Proof. Set $ g=x+{A}/{2}, \, a=B-{A^2}/{4}. $ Then, $ H=g^2+a. $ Since A and B are small functions with respect to x,

$$ \begin{align*} T(r,g)=T(r,x)+S(r,x), \quad T(r,a)=S(r,g). \end{align*} $$

Moreover,

$$ \begin{align*} N(r,g) \leq N(r,x)+N(r,A) =S(r,g). \end{align*} $$

Suppose that $a\not \equiv 0$ and consider the meromorphic function $ F={g^2}/{a}. $ It follows from the standard properties of the Nevanlinna characteristic that

$$ \begin{align*} T(r,F)\leq 2T(r,g)+T\bigg(r,\frac{1}{a}\bigg)+O(1) =2T(r,g)+S(r,g). \end{align*} $$

However,

$$ \begin{align*} 2T(r,g)=T(r,g^2) \leq T(r,F)+T(r,a)+O(1) =T(r,F)+S(r,g). \end{align*} $$

Hence, $ T(r,F)=2T(r,g)+S(r,g) $ and, in particular, F is nonconstant.

The poles and zeros of F satisfy the estimates

$$ \begin{align*} \overline N(r,F) \leq \overline N(r,g) +N\bigg(r,\frac{1}{a}\bigg) =S(r,g) \end{align*} $$

and

$$ \begin{align*} \overline N\bigg(r,\frac{1}{F}\bigg) \leq \overline N\bigg(r,\frac{1}{g}\bigg) +N(r,a) = \overline N\bigg(r,\frac{1}{g}\bigg)+S(r,g). \end{align*} $$

Since $ F+1={(g^2+a)}/{a}={H}/{a}, $ the hypothesis of the lemma gives

$$ \begin{align*} \overline N\bigg(r,\frac{1}{F+1}\bigg) \leq N\bigg(r,\frac{1}{H}\bigg)+N(r,a) =S(r,g). \end{align*} $$

Applying Nevanlinna’s second main theorem to F and the three values $0$ , $-1$ and $\infty $ , we obtain

$$ \begin{align*} \begin{aligned} T(r,F) &\leq \overline N(r,F) +\overline N\bigg(r,\frac{1}{F}\bigg) +\overline N\bigg(r,\frac{1}{F+1}\bigg) +S(r,F)\\ &\leq \overline N\bigg(r,\frac{1}{g}\bigg)+S(r,g) \leq T(r,g)+S(r,g). \end{aligned} \end{align*} $$

Together with the identity $ T(r,F)=2T(r,g)+S(r,g), $ this yields

$$ \begin{align*} 2T(r,g)\leq T(r,g)+S(r,g), \end{align*} $$

which is impossible since g is a transcendental meromorphic function.

Hence, $a\equiv 0$ , that is, $ B={A^2}/{4}. $ Therefore,

$$ \begin{align*} H=x^2+Ax+\frac{A^2}{4} =\bigg(x+\frac{A}{2}\bigg)^2.\\[-41pt] \end{align*} $$

Proof of Theorem 1.1.

We divide the proof into four steps. First, we obtain an affine representation of the shift $f(z+c)$ with small coefficients and then reduce h to a perfect square. Next, we prove that the corresponding small function s is identically equal to $1$ , after which we rule out alternative (II) and complete the classification.

Step 1. An affine relation for the shift. By [Reference Laine and Latreuch5, Theorem 2.2], it suffices to rule out alternative (II). Suppose that this alternative holds.

Set

$$ \begin{align*} \alpha:=\frac{h'}{h}, \quad \phi:=2f'-\alpha f. \end{align*} $$

Since $h\in \mathcal A$ , [Reference Laine and Latreuch5, Lemma 2.1] implies that f is a transcendental meromorphic function and

$$ \begin{align*} N(r,f)=S(r,f), \quad T(r,h)\asymp T(r,f). \end{align*} $$

Since $h\in \mathcal A$ , the classical lemma on the logarithmic derivative [Reference Hayman4, Lemma 2.3] gives

$$ \begin{align*} T(r,\alpha) \leq N(r,h)+N\bigg(r,\frac{1}{h}\bigg)+S(r,h) =S(r,f). \end{align*} $$

Moreover, the difference analogue of Clunie’s lemma used in [Reference Laine and Latreuch5, Proof of Theorem 2.2] gives $ T(r,\phi )=S(r,f). $

For any meromorphic function u, we write $ u_c(z):=u(z+c). $ The estimate for the characteristic of a shifted function [Reference Chiang and Feng1, (1.2)], together with [Reference Halburd, Korhonen and Tohge3, Lemma 8.3], shows that, for $\rho _2(F)<1$ ,

$$ \begin{align*} T(r,F_c)=T(r,F)+S(r,F), \end{align*} $$

and that $T(r,v)=S(r,F)$ implies $ T(r,v_c)=S(r,F). $ In particular,

$$ \begin{align*} T(r,\alpha_c)+T(r,\phi_c)=S(r,f). \end{align*} $$

We next show that $\phi \not \equiv 0$ . If $\phi \equiv 0$ , then $ ({f^2}/{h})'=0, $ and hence, $f^2=A_0h$ for some constant $A_0\ne 0$ . Differentiating (1.1) and using $2f'=\alpha f$ , we obtain $f_c'=\alpha f_c$ , whence $f_c=B_0h$ for some constant $B_0\ne 0$ . Therefore,

$$ \begin{align*} 2T(r,f) = T(r,h)+O(1) = T(r,f_c)+O(1) = T(r,f)+S(r,f), \end{align*} $$

which is impossible. Thus, $\phi \not \equiv 0$ .

Differentiating (1.1) and using $h'=\alpha h$ , we obtain from the definition of $\phi $ that

(2.1) $$ \begin{align} f\phi=\alpha f_c-f_c'. \end{align} $$

Since $ \phi _c=2f_c'-\alpha _c f_c, $ eliminating $f_c'$ from this identity and (2.1) gives

(2.2) $$ \begin{align} 2\phi f+(\alpha_c-2\alpha)f_c+\phi_c=0. \end{align} $$

Set $ d:=\alpha _c-2\alpha. $ If $d\equiv 0$ , then (2.2) would imply $ f=-{\phi _c}/{2\phi }, $ and hence,

$$ \begin{align*} T(r,f) \leq T(r,\phi_c)+T\bigg(r,\frac{1}{\phi}\bigg)+O(1) =S(r,f), \end{align*} $$

which is impossible. Therefore, $d\not \equiv 0$ and

(2.3) $$ \begin{align} f_c=Lf+M, \quad L:=-\frac{2\phi}{d}, \quad M:=-\frac{\phi_c}{d}. \end{align} $$

Since

$$ \begin{align*} T(r,d) \leq T(r,\alpha_c)+T(r,\alpha)+O(1) =S(r,f), \end{align*} $$

we obtain

(2.4) $$ \begin{align} T(r,L)+T(r,M)=S(r,f). \end{align} $$

Step 2. Quadratic reduction. Substituting (2.3) into (1.1), we obtain

(2.5) $$ \begin{align} h=f^2+Lf+M. \end{align} $$

We have already established that f is a transcendental meromorphic function and that $ N(r,f)=S(r,f), $ while the smallness of L and M follows from (2.4). Moreover, $h\in \mathcal A$ and $T(r,h)\asymp T(r,f)$ imply that

$$ \begin{align*} N\bigg(r,\frac{1}{h}\bigg) = S(r,h) = S(r,f). \end{align*} $$

Therefore, Lemma 2.1 can be applied to (2.5) with

$$ \begin{align*} x=f,\quad A=L,\quad B=M\quad\text{and}\quad H=h. \end{align*} $$

It yields $M=L^2/4$ and

(2.6) $$ \begin{align} h=(f+s)^2, \quad s:=\frac{L}{2}. \end{align} $$

Moreover,

(2.7) $$ \begin{align} T(r,s)=S(r,f). \end{align} $$

Step 3. Rigidity of the auxiliary function. It follows from (2.7) and [Reference Gundersen2, Lemma 5] that, for every $\lambda>1$ ,

$$ \begin{align*} T(r,s)\leq T(\lambda r,f) \end{align*} $$

for all sufficiently large r. Therefore, if s is nonconstant, then

$$ \begin{align*} \rho_2(s)\leq\rho_2(f)<1. \end{align*} $$

Set $ U:=f+s. $ From (1.1) and (2.6), we have $ f_c=2sf+s^2. $ Hence,

(2.8) $$ \begin{align} U_c=2sU+t, \quad t:=s_c-s^2. \end{align} $$

Since $T(r,s)=S(r,f)$ ,

$$ \begin{align*} T(r,U)=T(r,f)+S(r,f). \end{align*} $$

Therefore, U is a transcendental meromorphic function, $\rho _2(U)<1$ and $S(r,f)=S(r,U)$ . Moreover, $h=U^2$ and $h\in \mathcal A$ imply that

(2.9) $$ \begin{align} N(r,U)+N\bigg(r,\frac{1}{U}\bigg)=S(r,U). \end{align} $$

Applying the shift estimate established in Step 1 with $F=U$ and $v=s$ , we obtain $T(r,s_c)=S(r,U)$ . Hence,

(2.10) $$ \begin{align} T(r,s)+T(r,t)=S(r,U). \end{align} $$

Choose $\eta \in (0,1-\rho _2(U))$ . The difference analogue of the lemma on the logarithmic derivative [Reference Halburd, Korhonen and Tohge3, (5.3)] gives

$$ \begin{align*} m\bigg(r,\frac{U_c}{U}\bigg) = o\bigg( \frac{T(r,U)} {r^{1-\rho_2(U)-\eta}} \bigg) = S(r,U). \end{align*} $$

However, by (2.8), (2.9) and (2.10),

$$ \begin{align*} N\bigg(r,\frac{U_c}{U}\bigg) \leq N\bigg(r,\frac{1}{U}\bigg)+N(r,U_c)\leq N\bigg(r,\frac{1}{U}\bigg) +N(r,U)+N(r,s)+N(r,t)=S(r,U). \end{align*} $$

Consequently,

(2.11) $$ \begin{align} T\bigg(r,\frac{U_c}{U}\bigg)=S(r,U). \end{align} $$

Dividing (2.8) by U, and using (2.10) and (2.11), we obtain

$$ \begin{align*} T\bigg(r,\frac{t}{U}\bigg)=S(r,U). \end{align*} $$

If $t\not \equiv 0$ , then $ 1/U=(t/U)(1/t) $ would imply that

$$ \begin{align*} T(r,U) \leq T\bigg(r,\frac{t}{U}\bigg) +T\bigg(r,\frac{1}{t}\bigg)+O(1) =S(r,U), \end{align*} $$

which is impossible. Hence, $t\equiv 0$ and

(2.12) $$ \begin{align} s_c=s^2, \quad U_c=2sU. \end{align} $$

We now show that s is constant. Suppose the contrary. Choose $\varepsilon \in (0,1)$ and set $ q:={2}/{(1+\varepsilon )}>1. $ By the first equation in (2.12) and the estimate for the characteristic of a shifted function [Reference Chiang and Feng1, (1.2)], we have, for all sufficiently large r,

$$ \begin{align*} 2T(r,s) = T(r,s_c) \leq (1+\varepsilon)T(r+|c|,s). \end{align*} $$

Consequently,

$$ \begin{align*} T(r+|c|,s)\geq qT(r,s) \end{align*} $$

for all sufficiently large r. Choose $r_0$ sufficiently large so that the preceding inequality holds for all $r\geq r_0$ and $T(r_0,s)>0$ . Iterating this inequality gives

$$ \begin{align*} T(r_0+n|c|,s) \geq q^nT(r_0,s), \quad n\in\mathbb N. \end{align*} $$

Hence,

$$ \begin{align*} \rho_2(s) \geq \limsup_{n\to\infty} \frac{\log\log T(r_0+n|c|,s)} {\log(r_0+n|c|)} \geq1, \end{align*} $$

which contradicts the previously established inequality $\rho _2(s)<1$ . Thus, s is constant.

The first equation in (2.12) now gives $s=s^2$ and, hence, $s\in \{0,1\}$ . The case $s=0$ is impossible: the second equation would then imply $U_c\equiv 0$ and, consequently, $U\equiv 0$ and $h=U^2\equiv 0$ , contrary to $h\in \mathcal A$ . Therefore, $s=1$ and

$$ \begin{align*} h=(f+1)^2, \quad (f+1)_c=2(f+1). \end{align*} $$

Step 4. Exclusion of the second alternative. We now have $U=f+1$ . Set $ W:={U'}/{U}. $ Since U is a transcendental meromorphic function, $W\not \equiv 0$ . By the classical lemma on the logarithmic derivative [Reference Hayman4, Lemma 2.3], $ m(r,W)=S(r,U). $ Every pole of W is a zero or a pole of U and, hence, by (2.9),

$$ \begin{align*} N(r,W) \leq \overline N(r,U) +\overline N\bigg(r,\frac{1}{U}\bigg) =S(r,U). \end{align*} $$

Therefore,

$$ \begin{align*} T(r,W)=S(r,U), \quad N\bigg(r,\frac{1}{W}\bigg)=S(r,U). \end{align*} $$

Since $f'=U'=UW$ ,

$$ \begin{align*} N\bigg(r,\frac{1}{f'}\bigg) \leq N\bigg(r,\frac{1}{U}\bigg) +N\bigg(r,\frac{1}{W}\bigg) =S(r,U) =S(r,f). \end{align*} $$

This contradicts (II), since it would imply that

$$ \begin{align*} T(r,f) = N\bigg(r,\frac{1}{f'}\bigg)+S(r,f) = S(r,f). \end{align*} $$

Thus, alternative (II) is impossible.

By [Reference Laine and Latreuch5, Theorem 2.2(1)], alternative (I) must hold; that is, $ f(z)=e^{az+b}-1, \, e^{ac}=2 $ for some $a,b\in \mathbb C$ . Then,

$$ \begin{align*} f(z)^2+f(z+c) = (e^{az+b}-1)^2 +e^{a(z+c)+b}-1= (e^{az+b}-1)^2 +2e^{az+b}-1 = e^{2az+2b}, \end{align*} $$

and hence, $ h(z)=e^{2az+2b}. $

Conversely, for any $a,b\in \mathbb C$ satisfying $e^{ac}=2$ , the preceding calculation shows that the functions $ f(z)=e^{az+b}-1, \, h(z)=e^{2az+2b}, $ satisfy (1.1). In this case, $a\ne 0$ , so h is a zero-free transcendental entire function and

$$ \begin{align*} N(r,h)+N\bigg(r,\frac{1}{h}\bigg)=0. \end{align*} $$

Therefore, $h\in \mathcal A$ . Moreover, f has finite order and, hence, $ \rho _2(f)=0<1. $