<s>[INST] 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.
 [/INST] 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 $(a,b)$. 
</s>