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Let $a,b \in \mathbb{Z}^+ $. If $b\mid a$ and $b\mid (a+2)$ , prove that $b=1$ or $b=2$. Now, we have $a = b\,n$ and $a+2 = b\,m$ for some $m$ and $n$ in $\mathbb{Z}$. Suppose that $b \ne 1$. So, we have $2 = (m-n) b$. This means that $b\mid 2$. Since $b \in \mathbb{Z}^+ $ and $b \ne 1$, we have $b \geqslant 2$. So, $b \geqslant (m-n) b$. Since $b > 0$, we get $(m-n) \leqslant 1$. Now, since $2 = (m-n) b$, we have $(m-n) > 0 $ and since $(m-n) \in \mathbb{Z}$, we get $ (m-n) \geqslant 1$. Together with $ (m-n) \leqslant 1$, this means that $(m-n) = 1$. So, we have $b = 2$.

Is this a good proof ?

Thanks

Bernard
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user9026
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    Correct, but quite complicated. Once you have $b|2$, it is immediate that the only possibilities are $1$ and $2$. – TheSilverDoe Sep 01 '20 at 10:28
  • @thesilverdoe It was not apparent to me. What theorem would we use from that point on ? – user9026 Sep 01 '20 at 10:29
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    You could just write $2=bc$ for an integer $c$, so $b = 2/c$ and because $c \geq 1$, you get $b \leq 2$. Generally, if $p|q$ then you must have $p\leq q$... – TheSilverDoe Sep 01 '20 at 10:29
  • @user9026, May I ask why is $\ b|2$ not sufficient to say that $\ b=1,2$? You have unecessarily written a long proof. And for your main question, it is not a good proof. – Vishnu Kadiri Sep 01 '20 at 10:33
  • Vishnu, in proofs, we justify using known theorems. So, I wanted to as precise as possible – user9026 Sep 01 '20 at 10:35
  • It would be foolish to justify that if $\ b|2$ and it is a positive integer, then $\ b=1, 2$ using known theorems instead of known facts. – Vishnu Kadiri Sep 01 '20 at 10:37
  • @vishnu-kadiri Yes I get it now – user9026 Sep 01 '20 at 10:43

2 Answers2

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I like the start up to $ b \mid 2$.

The following steps can be simplified. Either:

  • An integer $m$ that divides another one $n$ is smaller or equal to $n$. Therefore $b \in \{1,2\}$.
  • Or the only natural integers that divide $2$ are $1$ and $2$.
mathcounterexamples.net
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Your proof is fine but it's a bit complicated. It can be done in a more easy way.

Just use the fact that if $b|a$ and $b|c$, then $b|(a - c). $The same thing goes here.

$b|(a + 2)$ and $b|a$ implies :-

$\rightarrow b|(a + 2) - a$

$\rightarrow b|2$

Since $a,b$ are positive integers the only possible choices for $b$ are $1,2$ .

Anonymous
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