= Solution
We prove the Mahler alternative in full. Let $f(z)=\sum_{n\ge0}z^{l^n}$. Its series converges locally uniformly in the <unit disc> and satisfies
$$
f(z^l)=f(z)-z.
$$
It is transcendental as a function. At a <root of unity> $\zeta$ with $\zeta^{l^r}=1$, the radial tail $\sum_{n\ge r}t^{l^n}$ in $f(t\zeta)$ is real and tends to infinity as $t\uparrow1$, while the initial terms remain bounded. These roots are dense on the <unit> circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and <polynomial discriminant>: elsewhere its defining <polynomial> and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb C(z)$.
Assume now that $\alpha$ and $f(\alpha)$ are algebraic, with $0<|\alpha|<1$, and put both in a <number field> $K$ of degree $d$. For each $N$, the $(N+1)^2$ coefficients of a <polynomial> $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1)^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have <integer> coefficients, so a nonzero <integer> solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge(N+1)^2-1$.
At $\alpha_r=\alpha^{l^r}$ the <functional equation> gives $f(\alpha_r)=f(\alpha)-\sum_{j<r}\alpha^{l^j}\in K$. Thus $\eta_r=R_N(\alpha_r)\in K$ is nonzero for all sufficiently large $r$ and
$$
\log|\eta_r|\le-T_Nl^r\log(1/|\alpha|)+O_N(1).
$$
Use the absolute <logarithmic height> $h$. The elementary inequalities for sums and <polynomial> evaluation give
$$
h(\eta_r)\le N\left(1+\frac1{l-1}\right)l^rh(\alpha)+O_N(r+1).
$$
Here $h(f(\alpha_r))\le h(f(\alpha))+(l^r-1)h(\alpha)/(l-1)+r\log2$. The <Liouville height inequality> gives $\log|\eta_r|\ge-dh(\eta_r)$. Divide by $l^r$ and let $r\to\infty$ to obtain
$$
T_N\log(1/|\alpha|)\le dN\frac l{l-1}h(\alpha).
$$
The left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence \b[every such algebraic $\alpha$ has transcendental $f(\alpha)$]. This is the <auxiliary-value height argument for the Fredholm series>.
For the rational continuation, let $a_n=\sum_{j=0}^n(p/q)^{l^j}$ and $Q_n=q^{l^n}$. This is a <rational number> with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,
$$
0<f(p/q)-a_n\le2(p/q)^{l^{n+1}}=2Q_n^{-\kappa},\qquad\kappa=l\left(1-\frac{\log p}{\log q}\right)>l(1-\delta)>2.
$$
A rational value $a/b$ would instead have distance at least $1/(bQ_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $\epsilon>0$ with $2+\epsilon<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-\epsilon}$. This contradicts <Roth theorem>. Thus the requested rational values are transcendental by the <Roth criterion for rational Fredholm values>.
For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/\mathbb F_q$, then $\deg\Phi=q$ and $\deg(1-\Phi)=N=\#E(\mathbb F_q)$: the latter map is separable and its kernel is exactly the rational points. The <divisor> identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\deg(u+v)+\deg(u-v)=2\deg u+2\deg v$. Polarizing it yields
$$
\deg(m+n\Phi)=m^2+tmn+qn^2,\qquad t=q+1-N.
$$
All these degrees are nonnegative. In particular $\deg(-t+2\Phi)=4q-t^2\ge0$, so $|t|\le2\sqrt q$. The Frobenius characteristic <polynomial> is $X^2-tX+q$; its complex roots have modulus $\sqrt q$, including the double-root case. The zeta function has numerator $1-tT+qT^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the <degree-form proof of the Hasse bound>; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.
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