= Solution
For genuine degrees three and two, the <resultant> is the <Sylvester matrix> <determinant>
$$
\boxed{R(p,q)=\det\begin{pmatrix}
a&b&c&d&0\\0&a&b&c&d\\
\alpha&\beta&\gamma&0&0\\0&\alpha&\beta&\gamma&0\\0&0&\alpha&\beta&\gamma
\end{pmatrix}}.
$$
Equivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2\prod_iq(r_i)$. Thus \b[$R(p,q)\ne0$ means that the <polynomials> have no common root], or equivalently are coprime over $\mathbb C$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\deg A<2$, $\deg B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the <resultant> for the actual <polynomial> degrees rather than assuming the displayed fixed-degree criterion unchanged.
An <algebraic addition theorem> for a <meromorphic function> $f$ means a nonzero <polynomial> $P(U,V,W)$ with
$$
P(f(z),f(w),f(z+w))=0
$$
identically where the values are finite. A local relation extends meromorphically to the other regular arguments.
For a <polynomial> $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. Form
$$
P(U,V,W)=\prod_{i=1}^d\prod_{j=1}^d\bigl(W-f(z_i+w_j)\bigr).
$$
Its coefficients are separately <symmetric polynomials> in the two sets of roots. The <Fundamental theorem of symmetric polynomials> and the fixed nonzero leading coefficient of $f$ show they are <polynomials> in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by <polynomial> identity. A constant <polynomial> instead has the relation $W-f(0)=0$.
For the elliptic case, let $E=\mathbb C/\Lambda$. A nonconstant <elliptic function> $f$ defines a <finite morphism> $E\to\mathbb P^1$, so the <function field> of $E$ is a finite <algebraic extension> of $\mathbb C(f)$. The <elliptic function-field decomposition> expresses it as $\mathbb C(\wp,\wp')$, with
$$
(\wp')^2=4\wp^3-g_2\wp-g_3.
$$
The <Weierstrass addition formula> makes translation algebraic:
$$
\wp(z+w)=-\wp(z)-\wp(w)+\frac14\left(\frac{\wp'(z)-\wp'(w)}{\wp(z)-\wp(w)}\right)^2.
$$
Differentiating and using the cubic differential equation also makes $\wp'(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite <algebraic extension> of $\mathbb C(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required <polynomial> relation. This sketches why every <elliptic function>, not only $\wp$, has an <algebraic addition theorem>. Constants are already covered.
For the final assertion, consider $f(z)=e^{e^z}$, an <entire function>. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its <derivative> forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb Z$, making it a <simply periodic function>.
Suppose it had an <algebraic addition theorem> $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient <polynomial> in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^\alpha$ outside this set, and set $c=\log\alpha$. Specializing $w=c$ gives a nonzero <polynomial> $Q(U,W)=P(U,e^\alpha,W)$ and, with $t=e^z$,
$$
0=Q(e^t,e^{\alpha t})=\sum_{m,n}q_{mn}e^{(m+\alpha n)t}.
$$
All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to+\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that \b[simple periodicity does not imply an <algebraic addition theorem>]. It is the <simply periodic entire function without an algebraic addition theorem> example.
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