Timeline for answer to Does this specific polynomial identity generate infinitely many triples satisfying $c > \text{rad}(abc)$? by joro

8 events

when toggle format	what		by	license	comment
3 hours ago	comment	added	Đào Thanh Oai		@joro Could you combine the results from the identity here and your answer to create a new study related to this topic?
2 days ago	comment	added	joro		@ĐàoThanhOai I am not working with p,q. If you fix v=1 u=B^k, then deg(rad(abc))=33 indeed, but the radical is divisible by B^k and rad(B^k)<=B and we have degrees of freedom to change B,k and get the radical divisible by m^r for integers m,r.
2 days ago	comment	added	Đào Thanh Oai		@joro If we fixed $q=const$ or $p=const$, we have $deg(c)=32$ and $deg(rad(abc))=33$
Mar 28 at 17:47	comment	added	joro		@ĐàoThanhOai Look at the elliptic curve over the rationals: $X^4+k b^4=Y^2$
Mar 28 at 17:18	comment	added	Đào Thanh Oai		@Wojowu Empirical evidence from further iterations suggests that the ratio $\frac{\ln(b)}{\ln(\text{rad}(b))}$ asymptotically approaches 4. Magnitude-wise, this implies that the recurrence generates infinitely many identities behaving like $X^2 + \text{rad}(b)^4 = Z^4$. Is there a rigorous mathematical proof or a known theorem that formally explains this limit?
Mar 28 at 16:40	vote	accept	Đào Thanh Oai
Mar 28 at 15:03	comment	added	Wojowu		More explicitly, for your choice of $u,v$, we get $c\approx B^{32k}$ and $rad(abc)\lesssim rad(B^{4k}-1)B^{28k+1}$. Taking e.g. $B=2$ and $3\mid k$ we get $rad(B^{4k}-1)<B^{4k}/3$ so the inequality for large $k$.
Mar 28 at 14:07	history	answered	joro	CC BY-SA 4.0