Problem related to GCD I was solving a question on GCD.  The question was calculate to the value of $$\gcd(n,m)$$
where $$n = a+b$$$$m = (a+b)^2 - 2^k(ab)$$
$$\gcd(a,b)=1$$ 
Till now I have solved that when  $n$ is odd, the $\gcd(n,m)=1$.
So I would like to get a hint or direction to proceed for the case when $n$ is even.

Key idea:  $ $ employ $\bigg\lbrace\begin{eqnarray}\rm Euclidean\ Algorithm\ \color{#f0f}{(EA)}\!: &&\rm\ (a\!+\!b,x) = (a\!+\!b,\,x\ \,mod\,\ a\!+\!b)\\ \rm and\ \ Euclid's\ Lemma\ \color{blue}{(EL)}\!: &&\rm\ (a,\,b\,x)\ =\ (a,x)\ \ \,if\,\ \ (a,b)=1\end{eqnarray}$
$\begin{eqnarray}\rm So\ \ f \in \Bbb Z[x,y]\Rightarrow &&\rm (a\!+\!b,\, f(\color{#c00}a,b))\stackrel{\color{#f0f}{(EA)}} = (a\!+\!b,\,f(\color{#c00}{-b},b)),\ \ by\, \ \ \color{#c00}{a\equiv -b}\!\!\pmod{a\!+\!b}\\
\rm \Rightarrow &&\rm(a\!+\!b\!,\, (\color{#0a0}{a\!+\!b})^2\! \color{#c00}{- a}bc) = (a\!+\!b\!,{\color{#0a0}0}^2\!+\!\color{#c00}bbc)\!\stackrel{\color{blue}{(EL)}}= \!(a\!+\!b,c)\ \ by\ \, \bigg\lbrace\begin{array}((a\!+\!b,b)\\\rm\, = (a,b)=1\end{array} \end{eqnarray}$
