Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is

Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.

I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.

Hmmm...

Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $x, y$ are zero.

Where do $x$ and $y$ come from? Is the following what was meant?

Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $a, b$ are zero.

I'll proceed along the lines of the latter. Clearly $a$ must be even, because if it's odd then $a^2 + a$ is singly even but $4b^2 + 4b$ is clearly doubly even. Therefore, with $a$ even, it follows that $$\frac{a^2 + a}{4}$$ is an integer, and so the equation can be rephrased as $$\frac{a^2 + a}{4} = b^2 + b.$$ Therefore $\sqrt{a^2 + a} = 2 \sqrt{b^2 + b}$. Trouble is, this requires both $\sqrt{a^2 + a}$ and $\sqrt{b^2 + b}$ to be integers.
Okay, I'm sorry, this is looking to be much more complicated than what you have...
