12 x^2 - 7x + 1 = 0 (mod p) has solution for every prime p
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From: John on 29 Jun 2010 15:21 Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod p) has solution. It seems that quadratic residue theory should help here but first we need to convert it to other form... Any hints ?
From: Pubkeybreaker on 29 Jun 2010 15:38 On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > p) has solution. > It seems that quadratic residue theory should help here but first we > need to convert it to other form... > > Any hints ? Are you sure that your polynomial is irreducible??????..........
From: John on 29 Jun 2010 16:14 On Jun 29, 10:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > > > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > > p) has solution. > > It seems that quadratic residue theory should help here but first we > > need to convert it to other form... > > > Any hints ? > > Are you sure that your polynomial is irreducible??????.......... No. It equals (3x - 1)(4x - 1)...
From: Pubkeybreaker on 29 Jun 2010 16:38 On Jun 29, 4:14 pm, John <to1m...(a)yahoo.com> wrote: > On Jun 29, 10:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > > > > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > > > p) has solution. > > > It seems that quadratic residue theory should help here but first we > > > need to convert it to other form... > > > > Any hints ? > > > Are you sure that your polynomial is irreducible??????.......... > > No. It equals (3x - 1)(4x - 1)... So now you have your answer.
From: John on 29 Jun 2010 16:50
On Jun 29, 11:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > On Jun 29, 4:14 pm, John <to1m...(a)yahoo.com> wrote: > > > On Jun 29, 10:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > > > > > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > > > > p) has solution. > > > > It seems that quadratic residue theory should help here but first we > > > > need to convert it to other form... > > > > > Any hints ? > > > > Are you sure that your polynomial is irreducible??????.......... > > > No. It equals (3x - 1)(4x - 1)... > > So now you have your answer. Thanks. Now I understand. One of the linear congruences has solution. |