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.

>	(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

    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

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