This problem is Exercise 11.3 in Atiyah/Macdonald Commutative Algebra. They ask to prove every finitely generated ideal in a Boolean ring is in fact a principal ideal.
The question has been answered already on StackExchange: note that $(x,y) = (x+y+xy)$ and then use induction. However, is there any clear motivation for using $x+y+xy$ as the generator of this ideal? (apart from pulling it out of a hat and noticing it has nice properties)
Best Answer
It's actually pretty nice to reason through! This is all possible thanks to two lattice theoretic properties of Boolean rings. It's good to know about these two properties, and it turns out they're pretty instructive in this proof.
Ingredients
One thing we need is the involutory map $(x)\mapsto (1-x)$ establishing an inclusion reversing bijection on the principal ideals of the Boolean ring $R$, which I will call $L$. So $L((a)):=(1-a)$, and $(a)\subseteq (b)$ implies $L((b))\subseteq L((a))$. (Hint: if $(a)\subseteq(b)$, take $a=br$ and do something with $1-b=(1-b)(1-a)+(1-b)a$.)
The other thing afforded us in Boolean rings that we need is that $(a)\cap (b)=(ab)$. The $\supseteq$ is obvious, and I leave the $\subseteq$ as an exercise.
Using these two things, we can cook up the formula $a+b+ab$.
Searching for $(a,b)$
Let's seek the least upper bound of $(a)$ and $(b)$. A natural candidate is to find the greatest lower bound of $L((a))$ and $L((b))$, and then map it back above $(a)$ and $(b)$ using the inclusion reversing map.
Explicitly, since we know the greatest lower bound of $(1-a)$ and $(1-b)$ is $((1-a)(1-b))$ we have $(1-a)\supseteq ((1-a)(1-b))$, and after applying $L$ we have $(a)\subseteq (1-(1-a)(1-b))$, and similarly $(b)\subseteq (1-(1-a)(1-b))$.
But look at $1-(1-a)(1-b)=a+b-ab$. We've just shown that $(a+b-ab)$ is an upper bound of $(a)$ and $(b)$. Finally, it's obviously contained in $(a,b)$, so it is actually equal.
Here's a picture of that proof:
Remember that all of the subtraction signs may as well be addition because the characteristic is 2.
Summary
So in hindsight, the "mysterious" character $a+b+ab$ is just a product of the interplay of this involution $a\mapsto 1-a$ with the way principal ideals intersect. We can retrieve it via a Galois connection on the principal ideals of the ring.