Show $f=f_1+f_2$ is irreducible in $K[x,y]$ 
Let $K$ be a field and $f_1,f_2\in K[x,y]$ two homogeneous polynomials of degrees $d\ge 1$ and $d+1$, respectively. Suppose that $f_1,f_2$ don't share a common factor. Show that the polynomial $f:=f_1+f_2$ is irreducible in $K[x,y]$.

I proceeded by reduction to absurdity, so I assumed that $f$ is reducible in $K[x,y]$ thus there exist polynomials $g$ and $h$ such that $f=gh$. Now, since every polynomial can be decomposed uniquely as the sum of its homogeneous components, we can write 
$$g=\sum_{i=k}^lg_i$$
and
$$h=\sum_{j=r}^sh_j$$
where each $g_i$ and each $h_j$ is an homogeneous polynomial of $K[x,y]$ of degree $i$, $j$ respectively, or a null polynomial.
Now, it seems clear that the homogeneous component of $f$ of least degree should be $g_kh_r$, and the one with most degree sould be $g_lh_s$, since the product of homogeneous polynomials with degrees $e,d$ is an homogeneous polynomial of degree $e+d$. Now it seems natural to conclude that, since $f=gk=f_1+f_2$, then
$$f_1=g_kh_r$$ and $$f_2=g_lh_s$$
because the homogeneous components of a polynomial $f$ are unique, and $f$ has only two of those components, one with the highest degree, and other one with the lowest. And here (if I've not made any mistakes yet) is where I'm stuck. I should get something that would led me to conclude that $k=l$ or $r=s$, because then $g_k$ or $h_r$ would be a common factor of both $f_1$ and $f_2$, and the we would get the contradiction that $f_1$ and $f_2$ are not each other primes.
The only hypothesis I've not used yet is that $deg(f_1)=d\ge 1$ and $deg(f_2)=d+1$, so I guess there's something that right now I can't see. Any help would be appreciate, and thanks in advance!
 A: Suppose $f = a b$, with $a, b$ both of positive degree. Write $a$ and $b$ as sums of homogeneous polynomials $a_{i}, b_{i}$ of degree $i$.
If $l$ is the smallest number such that $a_{l} \ne 0$, then $d- l$ must be the smallest number such that $b_{d-l} \ne 0$. 
Then $a = a_{l} + a_{l+1}$ and $b = b_{d-l} + b_{d-l+1}$. 
If $l = 0$ (or, symmetrically, $d - l = 0$), then $a_{1} \ne 0$. We get that $a b$ has degree $d+2$, unless $b_{d+1} = 0$, but then $f_{1}$ and $f_{2}$ have the common factor $b_{d}$.
So let $0 < l < d$. If one of $a_{l}, b_{d-l+1}$ are non-zero, we obtain a common factor for $f_{1}, f_{2}$. If none of them is zero, we get that $a b$ has degree $d+2$, a contradiction. 
A: Write (as you did) $$f_1+f_2=(g_k+\cdots+g_l)(h_r+\cdots+h_s)$$ with $g_k\ne0$, $g_l\ne0$, $h_r\ne0$, $h_s\ne 0$. (Here $g_i,h_j$ are homogeneous of degree $i,j$, respectively.) Then $k+r=d$. If $l\ge k+1$ and $s\ge r+1$ (that is, there are at least two terms in each sum) then $l+s\ge d+2$ and hence we must have $g_lh_s=0$, a contradiction. Thus $l=k$ or $s=r$, and in each case the two polynomials have a common factor, a contradiction.
