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Problem

Let $f:Z \to Z$ that for all $a\geq b\in Z$ exists at least one $k\in Z$ where $f(a) - kf(b) = f(a-b)$. Prove that $gcd(f(a), f(b)) = f(gcd(a,b))$.

Where did it come from?

I was working with $gcd(2^{100} -1, 2^{120}-1)$ and I conjectured $gcd(2^a-1, 2^b-1) = 2^{gcd(a,b)}-1$. So I thought about the problem.

My "solution"

I don't know if that is correct. Doing operations $gcd(f(a), f(b)) = gcd(f(a-b), f(b))$, by Euclidean Algorithm we get to $gcd(f(gcd(a,b)), f(gcd(a,b))) = f(gcd(a,b))$.

Bill Dubuque
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Kotarou
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  • Your "solution" doesn't immediately work. While you can add arbitrary amounts of $b$ resulting in $\gcd( f(a), f(b) ) = \gcd ( f(a +mb), f(b))$, you are unable to add arbitrary amounts of $a$ for $ = \gcd( f(na+mb), f(b)$. EG For fixed $a, b$, if $f(a) = 1, f(2a) = f(b)$, then $\gcd( f(2a), f(b) ) \neq \gcd(f(a), f(b))$. However, it's not clear to me that such starting values could be extended to a function on $Z$. $\quad$ We can show that $f( \gcd(a, b) ) \mid \gcd ( f(a), f(b) ) $. The reverse direction is less clear to me, because we might be able to introduce additional factors. – Calvin Lin Aug 01 '24 at 09:36
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    FYI Instead of saying "at least one $k \in Z$", could you phrase it as $ f(a) - k_{a, b} f(b) = f(a - b)$? This allows us to avoid misconceptions like Riyan's solution below. – Calvin Lin Aug 01 '24 at 09:40
  • I had a misconception and needed to delete the post by myself. Looking for an answer that justifies the question. – M.Riyan Aug 01 '24 at 09:59
  • Worth noting: your underlying question is addressed here – lulu Aug 01 '24 at 10:03
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    Here's a potential wrinkle: If $ f(0) = x \neq 0 $ and otherwise $f(n) = 0$ for all non-zero $n$, then the functional equation is satisfied by checking cases. However, $\gcd( f(0), f(1) ) = \gcd(x, 0) = x \neq f( \gcd(0, 1) ) = f(1) = 0 $, do the desired conclusion does not hold. $\quad$ As it turns out, if $f(0) = 0 $, then the desired conclusion holds (along the lines of your solution). – Calvin Lin Aug 01 '24 at 10:19
  • @CalvinLin That's true, and I tested for $f(x) = 2^x + 1$. Thank you all for the help. – Kotarou Aug 01 '24 at 13:08

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