Putnam 1985 B-1 
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 the minimum $k$ is achieved.

I expect this polynomial to be in the form:
$$ax^5+bx^4+cx^3+dx^2+ex+f$$
$$a,b,c,d,e,f\in\mathbb{Z} $$
So at most $k$ has $6$ values so:
$$1\le k\le6 $$
Since I want the lowest positive integer for $k$. I start with $k=1$.
But if $k=1$ then $p(x)$ should only be equal to $p(x)=x^5$. But because $m_1$, $m_2$, $m_3$, $m_4$, $m_5$ are distinct it will not lead to a situation where the rest of the coefficients are zeros.
Basically what I want to say is if:  
$$k=1$$ $$p(x) = x^5 +0x^4+0x^3+0x^2+0x+0 $$
This means that
$$p(x)=(x−0)(x−0)(x−0)(x−0)(x−0)$$
Meaning all $m_1$, $m_2$, $m_3$, $m_4$, $m_5$ are equal to zero.
But since the roots are distinct $m_i$ cannot have more than value that is zero
Thus $k\neq1$
Next I try $k=2$. If $k=2$ then $p(x)= x^5+ax^j$ where $0\le j\le 4$.
When I choose $j$ from $0$ to $4$ and try to find the roots they are either non distint or complex numbers
I need help for $k=3$.
 A: For $k=3$ you can get an example.
Indeed
$$(x-1)(x+1)(x-2)(x+2)(x-0)=(x^2-1)(x^2-4)x=x^5-5x^3+4x$$
A: If $k=1,$ then all the zeros are coincides, which is not the case we are looking at. Considering $P(x+m_5),$ we can assume that $m_5=0.$ Also by allowing rational coefficients we can assume that $m_4=1.$
Now lets show that $k\neq2.$
If this is the case we have a rational solution set for this system of linear equations.
A: Too long for a comment, so I'll make it an answer
Starting with you polynomial
$$x^5+bx^4+cx^3+dx^2+ex+f = (x-m_1)(x-m_2)(x-m_3)(x-m_4)(x-m_5)$$
For $k=2$, you could simply show that $e$ and $f$ can't be zero at the same time ($m_i$ being different)
If $f=0$, then $m_i=0$ for some $i$. (Since $f = -m_1\times m_2 \times m_3 \times m_4 \times m_5$). WOLOG $m_1 = 0$
$e$ is the sum of all product of 4 roots
$$e = m_1m_2m_3m_4+m_1m_2m_3m_5+m_1m_2m_4m_5+m_1m_3m_4m_5+m_2m_3m_4m_5$$
If $m_1=0$ we have
$$e = 0\cdot m_2m_3m_4+0\cdot m_2m_3m_5+0\cdot m_2m_4m_5+m0\cdot m_3m_4m_5+m_2m_3m_4m_5$$
$$e = m_2m_3m_4m_5$$
If order for $e$ to be egal to $0$, there need to be an other $m_i=0$, which is in contradiction to the $m_i$ all being differents.
You just need to check the possibility of $x^5+ex$ and $x^5+f$.
If $x^5+ex$, the coefficient of $x$ need to be negative in order to factorise,
$$x^5-ex = x(x^2-\sqrt{e})(x^2+\sqrt{e})\qquad e>0$$
$x^2+\sqrt{e}$ have complex roots. So it is impossible.
$x^5+f$ has only one real root.
For $k = 2$, @Alberto Saracco gave a great answer, I'll just add it here.  If you pair roots as $(x - x_1)(x-x_2)$, you'll only have even exponants.
$$x(x-1)(x+1)(x-2)(x+2) = x(x^2-1)(x^2-4)=x^5-5x^3+4x$$
This is an example where $k=3$ and is minimal.
