# Putnam 1985 B-1 Polynomial Problem

Problem: Let $k$ be the smallest positive integer for which there exist distinct integers $m_1, m_2, m_3, m_4, m_5$ such that the polynomial $$p(x)=(x-m_1)(x-m_2)(x-m_3)(x-m_4)(x-m_5)$$ has exactly $k$ nonzero coefficients. Find, with proof, a set of integers $m_1, m_2,m_3,m_4,m_5$ for which this minimum $k$ is achieved.

My attempt: Let $\quad \!\!i,j,k,l, \in\{1,2,3,4,5\}$ $$p(x)=x^5-\sum_im_ix^4+\sum_{i\neq j} m_i\quad\!\!\!\!m_j \quad\!\!\!\!x^3-\sum_{i\neq j\neq k}m_i\quad\!\!\!\!m_j \quad\!\!\!\!m_k\quad\!\!\!\!x^2$$ $$+\sum_{i\neq j \neq k \neq l} (m_i\quad\!\!\!\!m_j \quad\!\!\!\!m_k\quad\!\!\!\!m_l \quad\!\!\!\!x)-m_1m_2m_3m_4m_5$$

(Is there a simpler way to write this out using permutations/sigma notation?)

Regardless of the selection of the $m_i$, there is at least $1$ nonzero coefficient. If we want to argue that the minimum number of nonzero coefficients is precisely $1$, we note that all other coefficients than the first term $x^5$ has to be 0. This leads to the following system of equations.

\begin{align} m_1+m_2+m_3+m_4+m_5&=0\\ m_1m_2+\cdot\cdot\cdot+m_4m_5&=0\\ m_1m_2m_3+\cdot\cdot\cdot+m_3m_4m_5&=0\\ m_1m_2m_3m_4+\cdot\cdot\cdot+m_2m_3m_4m_5&=0\\ m_1m_2m_3m_4m_5&=0 \end{align}

Since the last coefficient has to equal $0$, the product $m_1m_2m_3m_4m_5=0$. So, $\exists \quad\!\!\!\!a\in\{1,2,3,4,5\}$ such that $m_a=0$.

But, in the $4th$ equation of the system, all terms including $m_a$ would be $0$ and the $4$-cycle not including $m_a$ would remain and equal $0$. So, this implies that $\exists \quad\!\!\!\! b \neq a\in\{1,2,3,4,5\}$ such that $m_b=0$. However, the roots of the polynomial have to be unique and so the coefficient of $x$ is nonzero, which contradicts our assumption.

Thus, we now proceed to showing that the number of nonzero coefficients is exactly 2, but I am not sure how to show this.

• Welcome to math.stackexchange! And thank you for showing your work in such a clear way. The case $x^5$ can be ruled out much more easily, since polynomials have unique factorizations and any solution of the problem must lead to a polynomial with 5 different linear factors. Try the case $(x-m_1)(x-m_2)(x-m_3)$ first. What is required to make the constant term disappear? What is required to make also the $x^2$ term disappear? Can that be extended to the problem at hand? – Hans Engler May 14 '16 at 4:30
• For the constant term $m_1m_2m_3$ to be $0$, $0$ has to be a root of the polynomial. For the coefficient of $x^2$ to be $0$, the sum of the roots have to be $0$. For these two cases to be simultaneously true, we have a root of the polynomial equal to zero, an arbitrary root and the opposite of the arbitrary root. But since the $x$ term has the coefficient $m_1m_2+m_1m_3+m_2m_3$, it has to be a nonzero coefficient. – Richard May 14 '16 at 4:54
• For $k=2$, you can just write the polynomial and show that its number of distinct roots cannot be $5$ by considering all the cases. – Alistair May 14 '16 at 6:11
• @Rebecca - so now you know how to make the even powered terms disappear for a cubic. Can you do that for a quintic polynomial? – Hans Engler May 14 '16 at 14:18

Clearly $k > 1$; otherwise $p(x) = x^5$ and $m_1, \ldots, m_5$ are not distinct. Assume $k = 2$, so $p(x) = x^5 + ax^j$ with $0 \le j \le 4$. We can not have $j \ge 2$ since then at least two of the $m_l$'s are equal to $0$. Hence $p(x) = x^5 + a$ or $p(x) = x(x^4 + a)$ with $a \neq 0$. But $x^5 + a$ and $x^4 + a$ have at most two real zeros. Therefore $k \ge 3$.
Set $m_1 = -2$, $m_2 = -1$, $m_3 = 0$, $m_4 = 1$, $m_5 = 2$. Then$$p(x) = x(x^2 - 1)(x^2 - 4) = x^5 - 5x^3 + 4x.$$Hence $k = 3$, and this value of $k$ is achieved for the given $m_l$'s.