| .1 contained a typo. Here is the corrected version:
Either of the following must hold:
1) Four of the numbers are multiples of 16 and the last number is
either a multiple of 16, 1 mod 8, or 3 mod 16
or
2) Four of the numbers are 8 mod 16 and the last number is either
8 mod 16, 1 mod 8, or 11 mod 16.
Topher
|
| > Find five different whole positive numbers a,b,c,d,e, any two of which
> sum to a square number.
First, note that any solution in rationals can be converted to an integer
solution by multiplying a,b,... by (lcd(a,b,...))^2, where lcd() is the
least common divisor.
Aside: We might multiply by a smaller number -- let
lcd(a,b,...) = p1^e1 * p2^e2 * ... * pn^en
be the prime decomposition of lcd(a,b,...). Let bi = (ei mod 2), and
lcd2(a,b,...) = p1^b1 * p2^b2 * ... * pn^bn
Then we can convert a rational solution to an integer solution by
multiplying a,b,... by (lcd2(a,b,...))^2.
To find the general form of three rationals a,b,c any two of which sum
to the square of a rational,
b + c = A^2
c + a = B^2
a + b = C^2
we can choose A, B, and C, and let
a = ( -A^2 + B^2 + C^2 )/2
b = ( A^2 - B^2 + C^2 )/2
c = ( A^2 + B^2 - C^2 )/2
To find the general form of four rationals a,b,c,d any two of which sum
to the square of a rational,
b + c = A^2
c + a = B^2
a + b = C^2
a + d = D^2
b + d = E^2
c + d = F^2
Suppose we can choose A and B. Then since
B^2 - A^2 = a - b = D^2 - E^2
(B - A)(B + A) = (D - E)(D + E)
This suggests the substitution
v = (B - A)/(D - E)
where v is a rational. Substituting this into the above gives:
(B - A) = (D - E) * v
(B + A) = (D + E) / v
and solve for D and E.
We proceed similarly, with F and E, and introduce r.
This gives two equations for E; equating these gives an equation for C.
We choose A, B, v, and r, and let
C = ((B^2 - A^2 - v^2)*r + (r^2-1)*B*v)/(v*(r^2+1))
a = ( -A^2 + B^2 + C^2 )/2
b = ( A^2 - B^2 + C^2 )/2
c = ( A^2 + B^2 - C^2 )/2
d = ( (C-B)^2*r^2 - 2*A^2 + (C+B)^2/r^2 )/4
This is a general solution for four rationals, any two of which sum to
the square of a rational.
|
| Note that the formulae in .3 express four variables with the desired
property as functions of A, B, r, and v. In a set of five variables,
three of them are common to two different sets of four. Those sets of
four can be expressed as functions of A, B, r0, v0, r1, and v1. The Cs
they generate will be equal, so we can solve for one of the variables
in terms of the others. Then the problem will be solved by finding a
value of that solution that is rational when the other variables are
rational and such that the sum of the generated "fourth" variables is a
square.
I've reduced this to a messy quartic in Maple. Can anybody
characterize when k^2 = an^4 + bn^2 + c has a solution for n and k in
the integers, for given a, b, and c?
-- edp
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|
| > I've reduced this to a messy quartic in Maple. Can anybody
> characterize when k^2 = an^4 + bn^2 + c has a solution for n and k in
> the integers, for given a, b, and c?
I think this has a solution for rational n and k when the curve
in the (n,k) plane has a singular point. You can then paramterize
the curve by the slope of a line passing through that point, which will
intersect the curve at one other unique point, guaranteed to be rational
in (n,k).
Singularity can be tested for by calculating a discriminant for the
curve, which vanishes. At least this works for a cubic.
I can look this up in a book on algebraic geometry tonight - I don't
really remember the details, though I once worked things through
for cubics.
- Jim
|
| Re .3
Peter, could you elaborate a bit on the following?:
> We proceed similarly, with F and E, and introduce r.
> This gives two equations for E; equating these gives an equation for C.
The nearest I can get to your subsequent 5 formulae is if I introduce an
upside-down definition of r compared with that of v:
r = (F-E)/(B-C) (since B^2-C^2 = c-b = F^2-E^2)
But then, for the first of the 5 formulae I get:
C = (r((B+A)v^2-(B-A))+(r^2-1)Bv)/v(r^2+1)
but with agreement for the other 4. I'm sure I've slipped up somewhere but I
can't see where. How exactly did you define your r?
We also have to bear in mind that we have some constraints on v and r, given
that we want a,b,c and d to be positive.
Re .5
edp, did you get any further along that path?
Dick
|