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If $(a,n)=(b,n)=1$ and $(c,n)=d$,$d|ab$ then is $d=1$ is only solution ? $\forall a,b,c \in \mathbb{N}$

Also $(a,b)$ denotes $gcd(a,b)$ My approach was like say if $n=2k$ then $(a,2k)=(b,2k)=1$ So $a,b$ has to be odd say $2k_{1}+1,2k_{2}+1$ then also $d=(c,2k)$

But i stucked,and i dont think it is the correct way to proceed like this.

Random values of $a,b,c$ always show $d=1$ ,but how to prove it ?

M Desmond
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4 Answers4

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If $d\mid ab$, then $(d,n)=1$ (any prime factor that $n$ has in common with $d$, it must also have in common with either $a$ or $b$) and since $d\mid n$, the only possibility is $d=1$.

Arthur
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Yes, $d = 1$ is the only solution. If $d > 1$, then $d$ in particular contains a prime factor $p > 1$, and $p | d | n$. Since $p | ab$, either $a$ or $b$ has $p$ as a factor, which is impossible, since $p | n$, meaning gcd($a,n$) or gcd($b,n$) is greater than 1.

Richard Jensen
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A variant by Bezout: $\begin{cases}au+nv=1\\bw+nx=1\\cy+nz=d\end{cases}$

$d\mid ab\implies ab=\alpha d$

Thus multiplying the third line by $(uw\alpha)$ we get

$cyuw\alpha+nzuw\alpha=duw\alpha=(au)(bw)=(1-nv)(1-nx)\iff \\c(yuw\alpha)+n(zuw\alpha+v+x-vxn)=1$

So $(c,n)=1$ and $d=1$.

zwim
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  • Simpler to use the GCD Universal Property (which can be proved by Bezout), e.g. see my answer. This nicely encapsulates the Bezout property so we don't need to work with crufty Bezout coef's every time we invoke such gcd laws, and it has the benefit of being more general - working in rings like $, \Bbb Z[x],$ and $, \Bbb Q[x,y],$ where Bezout fails. – Bill Dubuque Sep 13 '19 at 16:41
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$(c,n)\mid ab,n\,\Rightarrow (c,n)\mid (ab,n)\color{#c00}{=1},\ $ by $\ (a,n)\! \color{#c00}{=\! 1}\! =\! (b,n)\ $ and Euclid

Bill Dubuque
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