$\left|m^2 -\frac{n\cdot(n+1)}{2}\right|=1$ and A001110

Question

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?

Answer 1

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 $(1)$

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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 $n \mapsto \frac{n(n+1)}{2}$: $$(0,1)\mapsto (9,10)\mapsto (324,325)\mapsto (11025,11026)\mapsto ...$$