Can the ABC conjecture be expanded?

Has anyone considered expanding the range of terms $$a$$ and $$b$$ for each $$c$$?

I have generated triples $$(a, b, c)$$ that form integer triangles including the degenerate case of $$a + b = c$$ such that $$a \le b \le c$$, all terms are pairwise coprime and $$a + b \ge c$$. I have tested the Quality $$q$$ (see Wikipedia for definition: https://en.wikipedia.org/wiki/Abc_conjecture) of all these triples (~ 1.5M) for $$c$$ up to 400 and found that $$q < 2$$.

I conjecture that all such triples extend the terms of the ABC conjecture. Has anyone got the computing power to test if true for $$c \ge 1000$$?

(Added as a post script following counter example comment) - The set of integer triangles can be reduced to just obtuse (C angle) triangles by setting a further constraint that $$a^2+b^2 \le c^2$$. However this is still not sufficient to exclude the family of counter examples commented by @Gerry Myerson ($$c = 2^x \cdot 3^y \cdot 5^z$$ for some x, y, z). Only by setting the further constraint to $$a^u+b^u \le c^u$$, where $$u=1+\frac{1}{\log{(a*b*c)}}$$ is there a chance of satisfying the ABC conjecture with additional terms $$a$$ and $$b$$ other than $$a+b = c$$.

For $$c$$ up to 200 the additional integer triangles with the further constraint above that satisfy $$1 (where $$q$$ is the Quality function) are the triples

(49, 81, 125), (9, 121, 128), (49, 81, 128)

compared to

(1, 8, 9), (5, 27, 32), (1, 48, 49), (1, 63, 64), (32, 49, 81), (1, 80, 81), (4, 121, 125), (3, 125, 128)

where $$a+b=c$$.

Can anyone find counter examples to the above with this further constraint?

• $2^{20}<3^{13}<5^9$, $2^{20}+3^{13}>5^9$, quality is $\log(5^9)/\log30$ is considerably bigger than 2. – Gerry Myerson Nov 30 '18 at 10:58
• Using your counter example there are infinitely many more - Thanks. – Frank M Jackson Nov 30 '18 at 11:58
• "Can anyone find counter examples to the above with this further constraint?" It'd be nice (to me) to have the questions formulated explicitly, without all these PRONOUNS? I'd prefer NOUNS. – Wlod AA Dec 10 '18 at 19:31