If $gcd(m,n) = gcd(m,k) = gcd(n,l) =1$ show that $gcd(kn+lm, mn) = 1$ I started by showing that $gcd(kn, m) = gcd(lm, n) = 1$, and with Bezout's lemma I wrote $knx + my = 1$ and $lmx' + ny' = 1$. Then I solved for my and $ny'$ and multiplied them together to get:
$$
mnyy' = (knx - 1)(lmx' - 1) = klmnxx' - knx - lmx' + 1.
$$
Rearranging gives:
$$
mn(yy' - klxx') + knx + lmx' = 1.
$$
So it appears all that is left to do is to show that $x = x'$ I think, however I am lost on how to do this. Any help is welcome!
 A: $\begin{align} &(kn+lm,\color{#c00}m) = (kn,m)=1\\ &(kn+lm,\color{#0a0}n)\, = (lm,n)\, = 1\end{align}$ $\,\Rightarrow\, (kn+lm,\color{#c00}m\color{#0a0}n) = 1\ $ by Euclid.
Remark $ $ If you don't know that form of Euclid we can prove it directly
$(a,m)(a,n) = (aa,am,an,mn) = (a(a,m,n),mn) = (a,mn)\ $ by $\ (a,m,n) = 1$
You could also use Bezout above instead of gcd laws (distributive, commutative, associative), e.g. see the comparison here.  But that yields a less general proof.
A: With pure Bezout, it can be done, but it's quite a gymnastic...
$\begin{cases}
mA+kB=1 & \times\ nE & \implies mnAE+knBE=nE=1-mF \\
                    && \implies m(nAE+F)+kn(BE)=1 \\
\\  
nC+lD=1 & \times\ mF & \implies mnCF+lmDF=mF=1-nE \\
                    && \implies n(mCF+E)+lm(DF)=1 \\  
nE+mF=1\end{cases}$

By adding $0$ 
$\begin{cases} 
0=lm(BE)-m(lBE) & \text{we get} & (kn+lm)(BE)+m(nAE+F-lBE)=1 \\
0=kn(DF)-n(kDF) & \text{we get} & (kn+lm)(DF)+n(mCF+E-kDF)=1 \end{cases}$

Now we multiply these lines together:
$\begin{cases}(kn+lm)W+mX=1\\(kn+lm)Y+nZ=1\end{cases}\implies (kn+lm)(WY+nZW+mXY)+mn(XZ)=1$
