Conference rusure::math

    All you have to do is to let x = 1-y-z and substitute into the equation
    xyz=1.  Well you end up with a degree two polynomial equation of y and z.  
    If you solve y in terms of z, you get the delta to be:
     4    3    2
    z - 2z  + z - 4z = 0. 
    
    The zero root doesn't count, and this polynomial obviously does not
    have any rational none-zero roots.  So the conclusion is that there
    ain't no such three number.
    
    Eugene
    

If you allow rational complex numbers,
	x = 1
	y = i
	z = -i
works.
				Jeff

re .1:

	It's not necessary for the `delta' to be zero.
	All that's required is that it be the square of
	a rational number.

    re .3:
    
    You are right.  That was what I meant.  I tried to find solutions of
            2
    f(z) = n  but couldn't find any, and I am pretty confident that it does
    not have rational solutions.
    
    Eugene

Suppose that

     4    3    2
    z - 2z  + z - 4z

is the square of a rational number.  Then wlg

     4    3    2         2        2
    z - 2z  + z - 4z = (z - z - b) .

So
         2             4    3    2
    b = z  - z � sqrt(z - 2z  + z - 4z)		(which is no help)
or
                        3   2
    z = (b + 2 � sqrt(2b + b + 4b + 4))/(2b).

Thus, we need to find a value for b such that

      3   2
    2b + b + 4b + 4

is the square of a rational number.

>Thus, we need to find a value for b such that

>      3   2
>    2b + b + 4b + 4

>is the square of a rational number.

This can only happen if b is also rational = p/q. 
The problem is then the same as finding integers p and q for which 

    3    2          2      3
(2 p  + p  q + 4 p q  + 4 q ) q

is the square of an integer.

I doubt it. - Lynn

	re .0

>>Find three real rationals such that both their sum and product 
>>is 1, or prove that no such numbers exist.
>>
>>That is, do there exist rationals x, y, and z such that x + y + z = xyz = 1?

	The last sentence fascinated me, because I recognized this
	as a "famous unsolved problem".  I didn't want to say anything
	until I could check my reference, which I did last night (the
	book is at home so I'll try to remember to enter the identifying
	info tonight).  The article also listed the equivalents:

	a)  Are there three (signed) integers l, m, n such that

		(l + m + n)^3 = lmn

	b)  Are there three (signed) integers a, b, c such that

		a/b + b/c + c/a = 1

	c)  Is there a rational number r such that the polynomial

		x^3 - x^2 + rx - 1

	    has three rational roots?

	Dan

>	The last sentence fascinated me, because I recognized this
>	as a "famous unsolved problem".  

Ok, I confess.  I knew it was an unsolved problem when I posted it here.  I 
did so for three reasons:

1) Because I thought there was a slight chance that some bright person who
didn't know it was unsolved whould take a fresh approach and come up with a
solution.  I have heard that one of the early topologists (I think it was
Moore) used to give unsolved problems as extra credit problems to his students,
and some were actually solved!  The students had the advantage of naivety,
i.e., they lacked all the extra baggage (preconceived notions) of the old
timers.

2) Because most unsolved problems require at least a couple of years of college
mathematcs even to understand the problem, but this one is so simple that it
can be understood by elementary school students.

3) Because I wanted to see the direction the discussion took.  

The mathematics professor who posed this problem to us related the following
accompanying story.

He and several of his graduate student friends where on there way to some math
conference by car.  It was a long drive, so to pass the time they asked each
other difficult mathe problems.  One of them was a real whiz, solving most of
the problems before the others could understand them.  So the professor who
related the story posed the above problem.  

The whiz kid said he had it, then, "no,  wait a minute," and thought some more
...  Then the storyteller said he would give the answer -- "NO!  I almost have
it ..."   "The answer is"  " NO!  don't tell me."  This went on into the night
even after they got to their hotel.   The next morning the whiz, who had been
up all night working on the problem agreed to listen to the answer.  He almost
killed the person who asked the question when he found out it was unsolved.

A related question:  Is it possible for a question of this type to be one of
those that falls through the cracks in the system -- you know, is neither true
nor false with the standard set of axioms and undefined terms?  In other words,
it is unresolvable because if you add one set of axioms consistent with the
peano axioms it is true, but with another set it becomes false.  How can an
existence type conjecture be unresolveable?

Avery

        re .7
        
        The reference was in the book Enrichment Mathematics for
        High School Twenty-eighth Yearbook, (c) 1963 by the
        National Council of Teachers of Mathematics, Inc.,
        Washington, D.C., second printing 1967, chapter 17, "Some
        Unsolved Problems of Arithmetic" by Waclaw Sierpinski,
        Warsaw, Poland.  The author credits the problem to Werner
        Mnich, then a student of the Warsaw University, "then"
        being apparently "a couple of years" before either a
        lecture in 1959 or an article in Scripta Mathematica,
        Vol. 25, No. 2.
        
        Hmmmm...Is that enough to make it "famous"? :-)
        
        re .8
        
A related question:  Is it possible for a question of this type to be one of
those that falls through the cracks in the system -- you know, is neither true
nor false with the standard set of axioms and undefined terms?  In other words,
it is unresolvable because if you add one set of axioms consistent with the
peano axioms it is true, but with another set it becomes false.  How can an
existence type conjecture be unresolveable?
        
        If a solution to the problem in .0 exists, then "plugging
        in the numbers and doing the arithmetic" will provide a
        proof that a solution exists.  That is, the standard
        axioms of arithmetic are strong enough, that if you
        formulate this problem as the "obvious" sentence of first
        order Peano arithmetic, then you can munge "plugging in
        the solution and doing the arithmetic" into a formal
        proof of said sentence.
        
        If no solution exists, then one of four things can
        happen.  First, the standard axioms may be inconsistent
        and prove both the sentence and its negation.  Second,
        the standard axioms are consistent but prove the sentence
        anyway, even though there is no solution.  In my humble
        opinion neither of the previous two will happen. :-)
        Third, the standard axioms are consistent and are strong
        enough to prove the negation of the sentence.  Fourth,
        the standard axioms are consistent but prove neither the
        sentence nor its negation.  This is the case you were
        asking about, and it may happen, but only if there are
        actually no solutions.  If there are any solutions, then
        the "obvious" sentence stating that will be provable from
        the standard axioms.
        
        In general, if a Turing machine can always decide, yes or
        no, whether the nonnegative integers n1,n2,...,nk have
        "property foo", then there will be a formula of first
        order arithmetic A(x1,x2,...,xk) such that
        
        	whenever the T.M. says "yes" for n1,...,nk,
        	then A(n1,...,nk) is a theorem of the standard
        	axioms, and therefore (Ex1)(Ex2)...(Exk)A(x1,x2,...,xk)
        	is also a theorem of the standard axioms
        
        	whenever the T.M. says "no" for n1,...,nk,
        	then ~A(n1,...,nk) is a theorem of the standard
        	axioms
        
                but if the T.M. always says "no", so that the
        	sentence (Ex1)...(Exk)A(x1,...,xk) is "false",
        	and if the standard axioms are consistent, then
        	it may be the case that neither the sentence just
        	mentioned nor its negation is provable from the
        	standard axioms
        
        	(in most usual cases the negation is provable,
        	though, e.g., "property foo" is "x1 is even,
        	prime, and greater than 17")
        	
        Dan

Let
	       3    2          2      3
   z =   q (2 p  + p  q + 4 p q  + 4 q )

We want to show z cannot be a square of an integer if p and q are integers
(not both even, since we assumed p/q was in lowest terms).

Case 1) Assume p, q are both odd. Substitute p = 2P+1, q=2Q+1, and we get 

  z =
      3       2             2                           2       2       3
 (16 P  + 28 P  + 24 P + 8 P  Q + 40 P Q + 42 Q + 32 P Q  + 64 Q  + 32 Q + 11 )

which is of the form 4n+3 and thus cannot be square, since every square is
of the form 4n or 4n+1. 

Case 2) p even, q odd.
Case 3) q even, p odd.

One should be able to continue the attack on the problem by treating these 
cases in a similar way, but we can begin to see why the problem becomes 
intractable: The analysis depends on *how many* factors of 2 there are in 
the even member of (p,q), and this is not known beforehand; what we need is
to find an induction proof on k where, say, p=2^k*m, where m is odd.


This problem reminds me of the following easier one: Given a reflective 
rectangular box with integer sides horiz=p,vert=q; if I fire a laser into 
the box from the lower right corner at 45 degrees, which corner does the
laser beam strike next? The answer is after the spoiler: 



(horiz=p,vert=q)

(1) p,q both odd: upper left
(2) p "more even" than q: lower left
(3) q "more even" than p: upper right
where "more even" means divisible by a higher power of 2.

Q. uestion

	Do there exist rationals x, y, and z such that x + y + z = xyz = 1?


A. nswer

	No.


N. otation:
	Q		rationals
	Z		integers
	Z+		positive integers
	(a,b)		hcf alias gcd of a & b
	a = []		a is a square over Z.

P. roof

Equivalently to the problem as stated:

	Does there exist non-zero z in Q, such that:

		(z^2-z)^2 - 4z 

is a square over Q?   For then {x,y,z} satisfy the criteria, with:


	    (z-z^2) + sqrt( (z^2-z)^2 - 4z)
        y = -------------------------------
			2z
	x = 1/(yz).

	Now, wlog, I can take z > 0, since at least one of {x,y,z} is
positive.   (They add to 1, don't they.)

	Replacing z by a/b, where a & b lie in Z+, (a,b) = 1.   Multiplying by 
b^4, we have:

	a^4 - 2a^3.b + a^2.b^2 - 4a.b^3 = []

where n lies in Z also.   The lhs can be written as a.c, and if d = (a,c), we
know that a/d = [] and c/d = [].

	(a,c) = (a, a^3 - 2a^2.b + a.b^2 - 4b^3) = (a, -4b^3) = (a,4)

	This means that d = 1,2 or 4.   So either a or 2a is a square over Z.

Case: 2a = []:	a = 2*e^2
	4e^6 - 4e^4.b + e^2.b^2 - 2b^3 = [],
	Now, e is odd, since otherwise d would be higher
	& b is odd, since (a,b)=1, and we know a to be even
(& since odd squares are 1 mod 8)
	=>	4 - 4b +1 -2b = [] mod 8
(& since ditto)
	=>	4 - 6b = 0 mod 8
(& since this is not possible for odd b)
	=>	contradiction

So, a = [], e^2 say, so:

	e^6 - 2e^4.b + e^2.b^2 - 4b^3 = []
with (e,b) = 1, and e,b > 0.

	We know that the lhs = [], so set it = n^2, where n >= 0.
Now, n^2 > e^6 - 2e^4.b + e^2.b^2 = (e^3 -eb)^2.

	There are two possibilities:

Possibility1:	n > e^3 - eb >= 0
	Set n = e^3 - eb -f, where f >= 0.
Note that e^2 > b, since (e,b)=1.

We find:
	4b^3 +2e^3.f -2ebf +f^2 = 0
(& since f is obviously even, = 2g, say)
	b^3 + e^3g +g^2 -ebg = 0

	Now, b and e are both in Z+.   The question is, what about g, which is
greater than 0 by construction.

Suppose g = 0
b = 0, and contradiction.

So g > 0:
ebg > e^3g	=> b > e^2 and contradiction.

Possibility2:	n > eb - e^3 >= 0
	Set n = eb - e^3 - f, where f >= 0.
Note that b > e^2, since (e,b)=1.

We find this time:
	4b^3 -2e^3.f +2ebf +f^2 = 0
(& since f is obviously even, = 2g, say)
	b^3 -e^3.g +g^2 +ebg = 0

Again, suppose g = 0
b = 0, and contradiction.

So g > 0:
This time e^3g > ebg, so e^2 > b, so contradiction.

Regards
Andrew.

    Re .11:
    
    To arrive at the following equation:
    
    	4b^3 +2e^3.f -2ebf +f^2 = 0
    
    I believe you squared this equation:
    
	n = e^3 - eb -f
    
    and set it equal to this, which is n^2 by definition:
    
    	e^6 - 2e^4.b + e^2.b^2 - 4b^3
    
    However, if you will check the algebra again, I think you changed some
    of the signs; the actual result is:
    
    	4b^3 - 2e^3.f + 2ebf + f^2 = 0
    
    And that of course supports e^2 > b, which is not a contradiction.
    
    
    				-- edp

	First, can I thank you Eric for taking the trouble to wade all
the way through .11, and for replying.

	Second, there *is* a genuine bug in the proof, and it seems
pretty unfixable.   I have pushed through the calculations a ways further,
however, and I enclose the "advance" here.

	I have also checked the calculations through MAPLE.

	Let's pick the proof up at the point where we have just shown

		e^6 - 2e^4.b + e^2.b^2 - 4b^3 = []

	where (e,b) = 1, and e,b > 0.

-------------------------------------------------------------------------------

	The lhs is not only a square itself (of n), but also is just less than
a square (of |e^3 - eb|).   There is a limit to how close two squares can be, so
it seems to be profitable to explore the difference between them.   So define:

	f = |e^3 - eb| - n > 0

(Getting the sign of this wrong led to problems in the earlier version.)

	Calculate the difference of squares:

	4b^3 = |e^3 - eb|^2 - n^2 = 2f|e^3 - eb| - f^2.

	By mod 2 considerations, f is even, f = 2g say:

	b^3 + g^2 - eg|e^2 - b| = 0

	*Now* we can explore the relationship between e^2 & b.   

CaseI:	e^2 = b
	=> b=g = 0, contradiction.

CaseII:b > e^2
	b^3 + g^2 + e^3g = egb
egb > b^3	=> eg > b^2	(')
egb > g^2	=> eb > g	(")
egb > e^3g	=> b > e^2	(^)

	However, multiplying (') & (") gives:
e^2bg > b^2g	=> e^2 > b
	in contradiction with the hypothesis & (^)

So by elimination, we arrive at:

CaseIII:e^2 > b
	b^3 + g^2 + egb = e^3g

	Let h = (g,b). g = Gh, b = Bh.
(G,B) = 1, since otherwise h would be larger
(h,e) = (B,e) = 1, since (b,e) = 1.
h,B,G,e > 0.

	B^3h^2 + G^2h + eGBh = e^3G

mod h, we find h | e^3G, and since (e,h)=1, h | G

	so G = hH, where (H,B)=1 (since (H,B) | (G,B)) 

	B^3h + H^2h^2 + eHBh = e^3H

	By taking this mod h & mod H, we find that H | h  & also h | H, so h=H.

	This gives us the following form as we pause for breath.   Note that
g has now disappeared from the equation.   Suprisingly, it turned out to be
a perfect cube!

-------------------------------------------------------------------------------

	So, the problem is recast in the following form:

Find B,e,H in Z+, pairwise coprime such that

	B^3 + H^3 + eHB = e^3

Regards,
Andrew

Regards
Andrew.

    What happened here? Was the proof in .15 retracted? If so, could you
    share the bug with us?

>Was the proof in .15 retracted? 

	Yes, I pulled it.

>If so, could you share the bug with us?

	I'm too ashamed:  it's a very unenlightening transcription mistake.
:-(   Suffice it to say that I am still reasonably sure of the correctness
of the cubic linking e,B & H at the end of the previous reply.   I no longer
think that this problem can be linked to Note 70.

Sighfully yours,
Andrew.

    Re .16:
    
    Don't worry about making mistakes; it happens.  One proof I entered in
    here used the "fact" that 57 is prime.
    
    
    				-- edp