From Euclid I know $\gcd(a,b) = \gcd(a, b-ka)$
I'm wondering if it is legal to replace $k$ by $a^{n-1}$ ?
I'm a bit skeptical here because $a$ is a variable... It's not like $k$ I guess.
Appreciate any help. Thank you!
$\gcd(a, b) = \gcd(a, b-a^n)$ where $n\ge 1$ . Is this statement true?
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elementary-number-theory
greatest-common-divisor
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1I like your worry and skepticism! Shows concern and careful thought. But it's not necessary here. a isn't a "variable" here so much as an "unknown value". $a^{n-1}$ is most certainly a valid integer. And as math is love's direct application shows... it is certainly a true result. – 2017-01-14
2 Answers
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If $k$ can be any integer, then it is valid, seeing $a^{n-1}$ is also an integer.
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0All are integers in my question. Thank you for the confirmation! – 2017-01-14
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Let $\gcd (a,b)= m$ and $\gcd (a,b-a^n) = d$. So, $$ d\big| a \implies d\big| a^n \tag 1 $$ Also, $$d\big|(b- a^n)\tag 2 $$ From $(1)$ and $(2)$, we get $d\big| b$. Thus , $d\big| a$ and $d\big| b$. This implies that $$d| gcd (a,b) =m \tag 3$$ We now show,$m\big| d$. As $m\big| a$ and $m\big| (b-a^n)$ and ,$gcd (a,b-a^n) = d$, thus, $$m\big|d \tag 4$$ From $(3)$ and $(4)$, we get $m=d$.