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Let $a,b,c$ be integers such that $\gcd(a,b,c)=1$. Does the equation

$ a(x y)+c (z w)+b (z x)=1$

with $ x,y,z,w$ integers always have a solution?.

I know it's a simple question but I haven't gotten anywere. I tried to set any $s , t , r \in Z $ such that $ as+ct+br=1$ and then solve $s=x y,\, t= z w,\, r=z x$. But I realized that sysmtem not always have a solution (e.g. when $s=1 ,t=p, r=p^{2}$ for p a prime ).

Any hint would be appreciated thanks.

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    Are $a,b,c$ pairwise coprime? or would a set like $(6,10,15)$ qualify? – 2017-02-21
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    No, two of them could have common factors but not all three. – 2017-02-21

2 Answers 2

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Let's see if we can make it work with the reasonably hard case of $(a,b,c)=(6,10,15)$.

$\gcd(a,b) = 2$, so Bézout's identity there is $(2+5k)\cdot a -(1+3k)\cdot b = 2$

Then $\gcd(2,c) = 1$, and the identity there is $(8+15m)\cdot 2 -(1+2m) \cdot c =1$

So overall we have $(2+5k)(8+15m)\cdot a - (8+15m)(1+3k)\cdot b-(1+2m) \cdot c = 1$

If we set $k=2, m=3$ we get $12\cdot 53\cdot a - 53\cdot 7 \cdot b -7\cdot c = 1$ and we can choose $(w,x,y,z)=(1,53,12,-7)$

In fact this shows me that there is enough freedom to choose $k$ and $m$ fairly freely. Setting them both to zero gives $2\cdot 8\cdot a - 8\cdot b-1 \cdot c = 1$ and $(w,x,y,z)=(1,8,2,-1)$ also works

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Thanks, the freedom of $k$ and $m$ does indeed solve the problem. Let $d=gcd(a,b)$ by Bézout's identity we find $u_{0},v_{0},s_{0}$ and $t_{0}$ such that

$a\,u_{0}+b \, v_{0}=d$

$d \, s_{0}+c \, t_{0}=1$

By following Joffan reasoning we are led to the equations

$ x\,y=(u_{0}+k \cdot b/d)(s_{0}+m \, c) $

$x\,z =(v_{0}-k \cdot a/d)(s_{0}+m \, c)$

$w \, z=t_{0}-m \,d$

where we may pick $k$ and $m$ as we need.

Taking $x=s_{0}+m \, c \,$,$\,\,y=u_{0}+k \cdot b/d$ and $z=v_{0}-k \cdot a/d $, all we need to show is that exists $k$ and $m$ such that $v_{0}-k \cdot a/d $ divides $t_{0}-m \,d$ and then take $w= (t_{0}-m \,d)/(v_{0}-k \cdot a/d)$

Now since $ v_{0}$ and $a/d$ are relatively prime by Dirichlet theorem on arithmetic progressions $p=v_{0}-k \cdot a/d$ for infinetly many primes $p$ in particular we can choose $p \nmid d$ so that $d$ is invertible mod $p$. Finally we choose any $m$ such that $d \, m \equiv t_{0} \;(\bmod\; p)$ and we are done.