First I will explain why a weaker form is needed. And then I formulate the conjecture (more precisely, the formulation will be clear).

It is related to the question https://math.stackexchange.com/questions/40945/triangular-factorials and several Mathoverflow questions from the comments to that question. A number $m$ is called a triangular factorial if $m=\frac{n(n+1)}{2}=k !$ for some $n,k$. It is an open problem whether the set of triangle factorials is finite. Moreover the only known such numbers are $1, 6, 120$.

But (somewhat surprisingly for me) it can be shown that the ABC conjecture implies that there are only finitely many triangular factorials. Indeed, suppose that for arbitrary large $k,m$ we have $ \frac{n(n+1)}{2}=k!$. Then $n+1=\frac {2k!}{n}$. Let $a=n, b=1, c= \frac {2k!}{n}$. Then by the ABC conjecture $\frac {2k!}{n}<rad(2k!)^2$ where $rad(x)$ is the product of primes dividing $x$. Note that $n\sim \sqrt{2k!}$ and $rad(2k!)=rad(k!)$ is the product of all primes $\le k$ which, by Erdos theorem $\sim e^{k}$. Thus we have $\sqrt{2k!}< e^{2k}$ which is impossible for big enough $k$. Recall that $2k!\sim 2\sqrt{2\pi k}\, e^{k\log k-k}$.

Question: In the proof above what seems to be a very weak version of the ABC conjecture is used (instead of $rad(abc)^{1+\epsilon}$ one can take a much bigger function in $rad(abc)$). Perhaps that version can be proved easier than the original ABC conjecture?

Edit: It is easy to see that in the version of ABC conjecture used here, $b=1$. Perhaps that makes the conjecture easier? So we can formulate

A conjecture For every constant $d<\frac 12$ there are only finite number of natural $a$ such that $$a>rad(a(a+1))^{d\log\log a}.$$ Note that the exponent in the right hand side may have to be a little different.

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    $\begingroup$ Yes, I do. Also, I need to write ten characters. $\endgroup$ Jun 15, 2017 at 2:53
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    $\begingroup$ I thought there is a dichotomy: a written proof can be either correct or not correct. It seems that there exists the third option. It is not related to the question, though. $\endgroup$
    – user6976
    Jun 15, 2017 at 3:03
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    $\begingroup$ The product of the primes up to $k$ is asymptotically $e^k$, not $e^{2k}$, and I think this is a simple consequence of the Prime Number Theorem and known before Erdos came along. $\endgroup$ Jun 15, 2017 at 4:08
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    $\begingroup$ So he left out two words: which, by Erdos theorem has square $\sim e^{2k}.$ $\endgroup$ Jun 15, 2017 at 5:10
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    $\begingroup$ Your conjecture simply means that rad$(a(a+1))$ being bounded has finitely many solutions. This is true by the finiteness of solutions to the $S$-unit equation. But this doesn't help with the question you started with -- your reformulated conjecture is not strong enough for that application. $\endgroup$
    – Lucia
    Jun 20, 2017 at 4:46

1 Answer 1


For $a> e$, we have $\log \log a > 0$ and hence $x^{\log \log a}$ is an increasing function, so $$a>\mathrm{rad}(a(a+1))^{\frac {\log a}{\log\log a}}$$

then implies that $a^{\log \log a} > \mathrm{rad}(a(a+1))^{\log a}$. Note that $a^{\log \log a} = (e^{\log a})^{\log \log a} = (e^{\log \log a})^{\log a} = (\log a)^{\log a}$.

So we have $(\log a)^{\log a} > \mathrm{rad}(a(a+1))^{\log a}$,
so $(\log a) > \mathrm{rad}(a(a+1))$, so with $b=1$ and $c=a+1$ we have $$c > a > \exp(\mathrm{rad}(abc))$$

but for every $\varepsilon > 0$ there is a $K$ such that $c < \exp(K\,\mathrm{rad}(abc)^{\frac13+\varepsilon})$ (Stewart & Yu 2001) , which gives a contradiction for sufficiently large $\mathrm{rad}(abc)$. So we can just show that for every constant $L$, we have that $L > \mathrm{rad}(abc) = \mathrm{rad}(a(a+1))$ for only finitely many $a$.

This again follows form this result by Stewart & Yu (2001). If there were an infinitude of $a$ such that $L > \mathrm{rad}(a(a+1))$, since every such triple has a different value of $c$, there are arbitrarily large values of $c$ such that $L > \mathrm{rad}(abc)$, so $c < \exp(K\,\mathrm{rad}(abc)^{\frac13+\varepsilon}) < \exp(K\cdot L^{\frac13+\varepsilon})$, for arbitrarily large values of $c$, but the last is a constant, contradiction.

  • $\begingroup$ Thank you for the answer. It does resolve the wrong A conjecture. The conjecture has been improved. $\endgroup$
    – user6976
    Jun 20, 2017 at 13:41

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