The only positive divisor of both $a$ and $a + 1 $ is $1$

Question

Prove that if $a \in \mathbb Z$ then the only positive divisor of both $a$ and $a + 1$ is $1$.

When I saw this statement I didn't understand it. The only way that I can see it being true is if a is a negative number but since a ∈ Z a could also be positive and thus it could have more than one positive divisor. If that's the case would I have to disprove this instead of proving? And what would be the best way to do that?

Answer 1

You seem to have misunderstood the question. As far as I can tell you are thinking about divisors of $a$ and $a+1$ separately, but the question is asking you to show that the only positive number which is a divisor of both $a$ and $a+1$ simultaneously is $1$.

This statement is true and a hint for the proof is: any number which is a divisor of two integers (simultaneously) is also a divisor of their difference.

Answer 2

If $n$ divides both $a$ and $a+1$, it would also divide $a+1-a$.

Answer 3

We want to show that if $a$ is an integer, then the only positive number that can divide both $a$ and $a + 1$ is $1$.

Let us assume that there exists a positive integer $d$ (other than $1$) such that $d$ divides both $a$ and $a + 1$. This means that there exist integers $k$ and $m$ such that $a = dk$ and $a + 1 = dm$.

Subtracting the first equation from the second one, we have that $(a + 1) - a = dm - dk \Rightarrow 1 = d(m - k)$.

This implies that d is a divisor of $1$. The only positive divisors of $1$ are $1$ and $-1$, but since we are looking for a positive divisor, $d$ must be $1$.

Therefore, we have shown that if $a$ is an integer, then the only positive number that can divide both $a$ and $a + 1$ is $1$. This completes the proof.