A4. How many positive integers $N$ of two digits are there with the following property: The sum of $N$ and the reverse of $N$ is a square of an integer (the reverse of $N$ is the number you get by writing $N$ backwards)?
A4 If you write the number $N$ as $a b$ then $N=10 a+b$. The reverse of $N$ is then equal to $10 b+a$. Together this gives $11 a+11 b=11(a+b)$. Since the latter must be a square, it must hold that $(a+b)$ is equal to 11 or a square times 11. $(a+b)=11$ gives the solutions $29,38,47,56,65,74,83$ and 92. If $(a+b)$ must be a square times 11, then it is at least 44 and that is not possible because $a$ and $b$ must be less than 10. Therefore, 8 solutions.