Let $P(x)=x^{n}+a\_{1}x^{n-1}+\cdots+a\_{n-1}x+a\_{n}$, where $a\_1, a\_2, \dots, a\_n$ are intergers.


Question 1. When does the polynomial $P(x)$ have its zeros all being pure imaginary or zero(here 0 is a root of the given polynomial)?


Question 2. Does there exist a characterization that $P(x)$ have its zeros all being pure imaginary or zero?


Please point me to some references if this has already been studied. Thanks for your time!



When does a polynomial have all pure imaginary roots?
A necessary and sufficient condition is that $P(x)$ is a power of $x$ times a product of terms $x^2+c$ with $c$ real and positive. Hence $P(x)$ is $x^{n-2m}Q(x^2)$ where $m\ge0$ and $Q(t)$ is a unitary polynomial in $t$ of degree $m$ with integer nonnegative coefficients. Finally, a necessary condition is that $a\_{k}=0$ for every odd $k$ and one should further require the underlying polynomial $Q$ to be a product of $t+c\_i$.




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**EDIT** It appears that the problem of locating the zeroes of a polynomial resurfaces regularly on MO, see for example [this question](https://mathoverflow.net/questions/9431/when-does-a-real-polynomial-have-a-pair-of-complex-conjugate-roots) or [that one](https://mathoverflow.net/questions/2837/what-are-conditions-on-real-coefficients-for-zeros-of-a-polynomial-to-be-on-the-u) or [that one](https://mathoverflow.net/questions/20946/criteria-to-determine-whether-a-real-coefficient-polynomial-has-real-root). The answers to these provide the following facts.


First, there are Newton's inequalities. For any real numbers $(t\_i)\_{1\le i\le m}$, their elementary symmetric means $S\_k$ are such that, for every $k$, $S\_k^2\ge S\_{k-1}S\_{k+1}$. Here $S\_k=\sigma\_k/{m\choose k}$ where $\sigma\_k$ denotes the $k$th elementary function in the real numbers $(t\_i)\_{1\le i\le m}$, see [here](http://en.wikipedia.org/wiki/Newton%27s_inequalities).


So, write $Q$ as $Q(t)=t^m+b\_1t^{m-1}+\cdots+b\_m$ (and recall that $b\_k=a\_{2k}$). Any $Q$ under consideration is such that $b\_k$ must be nonnegative for every $k$ and Newton's inequalities indicate that supplementary necessary condition are $S\_k^2\ge S\_{k-1}S\_{k+1}$ for every $k$, where the $S\_k$ are based on the $\sigma\_k=b\_k$. (At first sight, one should choose $\sigma\_k=(-1)^kb\_k$ but the $(-1)^k$ disappear.) Hence, a necessary condition is that, for every $k$,
$$
k(m-k)b\_k^2\ge (k+1)(m-k+1)b\_{k-1}b\_{k+1}.
$$
Second, a *complete characterization* of the polynomials $Q$ with only real negative roots is based on the notion of Hermite forms, see [this answer](https://mathoverflow.net/questions/20946/criteria-to-determine-whether-a-real-coefficient-polynomial-has-real-root/30533#30533). 


Recall that the Hermite form of a polynomial $Q$ of degree $m$ is a symmetric matrix, usually denoted by $H\_1(Q)$, of size $m\times m$ with entries $(h\_{ij}(Q))$, defined by 
$$
h\_{ij}(Q)=s\_{i+j-2}(Q),
$$
where, for every $k$, $s\_k(Q)$ is the sum of the $k$th powers of the roots of $Q$ (and $s\_0(Q)=m$), see [these lecture notes](http://www.math.ucsd.edu/~njw/Teaching/Math271C/Lecture_11.pdf). Recall that the $s\_k(Q)$ are well known functions of the elementary symmetric functions $\sigma\_k=(-1)^kb\_k$, see [here](http://en.wikipedia.org/wiki/Symmetric_polynomial). Recall also that the signature of a symmetric matrix is equal to the number of its positive eigenvalues minus the number of its negative eigenvalues. Then the signature of $H\_1(Q)$ is equal to the number of real roots of $Q$.


In the context of this question, one already knows that $Q$ has no positive real root because $b\_k$ is nonnegative for every $k$ and one wants $Q$ to have $m$ real roots. Hence the signature of $H\_1(Q)$ must be $m$, that is, $H\_1(Q)$ must be positive definite.


Finally, necessary and sufficient conditions are that the odd numbered $a\_k$ are zero and that the even numbered $a\_k$ are nonnegative and define a polynomial $Q$ such that the symmetric matrix $H\_1(Q)$ is positive definite.


