Proves b ≫ a (log a)^{1/8} / (log log a)^{12} whenever a²(a²+1) divides b²(b²+1) with b>a, plus a power-saving bound under the abc conjecture.
Number Theory 184 (2018), 473–484
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Gaps between divisible terms in $a^2 (a^2 + 1)$
Proves b ≫ a (log a)^{1/8} / (log log a)^{12} whenever a²(a²+1) divides b²(b²+1) with b>a, plus a power-saving bound under the abc conjecture.