Let $E/\mathbb{Q}$ be an elliptic curve. Recall that Szpiro's conjecture says that for every $\epsilon > 0$, there exists $C_\epsilon$ such that $$ |\Delta_E| \leq C_\epsilon(N_E)^{6 + \epsilon}, $$ where $\Delta_E$ is the minimal discriminant of $E$ and $N_E$ is the conductor of $E$.

One consequence of Szpiro's conjecture is Fermat's Last Theorem for sufficiently large exponents and the $ABC$-conjecture for the exponent $3/2$.

My question is are there any other known consequences of Szpiro's conjecture (references are appreciated)?

EDIT: Preferably a consequence of the Szpiro conjecture that is distinct from a consequence of the $ABC$-conjecture.

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    $\begingroup$ Fermat's Last Theorem and the abc-conjecture --- isn't that enough? You want it to lead to a cure for cancer? $\endgroup$ – Gerry Myerson May 20 '12 at 9:18
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    $\begingroup$ I wanted an application of Szpiro's conjecture that was distinct from the ABC conjecture in my write-up of the Szpiro's conjecture. Interestingly though I once attended a colloquium by Professor Julie Mitchell where her she was talking about her research that led to a new cancer drug. news.wisc.edu/14453 $\endgroup$ – Eugene May 20 '12 at 12:31
  • $\begingroup$ Cool---thanks for the link. $\endgroup$ – Gerry Myerson May 20 '12 at 13:00

The abstract of a paper by Joe Silverman at http://arxiv.org/abs/0908.3895 says,

"It is known that Szpiro's conjecture, or equivalently the ABC-conjecture, implies Lang's conjecture giving a uniform lower bound for the canonical height of nontorsion points on elliptic curves. In this note we show that a significantly weaker version of Szpiro's conjecture, which we call "prime-depleted," suffices to prove Lang's conjecture."

For what it's worth (probably less than epsilon), Szpiro's conjecture has a Facebook page, http://www.facebook.com/pages/Szpiros-conjecture/139143682780725?nr=133320400042160.

  • $\begingroup$ I especially liked your "less than epsilon" comment. Thanks. $\endgroup$ – Eugene May 20 '12 at 12:33
  • $\begingroup$ You may get more (and quite possibly better) answers if you hold off a bit on accepting this one. $\endgroup$ – Gerry Myerson May 20 '12 at 13:03
  • $\begingroup$ Ok then. I shall hold off a bit then. Thanks again for the help though. $\endgroup$ – Eugene May 20 '12 at 13:10

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