19.3.7 * Find positive integers $a, b$ satisfying $a \leqslant 2000 \leqslant b$, such that $2, a(a+1), b(b+1)$ form an arithmetic sequence.
If $2, a(a+1), b(b+1)$ form an arithmetic sequence, then $a(a+1)-2=b(b+1)-a(a+1)$. Multiplying both sides by 4 and rearranging gives $2(2a+1)^2-(2b+1)^2=9$. Considering $\bmod 3$, we find that $A=2a+1, B=2b+1$ are both multiples of 3.
Let $A=3x, B=3y$, then we have the Pell-like equation $2x^2-y^2=1$, so
$$
\begin{array}{c}
x=\frac{1}{2\sqrt{2}}\left((1+\sqrt{2})^{2n+1}-(1-\sqrt{2})^{2n+1}\right)=\left[\frac{(1+\sqrt{2})^{2n+1}}{2\sqrt{2}}\right]+1, \\
y=\frac{1}{2}\left((1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}\right)=\left[\frac{(1+\sqrt{2})^{2n+1}}{2}\right].
\end{array}
$$

where $[k]$ is the floor function. The first few values of $n$ are:
\begin{tabular}{llccccc}
$n: 0$ & 1 & 2 & 3 & 4 & 5 & $\ldots$ \\
$x: 1$ & 5 & 29 & 169 & 985 & 5741 & $\ldots$ \\
$y: 1$ & 7 & 41 & 239 & 1393 & 8119 & $\ldots$ \\
$a: 1$ & 7 & 43 & 253 & 1477 & 8611 & $\ldots$ \\
$b: 1$ & 10 & 61 & 358 & 2089 & 12178 & $\ldots$
\end{tabular}

We can see that when $n=4$, $a=1477, b=2089$ is the solution.