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It is asked to perform a proof by induction over a variable $k$, which is $k=m+n$ and to use a given equation: $\gcd(a,b)=\gcd(a+b,b)=\gcd(ac+b,a)$, which might help throughout the proof-writing. Usually the induction proofs with one variable are carried out this way: assume for a $k$, that something is true. Show that something for $k+1$. Here, in this question, it is asked to assume for $k-1$ and to show for $k$ ("a shifted-index-induction"). This is one difficulty. The second difficulty is that we have to deal with the two variables $n$ and $m$ instead of just one variable as usual...

I am not clear about, how to perform the induction proof over the variable $k = m+n$...

Any help is appreciated.

Best Regards,

Ahmed Hossam

PS This question ist not a duplicate of that question, because non of the methods shown there is a proof by induction over $m+n$ and the question here comes with a helping equation $\gcd(a,b)=\gcd(a+b,b)=\gcd(ac+b,a)$, that also should be used in the proof. The question might look similar, but the method of solving, which is asked here, is not shown anywhere.

Ahmed Hossam
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    Possible duplicate. – K B Dave Apr 11 '19 at 15:26
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    Possible duplicate of Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ – K B Dave Apr 11 '19 at 15:27
  • I'm not sure if it'll work in this case because I haven't checked, but to use induction with $2$ variables, an approach is to fix one of the values, e.g., $m$, and then show that it works for base case of the other value, e.g., $n = 1$. Then use induction to show if it works for some $n = k \ge 1$ that it works for $n = k + 1$. This shows it works for all $n$ for the given $m$. However, since $m$ was arbitrary, this means it works for all $m$ and $n$. – John Omielan Apr 11 '19 at 19:51
  • In this question, it's asked to perform the induction over a variable $k$, which is equal to $m+n$, so $k=m+n$. It's also asked to use the helping equation $\gcd(a,b)=\gcd(a+b,b)=\gcd(ac+b,a)$ in the proof by induction... – Ahmed Hossam Apr 12 '19 at 06:31

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