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My teacher was proving the identity $$\gcd(t^m -1 , t^n -1 ) = t^{\gcd(m,n)}-1 $$

he first assumes by induction this rule is true and uses the 'n-m' case, I think to prove it. $$\gcd(t^m -1 , t^{n-m} -1 ) =t^{\gcd(m,n-m)}-1 $$ and consequently we use that $\gcd(m,n-m)=\gcd(m,n)$ to show that $$\gcd(t^m -1 , t^{n-m} -1 ) =t^{\gcd(m,n)}-1 = \gcd(t^m -1 , t^{n} -1 ) $$ which should "be true" since $$\gcd(t^m -1 , t^{n-m} -1 ) =\gcd(t^m -1 , t^{n} -1 ) $$ from the proof he presented immediately beforehand in the lecture.

I am somewhat confused as to how this proves the identity, since I can build different rules that still " work", like $\log(\gcd(n,m))$ or something. like I could repeat the argument saying that $$\gcd(t^m -1 , t^{n-m} -1 ) = \log(\gcd(m,n-m)) = \log(\gcd(m,n)) = \gcd(t^m -1 , t^{n} -1 ) $$

Then he moves on.

I guess I'm kind of confused as to how the "$n+1$"-th step comes in here.

Is it still valid to have increments of "$m$" instead of $1$? But even if this is ok, I can still use my "log" identity which is incorrect.

I'm not sure what I am doing wrong or what I am not seeing; assistance would be appreciated.

edit: the base case for the identity is easy to prove so I forwent it in the above

edit 2: my issue to clarify was why the "guess" of $ t^{\gcd(m,n)} -1 $ is unique/special. After spending another minute thinking about this I realized that it clearly has to be another polynomial; I guess no other polynomial-guess satisfies the base condition of n=1?

edit 3: with another bit to think about it, it's clear that the guess has to be of the form $ t^k - 1$ and that the form of $k$ should be $\gcd(m,n)$ to fit the conditions above (I suppose there's some simple proof to show the only factors of $t^j - 1$ must be of that form, so that must be your guess) (thank you Sirous!)

the remaining doubt I have is how in the proof you can do induction in what appears to be steps of "$m$" instead of a step of "$1$".

Tesla Daybreak
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    https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Dec 29 '20 at 07:34
  • Wecome to MSE, You can simply say: we want a number who can divide $(t^n-1)$ and $(t^m-1)$.You know $t^n-1$ and $t^m-1$ can have a common factor as $(t-1)$,Now consider $t^{m.n}-1= s(n-1)$ and conclude that $t^{m.n}-1$ can also be a common factor of $t^n-1$ and $t^m-1$ – sirous Dec 29 '20 at 07:48
  • This is an argument by strong induction: https://brilliant.org/wiki/strong-induction/ – Qiaochu Yuan Dec 30 '20 at 02:26
  • Have you ever used the Euclidean algorithm to calculate $\gcd(m,n)$? The exact same process is being done here, except in the exponent of $t$. – anon Dec 30 '20 at 03:01
  • @Qiaochu Yuan what I remain confused about is that we appear to assume this is true for the case of $t^{ n-m} $ (and all "previous" cases I guess) but then we show it is valid for $t^n $; in the explanation of strong induction I would expect that we should show its valid for $t^{n-m+1}$, but I guess if its true for $t^n$ then it must be true for $t^{n-m+1}$? – anon.jpg Dec 30 '20 at 03:27
  • @anon: that's ordinary induction. When you prove a family of statements $P(n)$ by strong induction you prove the base case $P(1)$ and then to prove $P(n)$ you can freely assume $P(k)$ for all $k < n$. – Qiaochu Yuan Dec 30 '20 at 04:46
  • @QiaochuYuan ah that makes sense. Then, to prove any P(n) you can just select P(n-m) to prove it since all the previous cases are valid. Thanks for the clarification – anon.jpg Dec 30 '20 at 17:07

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