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Added a link to the Sage Math Cell
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Pedja
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This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

EDIT

Test implementation in PARI/GP without directly computing the sum.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

EDIT

Test implementation in PARI/GP without directly computing the sum.

Updated the link to the Sage Math Cell
Source Link
Pedja
  • 2.7k
  • 15
  • 26

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test herehere. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

Corrected PARI/GP implementation of the test
Source Link
Pedja
  • 2.7k
  • 15
  • 26

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test herehere. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

This question is related to my previous question.

Can you prove or disprove the following claim:

Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.

You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.

Source Link
Pedja
  • 2.7k
  • 15
  • 26
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