[Algebra]
Problem: A two-digit number has the property that the square of its tens digit plus ten times its units digit equals the square of its units digit plus ten times its tens digit. Determine all two-digit numbers which have this property, and are prime numbers.

Solution: Let the two digit integer have tens digit $a$ and units digit $b$. Then the given information tells us

$$
\begin{aligned}
a^{2}+10 b & =b^{2}+10 a \\
a^{2}-b^{2}-10 a+10 b & =0 \\
(a+b)(a-b)-10(a-b) & =0 \\
(a-b)(a+b-10) & =0
\end{aligned}
$$

and so $a=b$ or $a+b=10$.

So the possibilities for the integer are 11, 22, 33, 44, 55, 66, 77, 88, 99, 19, 28, 37, 46, 55, $64,73,82,91$. We now must determine which integers in this list are prime.

We can quickly reject all multiples of 11 bigger than 11 and all of the even integers, to reduce the list to $11,19,37,73,91$.

All of these are prime, except for $91=13 \times 7$.

Therefore, the required integers are 11, 19, 37, and 73 .

Final answer: 11,19,37,73