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Science Forum Index » Mathematics Forum » Binomial
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| Deep K. Deb |
 Posted: Sun Dec 28, 2003 10:55 am |
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I would greatly appreciate any comment upon the correctness of the
following assertion.
Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =
1
Assertion: The following equality is not possible.
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1) |
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| Ignacio Larrosa Cañestro |
| Posted: Sun Dec 28, 2003 4:39 pm |
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"Deep K. Deb" <deepkdeb@yahoo.com> escribió en el mensaje
news:4a3bc6b5.0312280755.f30b848@posting.google.com...
Quote: I would greatly appreciate any comment upon the correctness of the
following assertion.
Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =
1
Assertion: The following equality is not possible.
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)
sqrt(a) + sqrt(b) = c^(5/2) + 5c^2*d^(1/2) + 10d*c^(3/2)
+ 10c*d^(3/2) + 5c^(1/2)d^2 + d^(5/2)
= (c^2 +10c*d + 5d^2)sqrt(c) + (d^2 + 10c*d + 5c^2)sqrt(d)
Then,
a = c(c^2 +10c*d + 5d^2)^2
b = d(d^2 + 10c*d + 5c^2)^2
a and b aren't squares because c and d aren't.
And gcd(a, b, c, d) = 1 if gcd(c, d) = 1.
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com |
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| Robert Israel |
| Posted: Sun Dec 28, 2003 4:44 pm |
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In article <4a3bc6b5.0312280755.f30b848@posting.google.com>,
Deep K. Deb <deepkdeb@yahoo.com> wrote:
Quote: I would greatly appreciate any comment upon the correctness of the
following assertion.
Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =
1
Assertion: The following equality is not possible.
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)
sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2 |
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| The Ghost In The Machine |
| Posted: Mon Dec 29, 2003 12:00 am |
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In sci.math, Robert Israel
<israel@math.ubc.ca>
wrote
on 28 Dec 2003 21:44:40 GMT
<bsnis8$if1$1@nntp.itservices.ubc.ca>:
Quote: In article <4a3bc6b5.0312280755.f30b848@posting.google.com>,
Deep K. Deb <deepkdeb@yahoo.com> wrote:
I would greatly appreciate any comment upon the correctness of the
following assertion.
Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =
1
Assertion: The following equality is not possible.
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)
sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
(sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3
+ 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2
= sqrt(2)*(2^2+10*3*2+5*3^2)
+ sqrt(3)*(3^2+10*2*3+5*2^2)
= sqrt(2)*109+sqrt(3)*89
= sqrt(23762)+sqrt(23763)
The solution checks out, which leads me to believe that
there was a problem in the original specification.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless. |
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| Ronald Bruck |
| Posted: Mon Dec 29, 2003 12:39 am |
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In article <lnd4c1-dgm.ln1@lexi2.athghost7038suus.net>, The Ghost In
The Machine <ewill@sirius.athghost7038suus.net> wrote:
Quote: In sci.math, Robert Israel
israel@math.ubc.ca
wrote
on 28 Dec 2003 21:44:40 GMT
bsnis8$if1$1@nntp.itservices.ubc.ca>:
In article <4a3bc6b5.0312280755.f30b848@posting.google.com>,
Deep K. Deb <deepkdeb@yahoo.com> wrote:
I would greatly appreciate any comment upon the correctness of the
following assertion.
Given a, b, c, d are non-square integers > 0 such that (a, b, c, d) =
1
Assertion: The following equality is not possible.
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5 (1)
sqrt(23762) + sqrt(23763) = (sqrt(2) + sqrt(3))^5
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
(sqrt(2) + sqrt(3))^5 = 2^2.5 + 5*2^2*3^.5 + 10*2^1.5*3
+ 3^2.5 + 5*3^2*2^.5 + 10*3^1.5*2
= sqrt(2)*(2^2+10*3*2+5*3^2)
+ sqrt(3)*(3^2+10*2*3+5*2^2)
= sqrt(2)*109+sqrt(3)*89
= sqrt(23762)+sqrt(23763)
The solution checks out, which leads me to believe that
there was a problem in the original specification.
One never knows about what's posted to Usenet :-(
<shrug>
What's even stranger is that if gcd(c,d) = 1 then
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5
for a = c(c^2 + 10cd + 5d^2)^2, b = d(5c^2 + 10cd + d^2)^2,
and gcd(a,b,c,d) = 1. So the equality is ALWAYS possible.
</shrug>
Perhaps what Robert is doing is giving an example where gcd(a,b) = 1
too? I haven't checked for smaller examples...
--Ron Bruck |
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| Edwin Clark |
| Posted: Mon Dec 29, 2003 3:03 pm |
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"Ronald Bruck" wrote
Quote:
One never knows about what's posted to Usenet :-(
shrug
What's even stranger is that if gcd(c,d) = 1 then
sqrt(a) + sqrt(b) = (sqrt(c) + sqrt(d))^5
for a = c(c^2 + 10cd + 5d^2)^2, b = d(5c^2 + 10cd + d^2)^2,
and gcd(a,b,c,d) = 1. So the equality is ALWAYS possible.
/shrug
Maybe he meant to assume that a,b,c,d are square free?
--Edwin Clark |
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