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        riddles >> putnam exam (pure math) >> 3 Diophantine Equations
        
(Message started by: THUDandBLUNDER on Feb 1^st, 2005, 6:59am)

Title: 3 Diophantine Equations
Post by THUDandBLUNDER on Feb 1^st, 2005, 6:59am

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

Title: Re: 3 Diophantine Equations
Post by Kaotis on Feb 1^st, 2005, 2:33pm

from a wired calculating i recived that there is no selution...
???
but it just me...

Title: Re: 3 Diophantine Equations
Post by Grimbal on Feb 1^st, 2005, 2:43pm

::[hide]
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? ???

[/hide]::

Title: Re: 3 Diophantine Equations
Post by Kaotis on Feb 1^st, 2005, 2:46pm

Guess so, I've recived the same result.

Title: Re: 3 Diophantine Equations
Post by Aryabhatta on Feb 1^st, 2005, 3:22pm

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.

Title: Re: 3 Diophantine Equations
Post by Eigenray on Feb 1^st, 2005, 3:48pm

(1) This one is quite nice: [hide]
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)[/hide]
is a complete parameterization.

(2) and (3) are finite calculations. Suppose a[le]b[le]c.
(2)[hide]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)[/hide].

(3)[hide]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)[/hide].

#2 reminds me of an equation that's used in classifying finite subgroups of SO₃:
2 - 2/N = [sum]1-1/r_p,
where N=|G|, r_p=|G_p|>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).