32. Determine the odd prime number $p$ such that the sum of digits of the number $p^{4}-5 p^{2}+13$ is the smallest possible.
32. Answer: 5
Let $n=p^{4}-5 p^{2}+13$. When $p=3$, we have $n=49$ and so the sum of digits is 13 . When $p=5$, we have $n=513$ and the resulting sum of digits is 9 .
Now let $p>5$ be a prime. We have
$$
n=p^{4}-5 p^{2}+13=(p-2)(p-1)(p+1)(p+2)+9 .
$$

Since $p-2, p-1, p, p+1$ and $p+2$ are five consecutive integers, at least one of them is divisible by 5 . Since $p \neq 5$, we must have 5 divides one of the numbers $p-2, p-1, p+1$ and $p+2$, so 5 divides the product $(p-2)(p-1)(p+1)(p+2)$. Observe that at least one of the numbers $p+1$ and $p+2$ is even. Therefore we see that $(p-2)(p-1)(p+1)(p+2)$ is divisible by 10 . It follows that for any.prime $p>5$, the number $n$ has at least two digits and the units digit is 9 , so the sum of the digits is greater than 9 . Consequently the smallest possible sum of digits is 9 , and this value is attained when $p=5$.