toseeifitworksforream

2022-04-04

Let a and b be coprime integers, and let m be an integer such that a | m and b | m. Prove that ab | m

alenahelenash

To prove that $ab\mid m$, we need to show that $m$ is a multiple of $ab$, which means that $m=abn$ for some integer $n$.
Since $a\mid m$, we can write $m=ak$ for some integer $k$. Similarly, since $b\mid m$, we can write $m=bl$ for some integer $l$.
Since $a$ and $b$ are coprime, we have $gcd\left(a,b\right)=1$. Therefore, there exist integers $x$ and $y$ such that $ax+by=1$ (this is the Bezout's identity).
Multiplying both sides by $m$, we get $\left(ax+by\right)m=m$. Since $m=ak=bl$, we can substitute these values to get $\left(ax+by\right)ak=\left(ax+by\right)bl$.
Expanding both sides, we get $axak+byak=axbl+bybl$, which simplifies to ${a}^{2}xk+abyk=abxl+{b}^{2}yl$.
Since $a\mid m$ and $b\mid m$, we know that $m$ is a common multiple of $a$ and $b$. Therefore, $ab$ is a common multiple of $a$ and $b$.
Using the definition of divisibility, we can say that $ab\mid {a}^{2}$ and $ab\mid {b}^{2}$. Hence, $ab$ divides the left-hand side of the above equation.
Since $ab\mid {a}^{2}xk$, $ab\mid abyk$, and $ab\mid abxl$, we can conclude that $ab\mid {b}^{2}yl$.
Since $a$ and $b$ are coprime, $b$ does not divide $a$. Therefore, $b$ must divide $yl$, which means that there exists an integer $n$ such that $yl=bn$.
Substituting this value, we get $ab\mid {b}^{2}bn$, which simplifies to $abn=m$.
Hence, we have shown that if $a$ and $b$ are coprime integers, and if $m$ is an integer such that $a\mid m$ and $b\mid m$, then $ab\mid m$.

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