| In four dimensions, there is
A (0,0,0,0) BC = 14
B (10,0,0,0) AC = 6
C (-3,5,1,1) AB = 10
The angle at the origin is 120 degrees.
If you put the 120 degree angle at the origin (point A)
and point B at (a,b,c) and point C at (d,e,f) then you
get the four equations in integer variables:
r^2 = a^2 + b^2 + c^2
s^2 = d^2 + e^2 + f^2
t^2 = (a-d)^2 + (b-e)^2 + (c-f)^2
t^2 = r^2 + rs + s^2 [this from the law of cosines]
I am about to program a search for solutions with the absolute values
of a-f less than or equal to 100. I doubt that I will find anything.
Before I start, has anyone done this kind of search already?
Dan
|
| What about the following argument, using your notation in .1, Dan.
i.e. looking for integer soln to
(i) r^2 = a^2 + b^2 + c^2
(ii) s^2 = d^2 + e^2 + f^2
(iii) t^2 = (a-d)^2 + (b-e)^2 + (c-f)^2
(iv) t^2 = p^2 + p.q + q^2
with r,s,t all > 0.
If I've got 4 integers W, X, Y, Z such that
W^2 = X^2 + Y^2 + Z^2
then EITHER W,X,Y,Z all even
OR W and X (say) odd, and the other two integers, Y and Z even
In the problem in hand, by looking at (i) and (ii) in
this way, there are 4 cases without loss of generality:
Case 1) a,b,c,d,e,f all even. In this case we can divide everything
by 2 repeatedly until we reach one of the other cases (Or all of
a,b,c,d,e,f are zero, in which case r,s,t are all zero). So ignore
this case.
Case 2) a and d odd, b,c,e,f even
Case 3) a and e odd, b,c,d,f even
In either of these cases r and s are both odd. However, if we
combine (iii) and (iv) and then substitute out r^2 and s^2, we have
(v) r.s = -2(a.d + b.e + c.f)
so at least one of r and s must be even
Thus we are forced to conclude...
Case 4) a is odd b,c,d,e and f all even.
r is odd and s is even.
Substitute in (ii) and (v) with r = 2R + 1, s = 2S, d = 2D,
e = 2E, f = 2F...
(ii') S^2 = D^2 + E^2 + F^2
(v') -R.S -(S/2) = a.D + b.E + c.F
so by (v') S is even so by (ii') D,E and F are even also
so substitute S = 2S', D = 2D', E = 2E' and F = 2F'
(ii'') S'^2 = D'^2 + E'^2 + F'^2
(v'') -R.S' -(S'/2) = a.D' + b.E' + c.F'
But this is where we were last time round. i.e. S,D,E and F have
as a factor 2^n for all n. i.e. S,D.E.F are all 0.
CONCLUSION: The proposed construction is impossible in three
dimensions.
Cheers,
Andy Buchanan.
|
| I'm impressed!
In case anyone is confused by some of your points:
>> If I've got 4 integers W, X, Y, Z such that
>> W^2 = X^2 + Y^2 + Z^2
>> then EITHER W,X,Y,Z all even
>> OR W and X (say) odd, and the other two integers, Y and Z even
... because the only squares mod 4 are 0 and 1.
>> Case 2) a and d odd, b,c,e,f even
>> Case 3) a and e odd, b,c,d,f even
... because the odd one in d-e-f is either in the same
coordinate as the odd one in a-b-c or it isn't.
Everything else was very straightforward. That's neat. Not
only do we all learn something, but I don't have to write
the search code, either! :^)
Dan
|