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Home » Olympiad Section » Number Theory » Number Theory Unsolved Problems
 
easy pq|2^p+2^q
Moderators: High School Olympiad Moderators, amfulger, Arne, darij grinberg, freemind, harazi, Megus, N.T.TUAN, orl, pbornsztein, ZetaX
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_tequilla
P versus NP
P versus NP

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Poland
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#1
easy pq|2^p+2^q
find all pairs p,q prime number , if:
2^p+2^q = 0 \pmod {pq}

PostPosted: Wed Feb 16, 2005 12:08 pm  Back to top 
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Zorro
Hodge Conjecture
Hodge Conjecture

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Poland
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#2
ude this theorem
let r be the smallest solution of n^x \equiv -1 \pmod p, then the solutions of n^x \equiv -1 \pmod p are (2k+1)r, and all the solutions of n^x \equiv 1 \pmod p are 2kr

PostPosted: Wed Feb 16, 2005 12:49 pm  Back to top 
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pbornsztein
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer

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Location: Paris, France
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#3
Have a look here :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20918

Pierre.

PostPosted: Wed Feb 16, 2005 1:00 pm  Back to top 
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_tequilla
P versus NP
P versus NP

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Joined: 13 Jun 2004
Posts: 32
Poland
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#4
so we know that:
0=2^p+2^q=2+2^q \pmod p
2^{q-1}=-1 \pmod p
2^{p-1}=1 \pmod p

so:
q-1=(2k+1)r
p-1=2lr
on the other hand:
q-1=2ms
p-1=(2n+1)s

so
{p-1 \over q-1}={2l \over 2k+1}={2n+1 \over 2m}
Contradiction
so p=2,q=3 p=2,q=2
thx

PostPosted: Wed Feb 16, 2005 1:06 pm  Back to top 
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