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  • $\begingroup$ @Kálmán: I am not sure if you understood what kiskis said below. Heath-Brown proved that the existence of an exceptional zero implies the twin prime conjecture: Prime twins and Siegel zeros, Proc. London Math. Soc. (3) 47 (1983), no. 2, 193-224. $\endgroup$ Commented May 19, 2013 at 21:44
  • $\begingroup$ @GH: I mentioned this result in my question above, but "an" exceptional zero isn't quite right: H-B crucially relies on an infinitude of them. What kiskis was hinting at was a possible unconditional proof of the twin prime conjecture though (unless I misunderstood), to use the result of H-B for that you need to show that the conjecture holds also conditioned on the absence of (an infinitude of) Siegel zeros. $\endgroup$ Commented May 19, 2013 at 21:59
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    $\begingroup$ @Kálmán: The notion of exceptional zero does not make sense for a single zero (this is also what Terry Tao tried to say). At any rate, HB's result can be formulated as: if the twin prime conjecture is false, then there is no exceptional zero (with an appropriate constant in the definition of the exceptional zero). In other words, there is $c>0$ such that if there exist a quadratic $χ$ modulo q and a real zero $\sigma>1−c/\log q$ of $L(s,\chi)$, then the twin prime conjecture is true. I continue in next comment. $\endgroup$ Commented May 19, 2013 at 23:00
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    $\begingroup$ Also, I think your opening line "What is known to follow from the existence of Siegel zeros?" does not reflect what you want (in the light of your comments). You really want: What is known to follow from both the existence and the absence of Siegel zeros. That is, what can we prove unconditionally by using the notion of Siegel zeros on the way. $\endgroup$ Commented May 19, 2013 at 23:01
  • $\begingroup$ @GH: The constant $c$ in your statement above (essentially Theorem 2 of HB's paper) is not computable however and it depends on whether the infinite sequence of zeros HB used in his Theorem 1 exists (if it doesn't, we just put $c$ small enough that no zero as specified exists and the implication is trivially true). [HB has a warning about this right after stating Theorem 2.] For my question to make proper sense you should assume that $c$ is given once and for all (cf. my first comment to Tao's answer below), in that case there is no trouble with the notion of a single Siegel zero. $\endgroup$ Commented May 20, 2013 at 0:10