No Diophantine pair of consecutive integers

Consecutive positive integers never form a Diophantine pair: for $a\ge 1$, $a(a+1)+1$ lies strictly between $a^2$ and $(a+1)^2$ and so is not a perfect square. Hence a Diophantine quintuple (or triple) always has $b-a\ge 2$; together with Fujita’s result ruling out b-a=2—Y. Fujita, The extensibility of Diophantine pairs \{k-1,k+1}$, J. Number Theory 128 (2008)—this gives $b-a\ge 3—cf. M. Cipu and Y. Fujita, Bounds for Diophantine quintuples, Glas. Mat. 50 (2015), proof of Theorem 1.1.