a) $p(z)$ is an even degree polynomial (of degree $k = 2j$) with real coefficients;
b) $p(0) = 1$;
c) $p(z)$ and $p(-z)$ have no roots in common anywhere in the complex plane;
d) $f(z) = p(z)/p(-z)$ is a Pade approximant to $\exp(z) = e^z$, such that the Taylor expansion of $f(z)$ agrees with that of $\exp(z)$ up to $(2k)$th order.
Are there simple proofs of the following conjectures:
i) the coefficients of $p(z)$ are all positive
ii) $f(x) \le \exp(x)$ for all nonnegative $x$
iii) $p(x) p(-x) \exp(ax)$ has all positive coefficients in its Taylor expansion for any $a \ge 1$
iv) $p(x) p(-x)$ has no real roots.
Comments: By Descartes rule of signs, (i) implies $p(z)$ has no positive roots. By a theorem of Laguerre (Ouvres, Tome 1), (iii) would imply (iv)