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Re: Coprime Arithmetic
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 In math_club@yahoogroups.com, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:>
> Solve this prob: if a â" b = 2n where n is
> any integer then a and b is coprime.
>
>
>
> ________________________________
> From: Tanvir Prince <tanvirp123@...>
> To: math_club@yahoogroups.com
> Cc: Prince <tanvirp123@...>
> Sent: Tue, October 20, 2009 7:52:40 AM
> Subject: Re: [math_club] Modular arithmatic prob
>
>
> we will use the following theorem which is a generalization of Fermat's theorem:
>
> If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
> here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.
>
> So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
>
> Now since p>3 , p and 6 are relatively prime.
> So by the above theorem:
>
> p^(phi(6))=p^ 2 = 1 (mod 6)
> add 1 to both side:
> p^2+1 = 2 (mod 6)
>
> So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
> This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or
> 2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)
> but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)
> And we are done.
>
>  On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
>
>
> >From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
> >Subject: [math_club] Modular arithmatic prob
> >To: math_club@yahoogrou ps.com
> >Date: Monday, October 19, 2009, 9:15 AM
> >
> >
> >
> >Can anyone prove this assertion: 2p â¡ p2 + 1(mod 6), where p is any prime greater than 3.
> >
> >
> >
> >
> ________________________________
> From: avik roy <avik_3.1416@ yahoo.co. in>
> >To: math_club@yahoogrou ps.com
> >Sent: Fri, October 16, 2009 12:04:49 AM
> >Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
> >
> >
> >
> >I am not using that fact. The fact that 1+ik = 1ik leads ta calculation as follows...
> >
> >1+ik / 1ik =1
> >
> >=>1+ik^2 / (1+ik)(1ik)  =1
> >=> 1+ik^2 = 1 + k^2
> >
> > On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
> >
> >
> >>From: Tanvir Prince <tanvirp123@yahoo. com>
> >>Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
> >>To: math_club@yahoogrou ps.com
> >>Date: Thursday, 15 October, 2009, 10:42 PM
> >>
> >>
> >>
> >>Good proof. I have only one concern:
> >>You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
> >>
> >> On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
> >>
> >>
> >>>From: avik roy <avik_3.1416@ yahoo.co.. in>
> >>>Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
> >>>To: "mathclub clubmath" <math_club@yahoogrou ps.com>
> >>>Date: Thursday, October 15, 2009, 12:36 AM
> >>>
> >>>
> >>>
> >>>Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
> >>>
> >>>let,
> >>>z1 = x
> >>>z2 = ikx [k is any arbitrary real constant]
> >>>
> >>>z1^2 + z2^2 = (1+k^2)*x^2
> >>>[u need not use the Pythagoras here, the concept of length does the job]
> >>>
> >>>z1+z2 = x1+ ik
> >>>
> >>>A simple observation leads that z1+z2 = z1z2
> >>>=>1+ik=1ik
> >>>A simple calculation proves that, i+ik^2 = 1+k^2
> >>>
> >>>This gives us that z1^2 +z2^2 = z1 + z2^2
> >>>
> >>>I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
> >>>
> >>>
> >>>________________________________
> >>>From cricket scores to your friends. Try the Yahoo! India Homepage!
> >>
> >________________________________
> >Add whatever you love to the Yahoo! India homepage. Try now!
> >
>i cant seem to understand wd u written in da equation there r crazy notations 0 Attachment
The assumption is not right. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done. On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogroups.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps.com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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Yes Tanvir you are right. I forgot to quote a and b as odds. But the fact of letting d as the divisor of both a and b seems very contradicting to me. In my statement a and b is coprime if so then there cannot be any integer d that divides a and b at the same time and even if d = 1 because (a, 1) is always coprime where a > 1 and any integers because they don't have the greatest common divisors. Don't you think its absurd!!!!??!! I can give you a couter example of your deduction that you used to proof the assumption. You said that d divides both a and b and 2^n but 7  3 = 2^2 so do we find any d that divides both 7, 3 and 4????!!!!!!!??????!!!!?? Your proof is completely wrong and contradicting. You proof must need a far rectification. Mail me. Thanks.
From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Sent: Mon, November 2, 2009 1:41:47 AM
Subject: Re: [math_club] Coprime Arithmetic
The assumption is not right.. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done. On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps.com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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Tanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be coprime. Mail me the proof.
From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Sent: Mon, November 2, 2009 1:41:47 AM
Subject: Re: [math_club] Coprime Arithmetic
The assumption is not right. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done. On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps..com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact.. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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I think the proof is ok.
The problem is "for odd a and b, if ab=2^n, a and b are coprime"
The proof shows that assuming any common divisor d for a and b, we can show that d=1, thus a and b have to be coprime... On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
Subject: Re: [math_club] Coprime Arithmetic
To: math_club@yahoogroups.com
Date: Monday, 2 November, 2009, 2:08 PMYes Tanvir you are right. I forgot to quote a and b as odds. But the fact of letting d as the divisor of both a and b seems very contradicting to me. In my statement a and b is coprime if so then there cannot be any integer d that divides a and b at the same time and even if d = 1 because (a, 1) is always coprime where a > 1 and any integers because they don't have the greatest common divisors. Don't you think its absurd!!!!?? !! I can give you a couter example of your deduction that you used to proof the assumption. You said that d divides both a and b and 2^n but 7  3 = 2^2 so do we find any d that divides both 7, 3 and 4????!!!!!!! ??????!!! !?? Your proof is completely wrong and contradicting. You proof must need a far rectification. Mail me. Thanks.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Mon, November 2, 2009 1:41:47 AM
Subject: Re: [math_club] Coprime Arithmetic
The assumption is not right.. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done.
 On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps.com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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The same proof is valid in this case....
for any number d to divide a and b, it has to divide p, hence d=1 or p, but a,b<p thus d can't be p and thus d has to be 1.
Q.E.D. On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
Subject: [math_club] Another Coprime Arithmetic
To: math_club@yahoogroups.com
Date: Monday, 2 November, 2009, 2:18 PMTanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be coprime. Mail me the proof.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Mon, November 2, 2009 1:41:47 AM
Subject: Re: [math_club] Coprime Arithmetic
The assumption is not right. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done.
 On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps..com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact.. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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Thanks Avik. Now I understood the proof.
From: avik roy <avik_3.1416@...>
To: math_club@yahoogroups.com
Sent: Mon, November 2, 2009 5:06:45 PM
Subject: Re: [math_club] Another Coprime Arithmetic
The same proof is valid in this case....
for any number d to divide a and b, it has to divide p, hence d=1 or p, but a,b<p thus d can't be p and thus d has to be 1.
Q.E.D. On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Another Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Monday, 2 November, 2009, 2:18 PMTanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be coprime. Mail me the proof.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Mon, November 2, 2009 1:41:47 AM
Subject: Re: [math_club] Coprime Arithmetic
The assumption is not right. For example, 16  8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:Let d divide both a and b. We will show that d = 1 and this will prove our assumption.Since d divide both a and b and ab=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.So we must have k =1 and thus d = 2^0 = 1 and we are done.
 On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Coprime Arithmetic
To: math_club@yahoogrou ps.com
Date: Saturday, October 31, 2009, 6:36 AMSolve this prob: if a – b = 2^{n} where n is any integer then a and b is coprime.
From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: Prince <tanvirp123@yahoo. com>
Sent: Tue, October 20, 2009 7:52:40 AM
Subject: Re: [math_club] Modular arithmatic prob
we will use the following theorem which is a generalization of Fermat's theorem:If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)here phi(n) is the euler phi function which count the number of integer from 1 to n1 which are relatively prime to n.So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.Now since p>3 , p and 6 are relatively prime.So by the above theorem:p^(phi(6))=p^ 2 = 1 (mod 6)add 1 to both side:p^2+1 = 2 (mod 6)So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)This means we need to show 2^p2 is divisible by 6 or 2(2^(p1) 1) is divisible by 6 or2^(p1)  1 is divisible by 3 or 2^(p1) = 1 (mod 3)but 2 = 1 (mod 3). So 2^(p1) = (1)^(p1) = 1 (mod 3) (since p being odd, p1 must be even)And we are done.
 On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: [math_club] Modular arithmatic prob
To: math_club@yahoogrou ps.com
Date: Monday, October 19, 2009, 9:15 AMCan anyone prove this assertion: 2^{p} ≡ p^{2} + 1(mod 6), where p is any prime greater than 3.
From: avik roy <avik_3.1416@ yahoo.co. in>
To: math_club@yahoogrou ps..com
Sent: Fri, October 16, 2009 12:04:49 AM
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
I am not using that fact.. The fact that 1+ik = 1ik leads ta calculation as follows...
1+ik / 1ik =1
=>1+ik^2 / (1+ik)(1ik)  =1
=> 1+ik^2 = 1 + k^2
 On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
From: Tanvir Prince <tanvirp123@yahoo. com>
Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: math_club@yahoogrou ps.com
Date: Thursday, 15 October, 2009, 10:42 PMGood proof. I have only one concern:You are using the fact that zz* = z^2 where z* is the conjugate of z and the fact that zz*=z^2 depends on pythagorean theorem
 On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
From: avik roy <avik_3.1416@ yahoo.co. in>
Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
To: "mathclub clubmath" <math_club@yahoogrou ps.com>
Date: Thursday, October 15, 2009, 12:36 AMHere, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
let,
z1 = x
z2 = ikx [k is any arbitrary real constant]
z1^2 + z2^2 = (1+k^2)*x^2
[u need not use the Pythagoras here, the concept of length does the job]
z1+z2 = x1+ ik
A simple observation leads that z1+z2 = z1z2
=>1+ik=1ik
A simple calculation proves that, i+ik^2 = 1+k^2
This gives us that z1^2 +z2^2 = z1 + z2^2
I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
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