I had some difficulty in writing this down.. Let $ [n]=\{1,2,\ldots n \} $, $ Z $ be the set of roots of $ f(x) $. Note that by condition iii, $ 0 $ is not a root and if $ \alpha i $ is a root, then so is $ -\alpha i $. So let the roots be $ \alpha_1 i,\alpha_2 i, \ldots \alpha_n i, -\alpha_1 i, -\alpha_2 i, \ldots -\alpha_n i $. We can assume that $ \alpha_j >0 $ $ \forall j \in [n] $. Let $ Z'=\{ \alpha_1 i,\alpha_2 i, \ldots \alpha_n i \} $ and also define $ A_k=\{ S : S \in Z, |S|=k \} $. Then note that by Vieta's formulae, we have \[ \frac{a_{2k}}{a_0}= \sum_{S \in A_{2k}} \left( \prod_{a \in S} a \right) \]. We will show that \[ \frac{a_{2k}}{a_0}= \sum_{|S|=k,S \in Z'} \left( \prod_{a \in S} a \right)^2 \forall k \in [n] \] We have \[ \frac{a_2}{a_0}= \sum_{S \in A_2} \left( \prod_{a \in S} a \right) \] \[ =\sum_{j=1}^{n} \alpha_j^2 - \sum_{k,l \in [n], k \neq l} ( \alpha_k \alpha_l + (-\alpha_k)(-\alpha_l)+ (-\alpha_k)\alpha_l + \alpha_k(-\alpha_l) ) \] \[ =\sum_{j=1}^{n} \alpha_j^2 \] Therefore, the assertion is true for $ k=1 $. Suppose that it is true for $ 1,2, \ldots, k-1 $. We have, \[ \frac{a_{2k}}{a_0}= \sum_{S \in A_{2k}} \left( \prod_{a \in S} a \right) \] Consider a term of the above summation. If there exists an index $ j $ such that both $ \alpha_j, -\alpha_j $ are present in the term, then $ j $ has property $ P $. Then our sum is, \[ \sum_{S \in A_{2k}} \left( \prod_{j_{\mid P}} \alpha_j^2 \left ( \sum_{S \in A_{2l}} \prod_{a \in S} a \right) \right) \] where the index $ 2l $ is determined by the number of $ j $ satisfying $ P $ and obviously, $ l \le k $. By the induction hypothesis, the sum $ \sum_{S \in A_{2l}} \prod_{a \in S} a $ is always zero by induction hypothesis for $ l<k, l \neq 0 $. The case $ l=0 $ gives the sum, \[ \sum_{|S|=k,|S| \in Z'} \left( \prod_{a \in S} a \right)^2 \forall k \in [n] \] So, we are only left with the case $ l=k $. But then, we are only left with terms of the kind $ i^{2k}\alpha_1 \alpha_2 \ldots \alpha_{2k} $ in which $ | \alpha_i | \neq | \alpha_j | $, therefore, for fixed first $ (2k-2) $ terms, we have the sum, \[ i^{2k} \sum \left(\alpha_1 \alpha_2 \ldots \alpha_{2k-2} \left( \sum_{i \neq j} \alpha_i \alpha_j \right) \right)=0 \] since the sum in the brackets is $ 0 $ by hypothesis. Hence, our claim, \[ \frac{a_{2k}}{a_0}= \sum_{|S|=k,S \in Z'} \left( \prod_{a \in S} a \right)^2 \] is established. Condition ii gives \[ \binom{2n}{n} \prod_{j=1}^{n} \alpha_j^2 \ge \sum_{j=0}^{j=n} \frac{a_{2j}}{a_0} \frac{a_{2n-2j}}{a_0} \] \[ =\left( \sum_{|S|=j,S \in Z'} \left( \prod_{a \in S} a \right)^2 \right) \left( \sum_{|T|=n-j,T \in Z'} \left( \prod_{b \in S} b \right)^2 \right) \] By the Cauchy Schwarz inequality and noting that for an $ S $, there exists $ T=Z'/S $, the last is \[ \ge \sum_{j=0}^{n} \left( \sum_{|S|=j,S \in Z'} \left(\prod_{a \in S} a \prod_{b \in Z'/S} b \right) \right)^2 \] \[ =\sum_{j=0}^{n} \binom{n}{j}^2 \prod_{j=1}^{n} \alpha_j^2 \] \[ =\binom{2n}{n} \prod_{j=1}^{n} \alpha_j^2 \] by the Vandermonde identity. Equality holds, thus we get that all $ \alpha_j^2 $ are equal and hence all $ \alpha \in Z' $ are equal. Therefore the roots of $ f(x) $ are $ n $ repeated $ \alpha$'s and $ n $ repeated $ -\alpha$'s, thus \[ f(x)=a_0(x^2+\alpha^2)^n \] where $ a_0>0, \alpha \in \mathbb{R}/0 $ and it is easy to check that any polynomial of the above form also satisfies these conditions. $ \blacksquare $