2

Problem : Prove (or disprove) that, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ (greatest common divisor), when $a, b, c $ are prime numbers and $a, b, c \geq 3, b \neq c$. $(a^b-1), (a^c-1)$ are factors of $(a^{bc}-1)$.

Click here to see the related question .

Community
  • 1
Michael
  • 489
  • 1
    So what if $a=b=c=3$? Then gcd=26, no? Maybe I misunderstand the question? – Bram28 Mar 07 '17 at 02:49
  • Second part, likewise divide, e.g. $(a^{bc}-1)/(a^b-1) = a^{bc-b} + a^{bc-2b} + ... + a^{bc-cb}$, that last term being equal to $1$. – A. Webb Mar 07 '17 at 02:52

2 Answers2

1

Let $f(n) = (a^n-1)/(a-1).\,$ By this answer $\,(f(b),f(c)) = f((b,c)) = f(1) =1$

Community
  • 1
Bill Dubuque
  • 272,048
0

assume $c > b$

Also gcd $(a, 1 + a + ... + a^b) = 1$

Let $x = (a^c - 1)/(a-1) = 1 + a + a^2 + ... + a^{c-1}$

Let $y = (a^b - 1)/(a-1) = 1 + a + a^2 + ... + a^{b-1}$

$gcd(x,y) = gcd(x-y, y) = gcd((a^b + ... + a^{c-1}), y) = gcd ((a^b(1 + a + ... + a^{c-b-1}), y) = gcd((1 + a + ... + a^{c-b-1}), y)$

as we keep doing, we get $gcd(x,y) = gcd((1 + a + ... + a^{k-1}), (1 + a + ... + a^{b-1}))$ with $k < b$ eventually ending up with $gcd(1 + a + ... + a^m, a^n) = 1$ for some integer $m$ and $n$.

Therefore there is no common factor between $x$ and $y$.

sku
  • 2,643