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 \(\dbinom{u}{v}=0\). 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 \(\dbinom{u}{v}\) given by \[\binom{u}{v}=\dfrac{u(u-1)(u-2)\cdots(u-v+1)}{v!}\] 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 \(2n+1\) is a factor of \(p_k(n)\) for even \(k\), consider the values of \(p_k(-n)\) when \(n\) is a positive integer.