Prove GCD(a, a-1)=1

Question

I've seen the typical proof for GCD(a, a+1) = 1.

But how do you do this for GCD(a, a-1) = 1?

a must be a positive integer throughout the proof.

For example, the GCD for 6 and 7 is 1. For every consecutive numbers paired together, this is the case because a = 2k and a+1 = 2k +1. I think I'm getting somewhere...

Another thing to note is that we cannot use P, because a or a+1 is not always a prime number.

Hint: If $p|a$ and $p|b$ ,then $p|a-b$. Furthermore, if you just change "$a$" to "$a-1$", then $\gcd(a,a+1)=1$ changes to $\gcd(a-1,a)=1$, which is what you wanted. — Sarvesh Ravichandran Iyer, Nov 04 '16 at 00:27
The answer below says precisely what I wanted to say. $p$ was an ordinary number, not a prime, but that's my fault. Of course, GCD does not depend on order? The gcd of $a,b$ is the gcd of $b,a$, because the greatest common divisor is defined in a symmetric manner: it doesn't give precedence to either $a$ or $b$ in the definition. — Sarvesh Ravichandran Iyer, Nov 04 '16 at 00:29
Wouldn't you have to add one to the other side too, to make it = 2?? — knowledge_is_power, Nov 04 '16 at 00:30
Does this answer your question? Prove that for any integer $k \ne 0$, $\gcd(k, k+1) = 1$ — Bill Dubuque, Feb 26 '23 at 00:38

score 3 · Answer 1 · answered Nov 04 '16 at 00:28

3

Let $d$ be the greatest common divisor of $a$ and $a-1$. Notice that it must divide $a-(a-1)$ and so $d$ divides $1$. We are done.

answered Nov 04 '16 at 00:28

Asinomás

105,651

Oh, one thing though, how did you get that it must divide a - (a-1)? – knowledge_is_power Nov 04 '16 at 00:33
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if any number, say $x$ divides another two numbers, say $a$ and $b$ then $x$ must also divide their difference. In other words $x$ must divide $a-b$. – Asinomás Nov 04 '16 at 00:35

Prove GCD(a, a-1)=1

1 Answers1