Conference rusure::math

There are two degenerate cases immediately:
(a,2a,3a) and (a,a,2a) both with area = 0...

A Heronian triangle must have non-zero area.

There are no small cases - I checked for (a, 2a, b) for 1<a,b<1000.

The problem is: are there any non-degenerate triangles with one side twice
as long as another, and all three sides and the area being integers.

Since the problem was posed in 1984 and has not been solved,
perhaps no one cares.

I think there are no such triangles.  Attempted proof follows.

Let the sides be a, 2a, and b, the semiperimeter be s, and the area be A.

s    = (3a+b)/2
s-a  = (a+b)/2
s-2a = (b-a)/2
s-b  = (3a-b)/2
                       2       2   2   2  2
By Hero's formula, (4A) = ((3a) - b )(b -a ).
The left side is a perfect square, so the right side must be a perfect square.
We can remove all square factors (exceeding 1) from each of the binomials,
and the remaining factors in each case must be the same.

 2   2    2                  2   2    2           2      2
b - a = cE         and   (3a) - b = cF     so (4A) =(cEF)

These are standard Diophantine equations.  By treating 3a as an integer
the formula will include some false solutions, but we'll see they do no harm.
The solution depends on the parity of c.

For C even, we have:
E = 2pmn				F = 2pmn
        2    2                                  2    2
a = p(gm - hn )				b = p(gm - hn )
        2    2                                  2    2
b = p(gm + hn )			       3a = p(gm + hn )


For C odd, we have:
E = pmn					F = pmn
        2    2                                  2    2
a = p(gm - hn )/2			b = p(gm - hn )/2
        2    2                                  2    2
b = p(gm + hn )/2		       3a = p(gm + hn )/2

g and h are integers with gh = c. m and n are relatively prime and > 0.

p is any integer if c is even or if c, m, n are odd.
If c is odd and m and n are of opposite parity, then p is even.

The a and the b in each binomial must have the same value, so
     2   2        2   2                 2   2        2   2
3p(gm -hn ) = p(gm +hn )     and    p(gm -hn ) = p(gm +hn ).

The same equations arise for c even or odd.

The only solutions are for p=0 or the parenthetical terms = 0.
Since m and n are >0, g=0 and h=0.  Also, since gh=c, m=0 and n=0.

This leads immediately to E=0, F=0, a=0, b=0.

Therefore, there are no solutions other than the degenerate triangle.


Corrections welcome.  The solution to the Diophantine equation is
from L.E.Dickson, Intro to the Th. of Numbers.

>Since the problem was posed in 1984 and has not been solved,
>perhaps no one cares.

	I was stuck.   I sent it out to a couple of friends, and I got a
reply a little while ago.   Unfortunately the hand-writing is arachnoid,
and nanotechnological, so I haven't managed to digest it.   How can people
possibly write so much on such small pieces of paper?   And you should
see the topological examples he sends me sometimes:  they're like lifescale
drawings of fingerprints.   However, I digress.

> 2   2    2                  2   2    2           2      2
>b - a = cE         and   (3a) - b = cF     so (4A) =(cEF)

	All tickety-boo so far.

>The a and the b in each binomial must have the same value, so
>     2   2        2   2                 2   2        2   2
>3p(gm -hn ) = p(gm +hn )     and    p(gm -hn ) = p(gm +hn ).

	Yes, but the p,g,m,h,n don't have to be the same!   Sorry, but you'll
need to think again.   Good luck.

	Perhaps I should return to my own musings on this one (after all, 
given my nodename I should be a specialist in these triangles), or attempt
to decipher Herv�'s Gallic scratchings again.

Regards,
Andrew.

Q.	A Heronian triangle is one in which each side a,b,c & the area S is
a *positive* integer.   The problem here is to classify those triangles T for 
which c = 2a.

A.	There are such no triangles.

Y.	If d is the half the perimeter of the triangle, then a well-known
formula says:
	S = sqrt(d(d-a)(d-b)(d-c))

	In this case, we have d = (3a+b)/2:
	S = �sqrt((3a+b).(3a-b).(a+b).(b-a)) = �sqrt((9a�-b�)(b�-a�))

	So, our requirements on a & b are:

	(i) (9a�-b�).(b�-a�) is a perfect square...
	(ii) ...and a multiple of 16
	(iii) wlog (a,b) = 2^z (for some z, possibly = 0.  (a,b) denotes hcf)

(ii) implies b+a is even.   SUPPOSE that a & b are each odd, so (a,b) = 1.
Then a� == b� == 1 mod 8.   Thus 9a�-b� == b�-a� == 0 mod 8.   So (ii) is
satisfied.

(9a�-b�,b�-a�) = (8a�,b�-a�) which divides 8(a�,b�-a�) = 8(a�,b�) = 8.
Since we know that 8 actually does divide 9a�-b� & b�-a�, we can write:
		    9a�-b� = 8e�
		    b�-a�  = 8f�
where (a,b)=(e,f) = 1
Rearrange:

	a� = e� + f� 
	b� = e� + (3f)�

	Now, this suggests constructing a new triangle, T', with sides {a,b,4f}.
It will have a perpendicular to the side 4f, with length e, that divides T'
into two right-angled triangles: {a,e,f} & {a,e,3f}.   What is the area of T'?
It is 2ef.   But hang on, that's exactly the area of T.   So since both 
triangles have sides a&b, we have that T = T'.   So 4f = 2a.   So b� = 3a�.
Which is impossible.

	Now, all this hinged on a & b each being odd.   SUPPOSE that they are
both even:  a = 2x, b = 2y.   The requirements become:
	(9x�-y�).(y�-x�) is a perfect square.
	(x,y) = 2^z' for some z'

	I say that x+y is even, because if not, then:
	(9x�-y�).(y�-x�) == 3 mod 4, 
which is impossible for a square.   So x+y are both odd (impossible, by the 
above) or both even (repeat this argument many times => 2^z divides (a,b) for 
all z, which is impossible).

	Any time that a problem yields to a sudden breakthrough like this is
suspicious, so checks are welcome.

Regards,
Andrew.

>	(iii) wlog (a,b) = 2^z (for some z, possibly = 0.  (a,b) denotes hcf)

	I don't see this.

	Also, hcf == highest common factor == gcd == greatest common divisor.
	Right?

	Yes, hcf :== highest common factor.

	Suppose that p is an odd prime which divides (a,b).   Then,
let's take a' = a/p & b' = b/p.   (i) & (ii) hold true for a'&b' as
much as for a&b, so wlog, no odd prime divides (a,b).

	This is exactly what (iii) states.

Regards,
Andrew.

    Congratulations, a neat proof. The way you laid it out meant that the
    crux of it:
    
>   ... or both even (repeat this argument many times => 2^z divides (a,b) for 
>all z, which is impossible).
    
    was rather tucked away. This technique ("infinite descent") seems to
    work well on several quartic Diophantine eaquations.
    
    dick

>	Now, this suggests constructing a new triangle, T', with sides {a,b,4f}.
>It will have a perpendicular to the side 4f, with length e, that divides T'
>into two right-angled triangles: {a,e,f} & {a,e,3f}.   What is the area of T'?
>It is 2ef.   But hang on, that's exactly the area of T.   So since both 
>triangles have sides a&b, we have that T = T'.   So 4f = 2a.   So b� = 3a�.
>Which is impossible.

	Hang on.   Two triangles *can* be different, and still share the
same area, and the same two sides.   The included angle in one will be
acute with value x, in the other, obtuse, with value pi - x.

	Drat!

	I therefore declare this problem unsolved again!

Regards,
Andrew.

	An old opponent, which I hope I have just vanquished.   The key part
is Lemma 1, which I put first so people can validate it.

Q.	A Heronian triangle is one in which each side a,b,c & the area S is
a *positive* integer.   The problem here is to classify those triangles T for 
which c = 2a.

A.	There are such no triangles.

All variables are assumed to be natural numbers, except where stated.

Lemma 1: The only natural number solutions to the twin equations:
		q� = s�-p�
		r� = s�-(2p)�
are of the form [p,q,r,s] = [0,N,N,N].

Proof of Lemma:
	If p is 0, then we get the family of solutions proposed above.
Otherwise, if (p,s)=N ((x,y) denotes greatest common divisor in all that
follows) then (p,q,r,s)=N.   So divide the equations by N� and then we can
assume that (p,s)=1.

	If we assign:

	P = 2pqrs			
	S = s^4 - 4p^4

and define Q,R from P,S in the obvious way, then we find that [P,Q,R,S] is
a solution if [p,q,r,s] is.   Now, this suggests to us that given a solution 
[p,q,r,s] we should see if a *smaller* solution exists.   We have:

	q� = s�-p�
	r� = s�-(2p)�
	(p,s)=1

	A square number == 0,1 or 4 mod 8.   It's obvious that the only
consistent way to allocate p,q,r&s mod 2, is 4 | p (bar denotes "divides") 
while q,r & s are odd.

	Hence (s+p,s-p) = (s+p,2s) | (s+p,s)(s+p,2) = (p,s)(s,2) = 1
and similarly (s+2p,s-2p) = (s+2p,2s) | (s+2p,s)(s+2p,2) = (2p,s)(s,2) = 1.
Therefore, we can decompose:
	r = k|l|
	q = mn,
where k,l,m,n are all odd, and (k,l) = (m,n) = 1, k >= |l|, m >= n.   For
technical reasons, we permit l to be negative.   Its sign is chosen such
that 4 | k+l.

	s+p = m�
	s-p = n�
	s+2p = k�
	s-2p = l�

Hence:
	m�+n� = k�+l�			(=2s)
	2(m�-n�) = k�-l�		(=4p)	

	Let's focus on the second equation.   Now define:
	p' = ((k+l)/4,(m+n)/2)
	q' = ((k+l)/4,(m-n)/2)
	r' = ((k-l)/2,(m+n)/2)
	s' = ((k-l)/2,(m-n)/2)

[Note:
	((k+l)/4,(k-l)/2) = 1
	((m+n)/2,(m-n)/2) = 1

	p',q',r' & s' are pairwise coprime.]

Thus:
	k+l = 4p'q'
	k-l = 2r's'
	m+n = 2p'r'
	m-n = 2q's'

	Substitute this lot into the (=2s) equation:

	4p'�q'� + r'�s'� = p'�r'� + q'�s'�

=>	p'�(4q'�-r'�) = s'�(q'�-r'�)	

	Next, examine
	X = (4q'�-r'�,q'�-r'�) = (3q'�,q'�-r'�) | 3
	(p'�, s'�) = 1

	This means that there are two possibilities, according as X = 1 or 3:

X=1:
p'� = q'�-r'�
s'� = 4'q�-r'�

	There is no consistent way to solve the second equation mod 4, unless
2 | (r',s'), which is not possible.   So X = 3, and hence:

X=3:
3p'� = q'�-r'�
3s'� = 4'q�-r'�

	Or rearranging:
q'� = s'�-p'�
r'� = s'�-(2p')�

	If you check the details, you find that:

	p = 2p'q'r's'			
	s = s'^4 - 4p'^4

as desired.   If p > 0, we can define a monotonically strictly decreasing 
series, with p, p', p'', p'''..., which is bounded below by 0.   This is a
contradiction.   p = 0, and we already know all the solutions of that form.

End of Proof of Lemma 1.


Proof of Main Result:

	If d is the half the perimeter of the triangle, then a well-known
formula says:
	S = sqrt(d(d-a)(d-b)(d-c))

	In this case, we have d = (3a+b)/2:
	S = �sqrt((3a+b).(3a-b).(a+b).(b-a)) = �sqrt((9a�-b�)(b�-a�))

	So, our requirements on a & b are simply:

	(i) (9a�-b�).(b�-a�) is a non-zero perfect square...
	(ii) ...and a multiple of 16

	Let (x,y) denote the greatest common divisor of x & y.

	Substitute a = KA, b = KB, where (A,B) = 1.   We need:

	(i) (9A�-B�).(B�-A�) is a non-zero perfect square
	(ii) 16 | K�.(9A�-B�).(B�-A�)

	Suppose that A+B == 1 mod 2, then the expression in (i) is == -1 mod 4,
so is not a square.  Thus A == B == 1 mod 2, and (ii) is automatically
satisfied, for any value of K.   So we are searching for odd, coprime A & B
such that:

	(9A�-B�).(B�-A�) is a perfect square...

(9A�-B�,B�-A�) = (8A�,B�-A�) which divides 8(A�,B�-A�) = 8(A�,B�) = 8.
Since we know that 8 actually does divide 9A�-B� & B�-A�, we can write:
		    9A�-B� = 8e�
		    B�-A�  = 8f�
where (A,B) = (e,f) = 1, and both A & B are odd (hence e+f is odd).

	At this point, we diverge from the previous non-solution.

	(B+A)(B-A) = 8f�
	(3A+B)(3A-B) = 8e�

(B+A,B-A) = 2, using a similar argument to before.
(3A+B,3A-B) = 2(3,B).   If 3 divides B, then substitute B = 3B', and e = 3e',
gives us just the same equations as before (try it!).   So we can assume that
(3A,B) = 1, and then if we find a solution [A,B] then we can have another
[B,3A], and [3A,3B], and so on.

{B+A,B-A} = {2i�,4j�}
{3A+B,3A-B} = {2g�,4h�}

where gh=e, ij=f, (g,h)=(i,j)=1.

|g�-2h�| = i�+2j�		(= B)
g�+2h� = 3|i�-2j�|		(= 3A)

Since e+f == 1 mod 2, exactly one of (g,h,i,j) is even:  by elimination, j is 
the even one, and:

2h�-g� = i�+2j�
g�+2h� = 3i�-6j�

h� = i�-j�
g� = i�-(2j)�

	But this is exactly the form which has been demonstrated only to
exist with j = 0, g=h=i. => f = 0 => A�=B� => the triangle has zero area.

End of Proof.

Comments:
	If anyone has checked this result all the way through, I'd be grateful
if you could tell me since it would be confirmation that I haven't made a slip.
I'm reasonably confident that this is correct.

	My main comment would be that I am clearly operating at a lower
conceptual level than is best for tackling this kind of problem.   What is it
that enables the recursive structure to exist which I exploit to create the
Lemma?   In what other cases will it exist?

    Re .11:
    
    >	k+l = 4p'q'
    >	k-l = 2r's'
    >	m+n = 2p'r'
    >	m-n = 2q's'

    I don't see how these are derived.  Consider that there might exist a
    prime p (not the p you have used) and a natural number i such that p^i
    | (k+l)/4, but p^i does not divide (m+n)/2 or (m-n)/2.  Then p^i would
    not divide p' or q', and hence k+l would not equal 4p'q'.
    
    >	s = s'^4 - 4p'^4

    I also don't see how that is derived, although I haven't examined it
    carefully yet, pending resolution of the above.
    
    > Since e+f == 1 mod 2, exactly one of (g,h,i,j) is even:  by
    > elimination, j is  the even one, and:
    
    How do you eliminate h?
    
    
    				-- edp

	Firstly, thank you for ploughing through .11.   I greatly appreciate
it.   A negative result such as I have (hopefully) proved, is much more
difficult to verify than something more constructive.

	Now, to address the points you raise...


>    
>    >	k+l = 4p'q'
>    >	k-l = 2r's'
>    >	m+n = 2p'r'
>    >	m-n = 2q's'
>
>    I don't see how these are derived.  Consider that there might exist a
>    prime � and a natural number i such that �^i
>    | (k+l)/4, but �^i does not divide (m+n)/2 or (m-n)/2.  Then �^i would
>    not divide p' or q', and hence k+l would not equal 4p'q'.
    
	We know that k�-l� = 2(m�-n�).   ie: (k+l)/4 | (m+n)/2*(m-n)/2.
Hence no prime � can divide (k+l)/4 without dividing one of (m+n)/2 or (m-n)/2.

	To talk around the subject a little:  what we have essentially is an 
equation:
	WX = YZ
where (W,X) = (Y,Z).   The construction:
	W = pq
	X = rs
	Y = pr
	Z = qs
where p,q,r,s are pairwise coprime is a trick I find useful in many problems.
The additional wrinkle here is that if k & l are both odd, then either k+l
or k-l == 2 mod 4, so (k+l,k-l) is at most 2.   This enables us to define
W&X such that (W,X) = 1.   Similarly for Y & Z.


>    >	s = s'^4 - 4p'^4
>
>    I also don't see how that is derived, although I haven't examined it
>    carefully yet, pending resolution of the above.

	Go back to the (=2s) equation, 
		2s = m� + n�
		   = (p'r'+q's')� + (p'r'-q's')�
		   = 2p'�r'� + 2q'�s'�
		 s = p'�(s'�-4p'�) + (s'�-p'�)s'�
		     s'^4 - 4p'^4


>    > Since e+f == 1 mod 2, exactly one of (g,h,i,j) is even:  by
>    > elimination, j is  the even one, and:
>    
>    How do you eliminate h?
    
	|g�-2h�| = i�+2j�
	g�+2h� = 3|i�-2j�|
        
	From either of these, g == i mod 2, and since exactly one of g,h,i,j
is even g == i == 1 mod 2.   Now, if h is even (and j is odd) then the second
equation gives us:

	1 + 0 == �3(1 - 2) mod 8
=>	1 == �3 mod 8.
=>	Contradiction.

	So j is the even one.   Working mod 8 is another favourite trick
with these quadratic problems, since all odd squares are == 1 mod 8.


Thanks once again,
Andrew.

    re .11
    
    I've made it through the Lemma. It looks good to me.
    
    I need a clue for the following line from the main proof
    
>Since we know that 8 actually does divide 9A�-B� & B�-A�, we can write:
    
    Where did the 8 come from?

re .14

>    I've made it through the Lemma. It looks good to me.

	Thanks.   Much appreciated.
    
>    I need a clue for the following line from the main proof
>>Since we know that 8 actually does divide 9A�-B� & B�-A�, we can write:
>    Where did the 8 come from?

	A & B are known to be odd, therefore the square of each is 1 mod 8.


re .11

	Perhaps I can interpolate a whimsical aside, which gives a little
structure to what's going on here.   The proof is reminiscent of the story
of the Magical Chase, where the quarry keeps changing into a different form
to escape the pursuer.   For a long time, I felt like the thwarted hunter:
I had a series of equations, which kept changing into different equations,
and a blizzard of variables, but it didn't look as if I was actually getting
anywhere.

	But I did seem to be able to keep forcing the equations to change,
and so I persisted, looking to see if I was able to chase the equations
round to a form I'd *already* encountered.   This looping is the essence of 
Lemma 1.

	The introduction of recursion, with the possibility then of proof by
infinite descent, is a common way to tackle such problems, which makes me
confident that the approach is good.

	However, it's a hacker's attack, and some insight would be helpful...

Regards,
Andrew.	

    Re .13:
    
    Thanks, I've finished now.  It looks good to me.
    
    
    				-- edp

Q.	A Heronian triangle is one in which each side a,b,c & the area S is
a *positive* integer.   The problem here is to classify those triangles for 
which c = 2a.


A.	There are such no triangles.


All variables are assumed to be natural numbers, except where stated.


Lemma 1: The only natural number solutions to the twin equations:
		q� = s�-p�
		r� = s�-(2p)�
are ordered quadruples [p,q,r,s] = [0,M,M,M] for some natural M.

Proof of Lemma:
	If p is 0, then we get the family of solutions proposed above.
Otherwise, p > 0 and hence s > 0.    If (p,s) = N > 0 ((x,y) denotes greatest 
common divisor in all that follows) then (p,q,r,s) = N.   So divide both the 
equations by N� and then we can assume that (p,s)=1.

	If we assign:

	P = 2pqrs
	S = s^4 - 4p^4	(which > 0)

and define Q,R from P,S in the obvious way, then we find that [P,Q,R,S] is
a solution if [p,q,r,s] is.   Now, this suggests to us that given a solution 
[p,q,r,s] we should see if a *smaller* solution exists.   We have:

	q� = s�-p�
	r� = s�-(2p)�
	(p,s)=1

	A square number == 0, 1 or 4 mod 8.   It's obvious given this that the 
only consistent way to allocate p, q, r & s mod 2, is 4 | p (bar denotes 
"divides") while q, r & s are odd.   s > 2p > 0, so s-p & s-2p are
greater than 0.

	Hence (s+p,s-p) = (s+p,2s) | (s+p,s)(s+p,2) = (p,s)(s,2) = 1
and similarly (s+2p,s-2p) = (s+2p,2s) | (s+2p,s)(s+2p,2) = (2p,s)(s,2) = 1.
Therefore, we can decompose:
	r = k|l|
	q = mn,
where k,l,m,n are all odd, and (k,l) = (m,n) = 1, k >= |l|, m >= n.   For
technical reasons, we permit l to be negative.   Its sign is chosen such
that 4 | k+l.

	s+p = m�
	s-p = n�
	s+2p = k�
	s-2p = l�

Since p > 0, k > |l| and m > n

Hence:
	m�+n� = k�+l�			(=2s)
	2(m�-n�) = k�-l�		(=4p)

	Let's focus on the second equation.   Now define:
	p' = ((k+l)/4,(m+n)/2)
	q' = ((k+l)/4,(m-n)/2)
	r' = ((k-l)/2,(m+n)/2)
	s' = ((k-l)/2,(m-n)/2)

[Note:
	((k+l)/4,(k-l)/2) = 1
	((m+n)/2,(m-n)/2) = 1

	p',q',r' & s' are pairwise coprime.]

	We know that k�-l� = 2(m�-n�).   ie: (k+l)/4 | (m+n)/2*(m-n)/2.
Hence no prime � can divide (k+l)/4 without dividing one of (m+n)/2 or 
(m-n)/2.

Thus:
	k+l = 4p'q'
	k-l = 2r's'
	m+n = 2p'r'
	m-n = 2q's'

	Substitute this lot into the (=2s) equation:

	4p'�q'� + r'�s'� = p'�r'� + q'�s'�

=>	p'�(4q'�-r'�) = s'�(q'�-r'�)	

	Next, examine
	X = (4q'�-r'�,q'�-r'�) = (3q'�,q'�-r'�) | 3
	(p'�, s'�) = 1

	This means that there are two possibilities, according as X = 1 or 3:

X=1:
p'� = q'�-r'�
s'� = 4'q�-r'�

	There is no consistent way to solve the second equation mod 4, unless
2 | (r',s'), which is not possible.   So X = 3, and hence:

X=3:
3p'� = q'�-r'�
3s'� = 4'q�-r'�

	Or rearranging:
q'� = s'�-p'�
r'� = s'�-(2p')�

	Going back to the (=2s) equation:
		2s = m� + n�
		   = (p'r'+q's')� + (p'r'-q's')�
		   = 2p'�r'� + 2q'�s'�
		 s = p'�(s'�-4p'�) + (s'�-p'�)s'�
		     s'^4 - 4p'^4

	Hence:
	s = s'^4 - 4p'^4
and similarly:
	p = 2p'q'r's'			

as desired.   If p > 0, we can define a monotonically strictly decreasing 
series, with p, p', p'', p'''..., which is bounded below by 0.   This is a
contradiction.   p = 0, and we already know all the solutions of that form.

End of Proof of Lemma 1.


Proof of Main Result:

	If d is the half the perimeter of the triangle, then a well-known
formula says:
	S = sqrt(d(d-a)(d-b)(d-c))

	In this case, we have d = (3a+b)/2:
	S = �sqrt((3a+b).(3a-b).(a+b).(b-a)) = �sqrt((9a�-b�)(b�-a�))

	So, our requirements on a & b are simply:

	(i) (9a�-b�).(b�-a�) is a non-zero perfect square...
	(ii) ...and a multiple of 16

	Let (x,y) denote the greatest common divisor of x & y.

	Substitute a = KA, b = KB, where (A,B) = 1.   We need:

	(i) (9A�-B�).(B�-A�) is a non-zero perfect square
	(ii) 16 | K�.(9A�-B�).(B�-A�)

	Suppose that A+B == 1 mod 2, then the expression in (i) is == -1 mod 4,
so is not a square.  Thus A == B == 1 mod 2, and (ii) is automatically
satisfied, for any value of K.   So we are searching for odd, coprime A & B
such that (9A�-B�).(B�-A�) is a non-zero perfect square.

(9A�-B�,B�-A�) = (8A�,B�-A�) which divides 8(A�,B�-A�) = 8(A�,B�) = 8.
A & B are known to be odd, therefore the square of each is 1 mod 8, 
so we know that 8 actually does divide 9A�-B� & B�-A�, and we can write:
		    9A�-B� = 8e�
		    B�-A�  = 8f�
where (A,B) = (e,f) = 1, and both A & B are odd (hence e+f is odd).

	At this point, we diverge from the previous non-solution.

	(B+A)(B-A) = 8f�
	(3A+B)(3A-B) = 8e�

(B+A,B-A) = 2, using a similar argument to before.
(3A+B,3A-B) = 2(3,B).   If 3 divides B, then substitute B = 3B', and e = 3e',
gives us just the same equations as before (try it!).   So we can assume that
(3A,B) = 1, and then if we find a solution [A,B] then we can have another
[B,3A], and [3A,3B], and so on.

As sets:
	{B+A,B-A} = {2i�,4j�}
	{3A+B,3A-B} = {2g�,4h�}

where gh=e, ij=f, (g,h)=(i,j)=1, g,h,i,j all > 0.

|g�-2h�| = i�+2j�		(= B)
g�+2h� = 3|i�-2j�|		(= 3A)

	Since e+f == 1 mod 2, exactly one of (g,h,i,j) is even.

	From either of the two equations relating g, h, i & j, we have that 
g == i mod 2, and since exactly one of g,h,i,j is even g == i == 1 mod 2.   
Now, if h is even (and j is odd) then the (=3A) equation gives us.

	1 + 0 == �3(1 - 2) mod 8
=>	1 == �3 mod 8.
=>	Contradiction.

and so j is the even one, and:

2h�-g� = i�+2j�
g�+2h� = 3i�-6j�

which goes to:

h� = i�-j�
g� = i�-(2j)�

	But this is exactly the form which has been demonstrated in Lemma 1 
only to exist with j = 0, g=h=i.   This contradicts our knowledge that j > 0.
So there are no solutions to our problem.

End of Proof.