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‘GAPS BETWEEN DIVISIBLE TERMS IN $a^{2}(a^{2}+1)$ ’
Bulletin of the Australian Mathematical Society ( IF 0.7 ) Pub Date : 2020-04-24 , DOI: 10.1017/s0004972720000362
TSZ HO CHAN

Suppose $a^{2}(a^{2}+1)$ divides $b^{2}(b^{2}+1)$ with $b>a$ . We improve a previous result and prove a gap principle, without any additional assumptions, namely $b\gg a(\log a)^{1/8}/(\log \log a)^{12}$ . We also obtain $b\gg _{\unicode[STIX]{x1D716}}a^{15/14-\unicode[STIX]{x1D716}}$ under the abc conjecture.



中文翻译:

$ a ^ {2}(a ^ {2} +1)$中 可分词之间的间隙”

假设 $ a ^ {2}(a ^ {2} +1)$ $ b> a $除以$ b ^ {2}(b ^ {2} +1) $ 。我们改进了先前的结果并证明了间隙原理,没有任何其他假设,即 $ b \ gg a(\ log a)^ {1/8} /(\ log \ log a)^ {12} $ 。我们还根据abc猜想获得 $ b \ gg _ {\ unicode [STIX] {x1D716}} a ^ {15 / 14- \ unicode [STIX] {x1D716}} $

更新日期:2020-04-24
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