S-integral points on hyperelliptic curves Homero Renato Gallegos ...

Write ν 1 = ν ′ ε and ε 1 = ε −1 . Hence we have expressed τ 2 − τ 1 in the form ν 1 ε 1 whereε 1 is a S K1 -unit and h(ν 1 ) ≤ H ∗ . We can repeat the same process on τ 2 +τ 1 . Similarly,the equation τ3 2 = κ 1(x − α 3 ) shows that τ 3 , τ 1 + τ 3 and τ 1 − τ 3 are S K2 -integers, andwe can repeat the process to write now τ 1 ± τ 3 in the form ν 2 ε 2 where ε 2 is a S K2 -unitand h(ν 2 ) ≤ H ∗ .We now want to express τ 2 ± τ 3 in a similar way. Define ν ′′ = √ κ 2 /κ 1 (τ 2 − τ 3 ).Note ν ′′ = κ 2 ξ 2 − √ κ 2 κ 3 ξ 3 is a S K3 -integer of K 3 . Similarly, √ κ 2 /κ 1 (τ 2 + τ 3 ) is aS K3 -integer of K 3 andν ′′√ κ 2 /κ 1 (τ 2 + τ 3 ) = κ 2 (α 2 − α 3 ).Hence, Norm SK3 (ν ′′ ) ≤ N. Using again Lemma 3.3.4 there is a S K3 -unit ε 3 such thatν ′′ = ν ′ ε andh(ν ′ ) ≤ c 6 (r 3 , d 3 )R + log N S K3(κ 2 (α 2 − α 3 ))[K 1 : Q]⎛⎞+ h ⎝ ∑ log p⎠ .p∈SLet ν = √ κ 1 /κ 2 ν ′ . Thus τ 2 − τ 3 = νε where h(ν) ≤ h(ν ′ ) + h(κ) ≤ H ∗ . We will applyProposition 3.6.1 to the unit equation(τ 1 − τ 2 ) + (τ 3 − τ 1 ) + (τ 2 − τ 3 ) = 0,which is indeed of the form ν 1 ε 1 + ν 2 ε 2 + ν 3 ε 3 = 0 where ε 1 is a S K1 -unit of K 1 , ε 2 is aS K2 -unit of K 2 and we consider ε 3 as a S L -unit of L. Observe that, since ∑ υ|p d υ = d,then there are at most d places in S L over a prime in S and hence S has at most d(s+1)elements. We obtain⎧c ∗ 10 ,( )τ1 − τ⎪⎨2c ∗ 13h≤ max,τ 2 − τ 3c ∗ 12 + c∗ 11 log(max{h(ε 1), h(ε 2 ), 1}),⎪⎩ c ∗ 15 + c∗ 14 log(max{h(ε 1), h(ε 2 ), 1}).40