$\left|m^2 -\frac{n\cdot(n+1)}{2}\right|=1$ and A001110 The pairs of squares and triangular numbers differing by $1$ are so far: $(0,1)$, $(3,4)$, $(9,10)$, $(15,16)$, $(120,121)$, $(528,529)$.  Do you think  there are more terms than are found in A001110?  If so, could there be a reason for this?
 A: Let's start with an existing solutuion $(m,n)$:
$$\left|m^2 -\frac{n\cdot(n+1)}{2}\right|=1 \tag{1}$$
Then
$$\left|(m+1)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)\right|=1$$
$$\left|(m+2)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)-(2m+3)\right|=1$$
$$\left|(m+3)^2 -\frac{n\cdot(n+1)}{2}-(2m+1)-(2m+3)-(2m+5)\right|=1$$
$$... \text{by induction}$$
$$\left|(m+k)^2 -\frac{n\cdot(n+1)}{2}-(2km+k^2)\right|=1$$
which is
$$\left|(m+k)^2 -(1+2+...+n+2km+k^2)\right|=1$$
but we want it to be
$$\left|(m+k)^2 -(1+2+...+n+(n+1)+...+(n+p))\right|=1 \tag{2}$$
leading to
$$2km+k^2=k(2m+k)=\frac{p(2n+p+1)}{2} \iff 2k(2m+k)=p(2n+p+1)$$
or, we could ask for
$$\left\{\begin{matrix}
2m+k=p\\ 
2k=2n+p+1
\end{matrix}\right. \Rightarrow 2k=2n+2m+k+1$$
leading to
$$\left\{\begin{matrix}
k=2m+2n+1\\ 
p=4m+2n+1
\end{matrix}\right. \tag{3}$$
Jumping back to $(2)$ which is
$$\left|(m+k)^2 -\frac{(n+p)(n+p+1)}{2}\right|=1 \tag{4}$$
and using $(3)$ we have
$$\left|(3m+2n+1)^2 -\frac{(4m+3n+1)(4m+3n+2)}{2}\right|=1 \tag{5}$$
which is easy to expand to check that it reduces to $(1)$.
To conclude:

The mapping/recurrence $$(m,n) \mapsto (3m+2n+1, 4m+3n+1)$$ leads to infinity of pairs satisfying $(1)$


Example:
$$(0,1)\mapsto (3,4)\mapsto (18,25)\mapsto (105,148)\mapsto ...$$
or, in style of the question, using $m \mapsto m^2$ and $n \mapsto \frac{n(n+1)}{2}$:
$$(0,1)\mapsto (9,10)\mapsto (324,325)\mapsto (11025,11026)\mapsto ...$$
