Polynomial with bounded coefficients and real root A polynomial with degree $2n$ has all coefficients in the range $[100,101]$ and has a real root. What is the minimum possible $n$?
Degree $0$ is clearly not possible. For degree $2$, the discriminant is $b^2-4ac<101^2-4(100)(100)<0$, so no real root exists. Having a real root $r$ means $a_{2n}r^{2n}+\ldots+a_1r+a_0=0$, so $$100|r|^{2n}\leq|a_{2n}r^{2n}|=|a_{2n-1}r^{2n-1}+\ldots+a_1r+a_0|\leq 101(|r|^{2n-1}+\ldots+|r|+1)$$ but if $r$ is very small this is possible.
 A: Let the polynomial $p(x)$ have coefficients be $c_j$, $j = 0 \ldots 2n$ with all $100 \le c_j \le 101 $.  Obviously there is no positive real root.  For negative $x$, 
$$ \eqalign{p(x) &\ge 100 + 101 x + 100 x^2 + 101 x^3 + \ldots + 100 x^{2n}\cr
&= 100 + (101 + 100 x)x(1 + x^2 + \ldots + x^{2n-2})\cr}$$
The only way to make this $0$ is to have $x < 0$ but $101 + 100 x > 0$, i.e. $-1.01 < x < 0$. It looks to me like the least possible $n$ is $100$, for which $x = -1$ is a root.
A: The polynomial
is of the form
$f(x)
=a\sum_{k=0}^{2n}x^k +\sum_{k=0}^{2n}b_kx^k
$
where
$a=100$
and
$0 \le b_k \le 1$.
$f(x)
=a\frac{x^{2n+1}-1}{x-1}+\sum_{k=0}^{2n}b_kx^k
$.
Since all the coefficients
are positive,
the real root must be negative.
Let $x = -y$.
$f(-y)
=a\frac{(-y)^{2n+1}-1}{-y-1}+\sum_{k=0}^{2n}b_k(-y)^k
=-a\frac{y^{2n+1}+1}{y+1}+\sum_{k=0}^{2n}b_k(-y)^k
$.
If
$f(-y) = 0$,
$\begin{array}\\
a\frac{y^{2n+1}+1}{y+1}
&=\sum_{k=0}^{2n}b_k(-y)^k\\
&=\sum_{k=0}^{n}b_k(-y)^{2k}+\sum_{k=0}^{n-1}b_k(-y)^{2k+1}\\
&=\sum_{k=0}^{n}b_{2k}y^{2k}-\sum_{k=0}^{n-1}b_{2k+1}y^{2k+1}\\
\end{array}
$
To make the right side
as large as possible
(this is my leap of faith),
set
$b_{2k} = 1$
and
$b_{2k+1} = 0$.
We then have
$a\frac{y^{2n+1}+1}{y+1}
=\sum_{k=0}^{n}y^{2k}
=\sum_{k=0}^{n}(y^2)^{k}
=\frac{(y^2)^{n+1}-1}{y^2-1}
$
or
$a(y^{2n+1}+1)
=\frac{y^{2n+2}-1}{y-1}
$.
So,
we want to find
the smallest value of $n$
for which this has a root.
At $y=1$
this is
$2a = 2n+2$
or
$n = a-1
=99$.
Don't know if this is right,
but I'll leave it at this.
