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Let $a$ be an integer. After looking at several examples, make a conjecture about the value of $\operatorname{gcd}(a-1,a+1)$ and prove it.

Ok. I found that:

  • if $a$ is even, $\operatorname{gcd}(a-1,a+1)=1$;
  • if $a$ is odd, $\operatorname{gcd}(a-1,a+1)=2$.

Is this conjecture right? Any advice to how to prove this conjecture?

Frenzy Li
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Wes
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3 Answers3

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Correct. Basically, just use the fact that $x-y$ is a multiple of $\gcd(x,y)$, i.e. $\gcd(x,y)|x-y$.

In both cases, $x-y=2$, so the gcd can only be 1 or 2.

user60177
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    Can I substitute like, $gcd(a-1,a+1)|(a-1)-(a+1)$ ? – Wes Oct 03 '13 at 01:48
  • Yes, definitely! – user60177 Oct 03 '13 at 02:02
  • Ok, so if $a$ is odd,$2n+1$, then $gcd((a-1,a+1)=2$. Let $a=2n+1$, want to show $gcd(a-1,a+1)|(a-1)-(a+1)$. In the other words,$gcd(a-1,a+1)|-2$ ? I don't know. – Wes Oct 03 '13 at 02:21
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    Acutally gcd(a−1,a+1)|−2 is the same as gcd(a-1, a+1)|2. So gcd(a-1,a+1) = 1 or 2. If a is odd, then a-1 and a+1 are even, so 2|gcd(a-1,a+1), so gcd(a-1,a+1)=2. If a is even, then a-1 and a+1 are odd, so the gcd cannot be 2, so it must be 1. – user60177 Oct 03 '13 at 02:56
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First, we are going to denote $p$ as the $gcd(a-1, a+1)$

This implies $$a-1 = pk, \ a+1 = pk'$$ with $k' > k$

Moreover $a+1 = (a-1)+2$. Thus $pk + 2 = pk' \Rightarrow p(k'-k) = 2$ with $k'-k \geq 1$ that implies that $0<p\leq 2$

Now you can see $p$ is 1 if $a$ is even and 2 if $a$ is odd

Rubén Ballester
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If a number, in this case GCD(a-1,a+1), divides two numbers, a-1 and a+1, it divides their difference, namely 2. If a is odd, both numbers are even, so the answer is 2. If a is even, a+1 is odd, so the answer is 1.

Airymouse
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