Proof. Without loss of generality $f_1,f_2\in \mathbb{Z}[X_1,X_2]$.

Upper bound: Write $P=(a_1:a_2)$, $a_1,a_2\in \mathbb{Z}$ coprime.

$$\begin{array}{llll}\hfill H(F(P))& \le \max (|f_1(a_1,a_2)|,|f_2(a_1,a_2)|)& \hfill & \\ \hfill & \le c_2\max (|a_1{|}^d,|a_2{|}^d)& \hfill & \end{array}$$

where

$$c_2=\underset{i=1,2}{\max }\text{(sum of absolute values of coefficients of }{f}_i\text{)}.$$

Therefore $H(F(P))\le c_{2}H{(P)}^d$.

Lower bound: We claim that there exists $g_{ij}\in \mathbb{Z}[X_1,X_2]$ homogeneous of degree $d-1$ and $\kappa \in {\mathbb{Z}}_{>0}$ such that

$$\sum _{j=1}^2{g}_{ij}{f}_j=\kappa X_i^{2d-1},i=1,2.\text{(†)}$$

Indeed, running Euclid’s algorithm on $f_1(X,1)$ and $f_2(X,1)$ gives $r,s\in \mathbb{Q}[X]$ of degree $<d$ such that

$$r(X)f_1(X,1)+s(X)f_2(X,1)=1.$$

Homogenising and clearing denominators gives ($†$) for $i=2$. Likewise for $i=1$.

Write $P=(a_1:a_2)$ with $a_1,a_2\in \mathbb{Z}$ coprime. ($†$) gives

$$\sum _{j=1}^2{g}_{ij}(a_1,a_2)f_j(a_1,a_2)=\kappa a_i^{2d-1},i=1,2.$$

Therefore $\gcd (f_1(a_1,a_2),f_2(a_1,a_2))$ divides $\gcd (\kappa \underset{1}{\overset{2d-1}{a}},\kappa a_2^{2d-1})=\kappa$.

$$|\kappa a_i^{2d-1}|\le \underset{j=1,2}{\max }\underset{\le \kappa H(F(P))}{\underbrace{|f_j(a_1,a_2)|}}\underset{\le {\gamma }_{i}H{(D)}^{d-1}}{\underbrace{\sum _{j=1}^2|g_{ij}(a_1,a_2)|}}$$

where

$${\gamma }_i=\sum _{j=1}^2\text{(sum of absolute values of coefficients of }{g}_{ij}\text{)}.$$

Therefore

$$\kappa |a_i{|}^{2d-1}\le \kappa H(F(P)){\gamma }_{i}H{(P)}^{d-1}$$

so

$$H{(P)}^{2d-1}\le \max ({\gamma }_1,{\gamma }_2)H(F(P))H{(P)}^{d-1}$$

so

$$\underset{=c_1}{\underbrace{\frac{1}{\max ({\gamma }_1,{\gamma }_2)}}}H{(P)}^d\le H(F(P)).\square$$

Proof. Recall (Lemma 5.4)

($=d$ say). Lemma 13.1 tells us that there exists $c_1,c_2>0$ such that

$$c_{1}H{(P)}^d\le H(\varphi (P))\le c_{2}H{(P)}^d\forall P\in E(\mathbb{Q}).$$

Taking logs gives

Proof. Put $n=0$ in above calculation to get

$$\left|\frac{1}{4^m}h(2^{m}P)-h(P)\right|<\frac{c}{3}.$$

Take limit $m\to \infty$. □

Proof. $ĥ(P)$ bounded means we have a bound on $h(P)$ (by Lemma 13.3). So only finitely many possibilities for $x$. Each $x$ gives $\le 2$ choices for $y$. □

Proof. By ?? 66, there exists $c>0$ such that

$$|h(\varphi P)-(\mathrm{deg}\varphi )h(P)|<c\forall P\in E(\mathbb{Q}).$$

Replace $P$ by $2^{n}P$, divide by $4^n$ and take limit $n\to \infty$. □

Proof. Let $E$ have Weierstrass equation $y^2=x^3+ax+b$, $a,b\in \mathbb{Z}$. Let $P,Q,P+Q,P-Q$ have $x$ coordinates $x_1,\dots ,x_4$. By Lemma 5.8, there exists $W_0,W_1,W_2\in \mathbb{Z}[x_1,x_2]$ of degree $\le 2$ in $x_2$ such that

$$(1:x_3+x_4:x_3{x}_4)=(\underset{={(x_1-x_2)}^2}{\underbrace{W_0}}:W_1:W_2).$$

Write $x_i=\frac{r_i}{s_i}$ with $r_i,s_i\in \mathbb{Z}$ coprime.

$$\begin{array}{llll}\hfill \underset{\gcd =1}{\underbrace{(s_3{s}_4:r_3{s}_4+r_4{s}_3:r_3{r}_4)}}& =({(r_1{s}_2-r_2{s}_1)}^2:\cdots )& \hfill & \\ \hfill H(P+Q)H(P-Q)& =\max (|r_3|,|s_3|)\max (|r_4|,|s_4|)& \hfill & \\ \hfill & \le 2\max (|s_3{s}_4|,|r_3{s}_4+r_4{s}_3|,|r_3{r}_4|)& \hfill & \\ \hfill & \le 2\max (|r_1{s}_2-r_2{s}_1{|}^2,\dots )& \hfill & \\ \hfill & \le c\max {(|r_1{|}^2,|s_1|)}^2\max {(|r_2|,|s_2|)}^2& \hfill & \\ \hfill & =cH{(P)}^{2}H{(Q)}^2& \hfill & \end{array}$$

where $c$ depends on $E$, but not on $P$ and $Q$. □

Proof. Lemma 13.6 and $|h(2P)-4h(P)|$ bounded gives that there exists $c\in ℝ$ such that

$$h(P+Q)+h(P-Q)\le 2h(P)+2h(Q)+c\forall P,Q\in E(\mathbb{Q}).$$

Replacing $P$, $Q$ by $2^{n}P$, $2^{n}Q$, dividing by $4^n$ and taking the limit $n\to \infty$ gives

$$ĥ(P+Q)+ĥ(P-Q)\le 2ĥ(P)+2ĥ(Q).$$

Replacing $P,Q$ by $P+Q$, $P-Q$ and $ĥ(2P)=4ĥ(P)$ gives the reverse inequality. Therefore $ĥ$ satisfies the parallelogram law, and hence $ĥ$ is a quadratic form. □