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Science Forum Index » Mathematics Forum » Hard diophantine equation
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| Arne Smeets |
 Posted: Mon Feb 21, 2005 11:38 am |
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Solve in positive integers: 2^x = 5^y + 3.
[Problem due to Charles Leytem] |
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| Mok-Kong Shen |
| Posted: Mon Feb 21, 2005 5:46 pm |
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Arne Smeets wrote:
Quote: Solve in positive integers: 2^x = 5^y + 3.
(3,1) (7,3) |
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| Guest |
| Posted: Mon Feb 21, 2005 5:56 pm |
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Arne Smeets wrote:
Quote: Solve in positive integers: 2^x = 5^y + 3.
[Problem due to Charles Leytem]
er...
x = 3, y = 1
x = 7, y = 3
?
was there a typo somewhere? |
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| Guest |
| Posted: Mon Feb 21, 2005 6:02 pm |
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matt271829-n...@yahoo.co.uk wrote:
Quote: Arne Smeets wrote:
Solve in positive integers: 2^x = 5^y + 3.
[Problem due to Charles Leytem]
er...
x = 3, y = 1
x = 7, y = 3
?
was there a typo somewhere?
Or ... afterthought ... I suppose maybe you are looking for a recipe
that will "easily" generate ALL solutions? |
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| Timothy Little |
| Posted: Tue Feb 22, 2005 1:52 am |
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matt271829-news@yahoo.co.uk wrote:
Quote: x = 3, y = 1
x = 7, y = 3
Or ... afterthought ... I suppose maybe you are looking for a recipe
that will "easily" generate ALL solutions?
That is all of them, unless you count y=0 in the "positive" integers.
- Tim |
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| Arne Smeets |
| Posted: Tue Feb 22, 2005 3:31 am |
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Of course I am looking for all solutions, and I know that (3, 1) and (7, 3) are the only solutions, but what about a
PROOF?
On 22 Feb 2005, Timothy Little wrote:
Quote: matt271829-news@yahoo.co.uk wrote:
x = 3, y = 1
x = 7, y = 3
Or ... afterthought ... I suppose maybe you are looking for a recipe
that will "easily" generate ALL solutions?
That is all of them, unless you count y=0 in the "positive" integers.
- Tim |
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| Guest |
| Posted: Tue Feb 22, 2005 6:15 am |
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Timothy Little wrote:
Quote: matt271829-news@yahoo.co.uk wrote:
x = 3, y = 1
x = 7, y = 3
Or ... afterthought ... I suppose maybe you are looking for a
recipe
that will "easily" generate ALL solutions?
That is all of them, unless you count y=0 in the "positive" integers.
- Tim
Right, thanks. I didn't get any further than the two obvious ones! |
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| Dave Rusin |
| Posted: Tue Feb 22, 2005 3:16 pm |
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In article <200502211759.j1LHx9N14608@proapp.mathforum.org>,
Arne Smeets <ernie_smeets@hotmail.com> wrote:
Quote: Solve in positive integers: 2^x = 5^y + 3.
[Problem due to Charles Leytem]
Among other possible solutions: consider the possible congruence
classes of x and y modulo 2 and 3 respectively; then you
are looking for integer solutions to equations of the form
{1 or 2} X^2 = {1 or 5 or 25} Y^3 + 3
with the additional constraints that X must be a power of 2
and Y must be a power of 5. Well, even without those additional
constraints, there are only finitely many integer pairs (X,Y) on
each of these -- or any other -- elliptic curves. Find the integer
points and retain the ones for which X and Y are powers as
appropriate.
Or in a similar way view the problem as a set of 9 Thue equations
to be solved: {1 or 2 or 4} X^3 - {1 or 5 or 25} Y^3 = 3 .
Some years back we had a Diophantine problem which likewise could
be approached in many ways: x^2+7=8p^n . See
http://www.math-atlas.org/99/exponent_dio
dave |
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| Timothy Little |
| Posted: Tue Feb 22, 2005 4:04 pm |
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Arne Smeets wrote:
Quote: Of course I am looking for all solutions, and I know that (3, 1) and
(7, 3) are the only solutions, but what about a PROOF?
My (inelegant) proof:
For any solution, we must have 5^y+3 = 0 mod 2^x. We know that x=7,
y=3 is a solution, and so in particular 5^y+3 = 0 mod 128. That means
y = 3 mod 64.
Now consider x>=8: we must have 5^y+3 = 0 mod 256 and we know that
y = 3 mod 64, so we must have either y = 3 or 67 mod 128. However in
both cases 5^y+3 = 128 mod 256, so no such y exists for x >= 8.
- Tim |
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| Gerry Myerson |
| Posted: Tue Feb 22, 2005 5:08 pm |
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In article <slrnd1n7hu.f4d.tim-via-n.i.net@soprano.little-possums.net>,
Timothy Little <tim-via-n.i.net@little-possums.net> wrote:
Quote: Arne Smeets wrote:
Of course I am looking for all solutions, and I know that (3, 1) and
(7, 3) are the only solutions, but what about a PROOF?
My (inelegant) proof:
For any solution, we must have 5^y+3 = 0 mod 2^x. We know that x=7,
y=3 is a solution, and so in particular 5^y+3 = 0 mod 128. That means
y = 3 mod 64.
Is that right, or does it only mean y = 3 mod 32?
The group of units in Z / 128Z is Z / 2Z + Z / 32Z, not Z / 64Z.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) |
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| Timothy Little |
| Posted: Tue Feb 22, 2005 6:00 pm |
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Gerry Myerson wrote:
Quote: Is that right, or does it only mean y = 3 mod 32?
The group of units in Z / 128Z is Z / 2Z + Z / 32Z, not Z / 64Z.
Oops, you're right. Scratch that "proof".
- Tim |
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