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Science Forum Index » Mathematics Forum » sqrt(8n^2+1) is integer
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| Gunnar |
 Posted: Fri Dec 19, 2003 4:02 pm |
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Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way? |
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| Rob Johnson |
| Posted: Fri Dec 19, 2003 4:02 pm |
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In article <ZHJEb.42905$dP1.167901@newsc.telia.net>,
Gunnar <gunnar@gunix.dk> wrote:
Quote: Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
Starting with a_0 = 0 and a_1 = 1, use a_n = 6 a_{n-1} - a_{n-2}.
0,1,6,35,204,1189,6930,40391,235416,1372105,etc.
This should give you all of them. This is derived from the continued
fraction for sqrt( = (2,1,4,1,4,1,4,...). In fact the sequence above
is the sequence of denominators of every second partial quotient. It
can be shown that if p/q - sqrt( < 1/(2q^2) then p/q is a partial
quotient of the continued fraction for sqrt(.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying |
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| Charlie Johnson |
| Posted: Fri Dec 19, 2003 4:33 pm |
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"Gunnar" <gunnar@gunix.dk> wrote in message
news:ZHJEb.42905$dP1.167901@newsc.telia.net...
Quote: Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
Mathematical Induction.
Lurch |
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| Guest |
| Posted: Fri Dec 19, 2003 5:07 pm |
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On 19 Dec 2003, Gunnar wrote:
Quote: Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
sqrt(8n^2+1) = m, where m is an integer.
8n^2 + 1 = m^2
8n^2 = m^2 - 1
n^2 = (m^2 - 1)/8
n = sqrt((m^2 - 1)/, let m vary over the integers
That gives all values of n to which m is an integer. |
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| Lynn Kurtz |
| Posted: Fri Dec 19, 2003 5:17 pm |
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On Fri, 19 Dec 2003 21:33:10 GMT, "Charlie Johnson"
<cj-bubba@bite.mindspring.com> wrote:
Quote:
"Gunnar" <gunnar@gunix.dk> wrote in message
news:ZHJEb.42905$dP1.167901@newsc.telia.net...
Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
Mathematical Induction.
Lurch
Induction???!! You are kidding of course.
I know it isn't responsive to your question, but I ran it through
MAPLE and up through 10,000, only n = 1, 6, 35, 204, 1189, 6930 work.
Not too many that's for sure.
--Lynn |
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| che phip |
| Posted: Fri Dec 19, 2003 5:26 pm |
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Charlie Johnson a écrit:
Quote: "Gunnar" <gunnar@gunix.dk> wrote in message
news:ZHJEb.42905$dP1.167901@newsc.telia.net...
Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
Mathematical Induction.
Lurch
Hello,
Pell Equation m^2 - 8*n^2 = 1 (search Google for details)
This one is used to find the square = triangle numbers.
n=1,6,35,204 ...
n_(i+2)=6*n_(i+1) - n_i with n_0 = 1 and n_1 = 6
or if you prefer getting directly n_i with
m + n*sqrt( = (3 + sqrt( )^i
Otherwise, just "Mathematical Induction" is not enough...
regards
--
chephip at free dot fr |
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| Hugo Pfoertner |
| Posted: Fri Dec 19, 2003 5:35 pm |
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Gunnar wrote:
Quote:
Hello.
Is there a way to find the values of n for which sqrt(8n^2+1) is an
integer? Or is testing n=1,2,3,4... the only way?
Look at http://www.research.att.com/projects/OEIS?Anum=A001109
a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2).
8*a(n)^2 + 1 is a perfect square.
0,1,6,35,204,1189,6930,40391,235416,1372105,7997214,....
with lots of formulas, links and references.
Hugo Pfoertner |
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| Jpr2718 |
| Posted: Fri Dec 19, 2003 11:42 pm |
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Gunnar gunnar@gunix.dk wrote:
Quote: Is there a way to find the values of n
for which sqrt(8n^2+1) is an integer?
See methods for solving the Pell equation given in
Solving the generalized Pell equation - PDF File
at
http://hometown.aol.com/jpr2718/
to solve x^2 - 8y^2 = 1.
Others have posted correct solutions.
John Robertson |
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| Gunnar |
| Posted: Sat Dec 20, 2003 3:44 am |
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Thanks for all the answers and ruining my X-mas (now I just have to study
this)
Quote: See methods for solving the Pell equation given in
Solving the generalized Pell equation - PDF File
at
http://hometown.aol.com/jpr2718/
to solve x^2 - 8y^2 = 1.
Others have posted correct solutions.
John Robertson |
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