Question

Let $a$ and $b$ be two integers. If $a|b$ and $b|a$, them show that $a=\pm b$

Answer 1

If $a|b$, then $b = ka$ for some integer $k$ . If $b|a$ , then $a = lb$ for some integer $l$ . Hence $b = ka=klb$, then we obtain $b-klb=0$. This implies $b(1-kl)=0$ so either $b = 0$ or $kl=1$. If $b = 0$ , then $a = 0$ and hence $a=\pm b$ and we are done.

If $kl=1$ , then either $k = −1$ and $l = −1$ or $k=1$ and $l=1$. In the first case $b=-a$, in the second case $b=a$.

Hence $b=\pm a$.