37th Iranian Mathematical Olympiad

The answer is $I=(1,1+\frac{1}{d})$.

Assume that the desired interval is of the form $(b,c)$. For some $q,r$ in $(b,c)$, put $a_i=q$ for odd and $a_i=r$ for even $i$, where $0\le i\le 2d-1$. Then $$P(-1)=1-dq+dr.$$ Since $P(x)$ has no real root, we must have $P(-1)>0$. Thus, $q-r<\frac{1}{d}$. We can assume that $c-b=\frac{1}{d}$. Therefore, consider the interval as $(b,b+\frac{1}{d})$. It is easy to find that $b>0$, since by choosing $a_0=b+ϵ$ should be positive. Now, assign the following numbers to the coefficients of the polynomial. $$a_{2d-1}=a_{2d-3}=\cdots =a_1=b+\frac{1}{d}+ϵ,a_{2d}=a_{2d-2}=\cdots =a_0=b+ϵ,$$ for some sufficiently small $ϵ>0$. It is clear that for all positive real $x$, $P(x)>0$. For all negative real $x$, putting $x=-t$, where $t>0$, then, $$P(-t)=t^{2d}-\left(b+\frac{1}{d}\right)t^{2d-1}+b{t}^{2d-2}-\cdots -\left(b+\frac{1}{d}\right)t+b+ϵQ(t),$$ for some polynomial $Q(t)$, $\mathrm{deg}Q(t)=2d-1$. As $ϵ$ tends to zero, it remains to find all $b>0$ such that $$R(t)=t^{2d}-\left(b+\frac{1}{d}\right)t^{2d-1}+b{t}^{2d-2}-\cdots -\left(b+\frac{1}{d}\right)t+b\ge 0.$$ Note that $R(1)=0$. Therefore, $R^{\prime }(1)$ should be zero. That is, $$2d-(2d-1)\left(b+\frac{1}{d}\right)+(2d-2)b-\cdots -\left(b+\frac{1}{d}\right)=0.$$ Hence, we claim the interval $I=(1,1+\frac{1}{d})$ works! It is obvious that $P(x)$ has no positive real roots. Now, we prove that $P(x)>0$ for all negative real $x$. Put $x=-t,t>0$. Then, $$P(-t)>t^{2d}-\left(1+\frac{1}{d}\right)t^{2d-1}+t^{2d-2}-\cdots -\left(1+\frac{1}{d}\right)t+1.$$ Thus it remains to prove that $$\frac{t^{2d}+t^{2d-2}+\cdots +t^2+1}{d+1}\ge \frac{t^{2d-1}+t^{2d-3}+\cdots +t^3+t}{d}.$$ Now, since $t^{2d}+1\ge t^{2d-2k+1}+t^{2k-1}$ and $t^{2k}+t^{2k-2}\ge 2t^{2k-1}$, we are done. Thus, $$b\le \frac{1}{d}\underset{t\to 1^{+}}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}.$$ and $$b\ge \frac{1}{d}\underset{t\to 1^{-}}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1},$$ since the function $$\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}$$ has a limit as $t$ approaches 1. Thus, we find that $$\underset{t\to 1^{+}}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}=\underset{t\to 1^{-}}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}.$$ Hence, $$b=\frac{1}{d}\underset{t\to 1}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}.$$ Finally note that $$\begin{array}{rl}& \frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}\\ & =\frac{(t^{2d}-t^{2d-1})+(t^{2d}-t^{2d-3})+\cdots +(t^{2d}-t)}{(t-1)(t^{2d-2}+t^{2d-4}+\cdots +1)}\\ & =\frac{(t-1)(t^{2d-1}+t^{2d-3}(t^2+t+1)+\cdots +t(t^{2d-2}+\cdots +1))}{(t-1)(t^{2d-2}+t^{2d-4}+\cdots +1)}\end{array}$$ Thus, $$\underset{t\to 1}{\lim }\frac{d{t}^{2d}-(t^{2d-1}+t^{2d-3}+\cdots +t)}{t^{2d-1}-t^{2d-2}+\cdots +t-1}$$ $$=\underset{t\to 1}{\lim }\frac{t^{2d-1}+t^{2d-3}(t^2+t+1)+\cdots +t(t^{2d-2}+\cdots +1)}{t^{2d-2}+t^{2d-4}+\cdots +1}=\frac{d^2}{d}=1.$$ That is, $b=\frac{1}{d}$ and $d=1$.