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Gerry Myerson
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Reasons behind assuming the existence of SigelSiegel zeros can be used to prove something stronger than assuming GRH?

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Johnny T.
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There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $P(a,q) \ll q^L$ where $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $P(a,q) \ll q^L$ where $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

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Johnny T.
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There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

There are few results that I am aware of where one can prove something stronger by assuming the existence of Siegel zeros than by assuming the GRH. For example Heath-Brown proved the existence of Siegel zeros imply the twin prime conjecture while it is unknown under GRH. Also there is: Let $P(a,q)$ be the least prime $\equiv a \pmod q$. Then assuming GRH we have $L < 2 + \varepsilon$, but assuming the existence of the Sigel zeros we have $L < 2$. (This I just took from Good uses of Siegel zeros?)

I was interested in this phenomenon. I would greatly appreciate if anyone could explain why or give some ideas on why this is the case?

Also this is a question with subjective answer, but when there is a result of this type (where by assuming Siegel zeros one gets even stronger results than assuming GRH), do we generally expect it to be the truth? Thank you. Any comments are appreciated.

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Johnny T.
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Johnny T.
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