irreducible polynomial in polynomial ring of 2 variables

Question

So I've been asked to prove that $x^2+y^2$ is irreducible over $\mathbb R[x,y]$, the polynomial ring over $\mathbb R$ in two indeterminates $x$ and $y$. I don't want the solution. I just don't know if I am heading in the right direction. What I've done so far is said suppose $$x^2+y^2=(a_0+a_1x+a_2y+a_3xy)(bo+b_1x+b_2y+b_3xy)$$ and simplifying and getting a system of many equations. All of them (except the two I mention) equal to 0 while I have $a_1b_1=1$ and $a_2b_2=1$. I just dont know where to go from here. I feel like the correct proof is probably something involving many tricks and very contrived. I am in an undergraduate abstract algebra course. Topics we have covered are integral domains, factor rings ahd homomorphisms (including First Isomorphism Theorem), prime and maximal ideals, and introductory theory of UFDs and PIDs, so please dont use anything advanced that would be beyond the scope of my understanding.

Answer 1

Your approach works. As you do not want a solution I will merely point out the steps you can do to get it for yourself:

1. Without loss of generality $b_0 = 0$ (equation from constant coefficient).
2. This implies $a_0 = 0$ (equation from coefficient $x$ and $a_1 b_1 = 1$).
3. Simplify equation from coefficient $xy$ and use $a_1 b_1 = 1$ and $a_2 b_2 = 1$ to get a contradiction.

I hope this is not too much solution for you.

There are of course other possible approaches like using $\mathbb{R}[x,y] \rightarrow \mathbb{R}[x], x \mapsto x, y \mapsto 1$ and the irreducibility of $x^2 +1$ in $\mathbb{R}[x]$ or like embedding $\mathbb{R}[x,y]$ into $\mathbb{C}[x,y]$ and using the way $x^2 + y^2$ decomposes there and the fact that $\mathbb{C}[x,y]$ is a UFD.