I am currently reading rational points on elliptic curves by Silverman. In exercise $3.2$ $(b)$ of Chapter $3$, it asks (regarding an elliptic curve of the form $C: y^2 = x^3+ax^2+bx+c$)
Prove that there is a constant $\kappa$ which depends only on $a,b,c$ so that for all rational points $P_1$ and $P_2$, $$h(P_1+P_2)+h(P_1-P_2) \leq 2h(P_1) + 2h(P_2) + \kappa$$
Where $h(P) = \log(H(P))$ and $H(P) = H(x(P)) = H(\frac{a}{b}) = \max\{|a|, |b|\}$ is the height of a rational number. I realize this amounts to proving $$H(P_1+P_2)H(P_1-P_2) \leq H(P_1)^2 H(P_2)^2 *\kappa^{\prime}$$ I attempted a proof modeled after Silverman's outlined proof of a Lemma used to prove Mordell's Theorem, but I reached a weird conclusion. This is my attempt: We have $$x(P_1 \pm P_2) = \bigg(\frac{\pm y_2 - y_1}{x_2-x_1}\bigg)^2-(a+x_1+x_2)$$ $$=\frac{y_2^2 \mp2y_1y_2+y_1^2-(a+x_1+x_2)(x_2^2-2x_1x_2+x_1^2)}{x_2^2-2x_1x_2+x_1^2}$$ $$=\frac{\mp2y_1y_2+(y_2^2-x_2^3-ax_2^2)+(y_1^2-x_1^3-ax_1^2)+x_1^2x_2+x_1x_2^2+2ax_1x_2}{x_2^2-2x_1x_2+x_1^2}$$ using $y_i^2 = x_i^3+ax_i^2+bx_i+c$ $$=\frac{\mp2y_1y_2+b(x_1+x_2)+2c+x_1^2x_2+x_1x_2^2+2ax_1x_2}{x_2^2-2x_1x_2+x_1^2}$$ replacing each coefficient with an arbitrary rational number, we have $$=\frac{Ay_1y_2+Bx_1^2x_2+Cx_1x_2^2+Dx_1x_2 + E(x_1+x_2)+F}{Gx_2^2+Hx_1x_2+Ix_1^2}$$ we can assume $A-I$ are integers, as multiplying across by a common denominator gives integers. Also, it was shown earlier in the text that each rational point satisfies $x = \frac{m}{e^2}$ and $y = \frac{n}{e^3}$, and they satisfy the inequalities $$e \leq H(P)^{1/2}, \text{ } n \leq KH(P)^{3/2}, \text{ } m \leq H(P)$$ for some constant $K$. Substituting these in and clearing denominators on the numerator and denominator by $e_1^4 e_2^4$ give (after simplifying) $$\frac{An_1e_1n_2e_2 + Bm_1^2m_2e_2^2+Cm_1e_1^2m_2^2+Dm_1e_1^2m_2e_2^2+Em_1e_1^2e_2^4+Ee_1^4m_2e_2^2+Fe_1^4e_2^4}{Gm_1^2e_2^4+Hm_1e_1^2m_2e_2^2+Ie_1^4m_2^2}$$ Thus $$H(P_1 \pm P_2) \leq \max\{|An_1e_1n_2e_2 + Bm_1^2m_2e_2^2+Cm_1e_1^2m_2^2+Dm_1e_1^2m_2e_2^2+Em_1e_1^2e_2^4+Ee_1^4m_2e_2^2+Fe_1^4e_2^4|, |Gm_1^2e_2^4+Hm_1e_1^2m_2e_2^2+Ie_1^4m_2^2|\}$$ (either the numerator and denominator are in simplest form or they reduce, in which case the inequality still holds). The first element of the max function is $$|An_1e_1n_2e_2 + Bm_1^2m_2e_2^2+Cm_1e_1^2m_2^2+Dm_1e_1^2m_2e_2^2+Em_1e_1^2e_2^4+Ee_1^4m_2e_2^2+Fe_1^4e_2^4|$$ $$\leq |An_1e_1n_2e_2| + |Bm_1^2m_2e_2^2|+|Cm_1e_1^2m_2^2|+|Dm_1e_1^2m_2e_2^2|+|Em_1e_1^2e_2^4|+|Ee_1^4m_2e_2^2|+|Fe_1^4e_2^4|$$ $$\leq H(P_1)^2 H(P_2)^2 \Big(|AK_1K_2| + |B| + |C| + |D| + 2|E| + |F| \Big)$$ The first step is by the triangle inequality, and the second by the aforementioned inequalities on $H(P_i)$ and $m_i, n_i, e_i$. Similarly, the second element of the max function is $$|Gm_1^2e_2^4+Hm_1e_1^2m_2e_2^2+Ie_1^4m_2^2|$$ $$\leq |Gm_1^2e_2^4|+|Hm_1e_1^2m_2e_2^2|+|Ie_1^4m_2^2|$$ $$\leq H(P_1)^2 H(P_2)^2 \Big(|G| + |H| + |I|\Big)$$ Therefore, if we let $\kappa^{\prime \prime} = \max\{|AK_1K_2| + |B| + |C| + |D| + 2|E| + |F|,|G| + |H| + |I|\}$ $$H(P_1 \pm P_2) \leq H(P_1)^2 H(P_2)^2 \kappa^{\prime \prime}$$ If this inequality holds, then we have $$H(P_1 + P_2)H(P_1 - P_2) \leq \Big(H(P_1)^2 H(P_2)^2 \kappa^{\prime \prime}\Big)^2$$ But this was not the desired inequality. My question is: if my proof was correct, where did I go wrong? Is there another set of inequalities that should have been followed or does it still somehow follow from what I have shown already? Thank you,