Problem of the Month
Hint for Problem 4: January 2023

What are the possible values of $c$ ?
How many distinct integers can occur in a triple in $S$ ?
Try to generalize the idea in part (c). The constants $a_{2}$ through $a_{k + 1}$ do not depend on $n$ .
For positive integers $u$ and $v$ with $u < v$ , the usual convention is that $(\binom{u}{v}) = 0$ . This convention makes sense for (at least) two reasons. First, there are zero ways to choose $v$ objects from $u$ distinct objects if $u < v$ , so " $u$ choose $v$ " should be equal to $0$ . Second, the formula for $(\binom{u}{v})$ given by $(\binom{u}{v}) = \frac{u (u - 1) (u - 2) \dots (u - v + 1)}{v!}$ will have a factor of $0$ in the numerator if $u < v$ .
Directly compute an expression for $p_{5} (n) - p_{5} (n - 1)$ . It should be a polynomial with coefficients depending on $a_{1}$ through $a_{6}$ . By equating coefficients with the polynomial $n^{5}$ , solve for $a_{1}$ through $a_{6}$ . After these coefficients are known, $a_{0}$ can be computed from $p_{5} (1) = 1$ .
A polynomial with infinitely many roots must be the constant zero polynomial. Using this fact, show that $p_{k} (n) - p_{k} (n - 1) = n^{k}$ for all real numbers, not just positive integers. This means you need to "extend" $p_{k} (n)$ to accept inputs that are not positive integers. Once this is done, determine the values of $p_{k} (0)$ and $p_{k} (- 1)$ . To show that $2 n + 1$ is a factor of $p_{k} (n)$ for even $k$ , consider the values of $p_{k} (- n)$ when $n$ is a positive integer.