with $2^{n+2}$, the other factor of $2$ come in with either the $k-1$ or the $k+1$ term. Since $k$ is odd both $k-1$ and $k+1$ are even and thus one must be divisible by $4$ (because of consecutive evens).

I don't know if this is a valid proof, as the above proof was far more complex than mine so can someone please verify this?

Aops Vol. 2
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    It's valid if you state and prove (by induction) that $k^{2^n+2} - 1= (2^{2n+1} + 1)(2^{2n} + 1)... (k^{2}+1)(k+1)(k-1)$. A statement of $...$ is an implication of a proof by induction and this follows. I wouldn't say the proof above yours is not complex. It's just more formal and sophisticated in its notation. – fleablood Oct 06 '19 at 04:37
  • What do you mean by proving by induction, i know what induction is but what do you want me to prove? Is my proof not complete? – Aops Vol. 2 Oct 06 '19 at 04:39
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    You need to prove $k^{2^n+2} - 1= (2^{2n+1} + 1)(2^{2n} + 1)... (k^{2}+1)(k+1)(k-1)$. You do so with induction. – fleablood Oct 06 '19 at 04:51
  • I think your misunderstanding my proof. I factored $k^{2^{n}}-1$ as the product (because its the difference of two squares) hence the second line above. Since all of those terms in the product are even (divisible by 2) and there are $n+1$ of them; $k^{2^{n}}-1$ is divisible by $2^{n+1}$ – Aops Vol. 2 Oct 06 '19 at 04:53
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    I understood that perfectly. $(k^{2^n+2} -1) = (k^{2^n+1}-1)(k^{2^n+1} + 1)$. That's one step. But you need to prove it for $n+1$ more steps! To do that you need induction. – fleablood Oct 06 '19 at 05:01
  • $... $ implies an inductive statement. It implies it is obviously true for small values and if it true for some small values it is obviously true for all the next values so it is obviously true for all values. That's a statement of inductive. I'm just pointing out it should be explicitely shown. – fleablood Oct 06 '19 at 05:05
  • Ok ok, i understand your confusion. I'm not using induction im just factoring further. What im saying is; as a base case: $k^{2^4}-1=k^{16}-1=(k^8+1)(k^4+1)(k^2+1)(k+1)(k-1)$ do you understand now (sorry it was confusing in the above proof). – Aops Vol. 2 Oct 06 '19 at 05:05
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    I understood that from the very beginning. To state that for all $n$ (not just $n=4$) but for $n=5$ and $n=6$ and any possible $n$ you must use induction. – fleablood Oct 06 '19 at 05:07
  • " I'm not using induction im just factoring further." If you are factoring an unknown ammount indefinitely then, yes, you are using induction. If you ever state anything using "$...$", "$...$" is shorthand for "and so on" and "and so on" IS induction. – fleablood Oct 06 '19 at 05:09
  • Im sorry, im only 15 so i might not be as advanced as you... but if i understand what you are saying: I must prove that i can factor the difference of two squares? – Aops Vol. 2 Oct 06 '19 at 05:11
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    .... no... not really. I'm say that you proof is valid and definitely the correct way to go. But I'm pointing out that for formality that the what you imply via "$...$" is actually an inductive statement of repeated factoring of squares. I was originally only trying to point out a subtle point. I got a little carried away in insisting upon it. I apologize. The proof you consider "more complex" is basically the same as yours but with more sophisticated notation. I only meant you should state something like "with repeated factoring of difference of squares..". I didn't mean to confuse you. – fleablood Oct 06 '19 at 05:23
  • Ok, thanks ill remember that next time. :) Im trying to get better at math and hopefully become a mathematician one day. I hope i end up just as smart (or even close) as you. – Aops Vol. 2 Oct 06 '19 at 05:28
  • @AopsVol.2 You might find of interest the comments about induction vs. intuition in / on this answer – Bill Dubuque Oct 08 '19 at 13:38
  • @billdubuque: was this a case of induction or intuition – Aops Vol. 2 Oct 09 '19 at 13:00
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    @AopsVol.2 Only you can answer that, since you haven't written enough for the reader to unambiguously determine the intended proof. – Bill Dubuque Oct 09 '19 at 13:11
  • @BillDubuque: ok, my intention in the proof was to write the difference of square as a product so intuition. Thanks for the help. – Aops Vol. 2 Oct 09 '19 at 13:18
  • @AopsVol.2 The issue concerns not that but rather the (inplicit informal) use of induction- see the prior comments. – Bill Dubuque Oct 09 '19 at 13:20
  • @BillDubuque: I understand, should I delete my answer since it uses induction? Or should I just edit it? – Aops Vol. 2 Oct 09 '19 at 13:28
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    There is no need to delete. This is a common question for those learning to formalize inductive intuition so others can learn from it. – Bill Dubuque Oct 09 '19 at 14:22
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The trick is $k^{2*m} - 1 = (k^m -1)(k^m+1)$ and if $2^z\mid k^m -1$ then $k^m$ is odd and $k^m + 1$ is even so $2^z|k^m-1$ and $2\mid k^m+1$ so $2^z*2\mid (k^m-1)(k^m+1)$.

So by induction:

Base case: If $k$ is odd the $k = 2m+1$ for some $m$ and $(2m+1)^2-1= 4m^2 + 4m = 4m(m-1)$ either $m$ or $m-1$ is even so $2|m(m-1)$ and $2^3 = 8 = 4*2\mid 4*m(m-1)$

Induction case: So if ${2^{k-1}}\mid (k^{2^{k+1}}-1)$ then $k$ is odd and $2^{2^k}=2^{2^{k-1}}*2\mid (k^{2^{k+1}}-1)(k^{2^{k+1}} + 1) = (k^{2^{k+2}} - 1)$.

Martin Sleziak
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fleablood
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