Let $ab+1=r^2$ with 0<a<b and b−a>2. Then a+b-2r\ge 1\: indeed, if a+b≤2r then (a+b)2≤4r2=4ab+4, so (b-a)^2\le 4\ and b−a≤2—acontradiction.ThisistheintegerformofLemma3.5ofM.CipuandY.Fujita,BoundsforDiophantinequintuples,Glas.Mat.50(2015),appliedintheproofofTheorem1.1tocontrola^{1/2}(b-a)^{-1}—type error terms.
Preamble
import Mathlib.Tactic
Formal statement
theorem diophantine_gap_lemma (a b r : Nat) (hab : a < b)
(hr : a * b + 1 = r ^ 2) (hgap : 2 < b - a) :
2 * r + 1 ≤ a + b := by sorry
Source
M. Cipu and Y. Fujita, Bounds for Diophantine quintuples, Glas. Mat. 50 (2015), 25-34, Lemma 3.5