This question came up when I was discussing Ex 1.23 of An Introduction to Analytic Number Theory (Apostol) with a user.
For positive integers $a,b$, when are the values of $a+b$ and $ab$ palindromic?
A simple example to demonstrate this is when $a=3$ and $b=24$. The integers $27$ and $72$ are palindromic. Moreover, the pair $(2,2)$ is trivial.
Is it possible to find all the solutions $(a,b)$?
UPDATE: The case $a=2$ has now been proven by myself and @Holo. The only solutions are indeed $(2, 5\cdot10^k-3)$ for a non-negative integer $k$. The case $a=3$ has also been shown that the only solution is $(3,24)$.
The following is a technical way to approach it for the general case.
In general, if $a+b=c_{m-1}c_{m-2}\cdots c_1c_0$ has $m$ digits and $ab=c_0c_1\cdots c_{m-2}c_{m-1}$ has $n$ digits then $$a+b=\sum_{i=0}^{m-1}10^ic_i\quad\text{and}\quad ab=\sum_{i=0}^{m-1}10^ic_{m-i-1}$$ so we wish to solve the quadratic $$a^2-a\sum_{i=0}^{m-1}10^ic_i+\sum_{i=0}^{m-1}10^ic_{m-i-1}=0$$ or $$2a=\sum_{i=0}^{m-1}10^ic_i\pm\sqrt{\left(\sum_{i=0}^{m-1}10^ic_i\right)^2-4\sum_{i=0}^{m-1}10^ic_{m-i-1}}.$$ What is important about this is that we want $$\left(\sum_{i=0}^{m-1}10^ic_i\right)^2-4\sum_{i=0}^{m-1}10^ic_{m-i-1}=k^2\implies \left(\sum_{i=0}^{m-1}10^ic_i\right)^2-k^2=4\sum_{i=0}^{m-1}10^ic_{m-i-1}$$ for some integer $k$, which kind of resembles this post.
In the paper On finding all positive integers $a,b$ such that $b±a$ and $ab$ are palindromic, written by I and @TheSimpliFire, there is a full solution for (more general) question.
It was solved by restricting $a$ and then doing case by case analysis.
The only solutions are $(2,5\cdot 10^k-3)$, $(3,24)$ and $(9,9)$ for $k=0,1,2,3,\ldots$