Author |
Topic: 3 Diophantine Equations (Read 1226 times) |
|
ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
3 Diophantine Equations
« on: Feb 1st, 2005, 6:59am » |
 Quote  Modify
|
Find all positive integers a,b,c such that
1) 1/a + 1/b = 1/c
2) 1/a + 1/b + 1/c = 1
3) (1 + 1/a) (1+ 1/b) (1 + 1/c) = 3
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
Kaotis
Newbie
Gender:
Posts: 18
|
|
Re: 3 Diophantine Equations
« Reply #1 on: Feb 1st, 2005, 2:33pm » |
Quote Modify
|
from a wired calculating i recived that there is no selution...
but it just me...
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: 3 Diophantine Equations
« Reply #2 on: Feb 1st, 2005, 2:43pm » |
Quote Modify
|
::
1) and 2) ==> 1/c+1/c = 1 => c = 2
1) becomes: 1/a + 1/b = 1/2
a=2 => 1/b would be 0
a=3 => 1/b = 1/6 => b=6
a=4 => 1/b = 1/4 => b=4
a=5 => 1/b = 3/10
a=6 => 1/b = 1/3 => b=3
In short (a,b,c) = (6,3,2), (4,4,2), (3,6,2)
3) becomes: (1 + 1/a) (1+ 1/b) = 2
(7/6) (4/3) = 14/9
(5/4) (6/4) = 15/8
(4/3) (7/6) = 14/9
There is no solution?
::
|
|
IP Logged |
|
|
|
Kaotis
Newbie
Gender:
Posts: 18
|
|
Re: 3 Diophantine Equations
« Reply #3 on: Feb 1st, 2005, 2:46pm » |
Quote Modify
|
Guess so, I've recived the same result.
|
|
IP Logged |
|
|
|
Aryabhatta
Uberpuzzler
Gender:
Posts: 1321
|
|
Re: 3 Diophantine Equations
« Reply #4 on: Feb 1st, 2005, 3:22pm » |
Quote Modify
|
Did T&B mean simultaneous equations or 3 separate ones to be solved? From the title of his other thread it looks like he wants 3 separate equations to be solved.
|
| « Last Edit: Feb 1st, 2005, 3:32pm by Aryabhatta » |
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: 3 Diophantine Equations
« Reply #5 on: Feb 1st, 2005, 3:48pm » |
Quote Modify
|
(1) This one is quite nice:
We need ab = c(a+b).
Write a=dx, b=dy, with gcd(x,y)=1, so this becomes
d2xy = cd(x+y).
As gcd(x,x+y)=gcd(x,y)=gcd(y,x+y)=1, (x+y)|d, say d=k(x+y). Thus
(a,b,c) = (kx(x+y), ky(x+y), kxy)
is a complete parameterization.
(2) and (3) are finite calculations. Suppose a[le]b[le]c.
(2)We must have 1<a<4.
If a=3, we must have b=c=3 as well, so we get (3,3,3).
If a=2, we have 1/b + 1/c = 1/2, so 2<b<5. This gives (2,3,6) and (2,4,4).
(3)Since 3 [le] (1+1/a)3, we must have a<3.
Case a=1: 3/2 = (1+1/b)(1+1/c) [le] (1+1/b)2, so b<5. We get the solutions (1,3,8 ) and (1,4,5).
Case a=2: 2 = (1+1/b)(1+1/c) [le] (1+1/b)2, so b<3. This gives (2,2,3).
#2 reminds me of an equation that's used in classifying finite subgroups of SO3:
2 - 2/N = [sum]1-1/rp,
where N=|G|, rp=|Gp|>1 is the order of the stabilizer of a pole p, and the sum is over 1 pole per orbit.
In particular this allows you to determine all Platonic solids, but there are other ways of doing that (e.g., V-E+F=2, which is yet another Diophantine equation).
|
| « Last Edit: Feb 1st, 2005, 4:11pm by Eigenray » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print