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.
 A: Left side looks almost like $(2b+1)^2$, indeed adding $1$ we can write $(2b+1)^2=a^2+a+1$. It is known that right side can be a perfect square only in few cases - see Integral value of $n$ that makes $n^2+n+1$ a perfect square. 
Specifically, when $a>0$, the right side lies between two consecutive squares: $a^2<a^2+a+1<(a+1)^2,$
so it cannot be itself a square. Similarly for $a<-1$, we have $(a+1)^2< a^2+a+1 < a^2$, same reasoning applies. So it only remains to check $a=-1$.
A: 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...
A: A varied version of @Oscar Lanzi's answer:
Rewrite the equation
$$a^2+a-(4b^2+4b)=0$$
$$\Delta=1+4(4b^2+4b)=(4b+2)^2-3\overset{def}{\equiv} K^2-3$$
If $a$ is an integer, then there exists some integer $M$ such that $\Delta=M^2$.
$$K^2-3=M^2\iff K^2-M^2=3$$ 
So there will only be a set of $(K,M)$ or $(a,b)$. 
A: Multiply by $4$ and complete the square on both sides.  This gives
$(4b+2)^2-4=(2a+1)^2-1$
$(4b+2)^2-(2a+1)^2=3$
What are the only two squares differing by exactly $3$?
