(Update, 8/7/09):
V. Equal Sums of Like Powers
The results in this section are by the author. See also the separate section on
a 3(a 6(a
[6, 23, 25]
Corollary: These relations can be woven together. Define
32(a
3 E 6 E
Square eq.1, multiply eq.2 by E
32 E 32 E
as stated. (
a
then n = 1/2 and the above becomes,
64(a
which is a concise version of the
F 4 F 8 F 24 F
Corollary: Define
25(a
or, equivalently, 25 F
G 5 G 10 G 30 G 120 G
Corollary: "Define
H 6 H 12 H 36 H 144 H 720 H
Corollary: "Define
And so on for other k = 2,4,..2n. While an infinite number of solns are known for Theorems 1 to 4,
32 E 50 F 72 G 98 H
This has the general form,
2m
starting with m = 4,5,6… so one can see the next in the series. However, by taking square roots of the variables, these theorems are also valid for systems with k = 1,2,3,…n, or those satisfying the
{a,b,c,d} = {m+p, m-p, m+q, m-q}
If p
{a,b,c,d} = {m+p, m-p, m+q, m-q} {e,f,g,h} = {m+r, m-r, m+s, m-s}
If,
p p
then 25(a
Also, for
(u
which can be solved by using the formula for Pythagorean triples separately on {u,v} and {x,y}.
{a,b,c,d,e,f} = {m+p, m-p, m+q, m-q, m+r, m-r} {g,h,i,j,k,l} = {m+s, m-s, m+t, m-t, m+u, m-u}
If,
p p
then, 49(a
And, for any
{p,q,r,s,t,u} = {2, 289, 610, 170, 223, 614} (by "Martin") {p,q,r,s,t,u} = {1575, 11522, 11850, 4647, 8890, 13230} (by James Waldby)
after this author posted the problem on the newsgroup
z 2(z
Eq. 1 can be completely solved as {z
(u
and it is easy to solve this using a variant of the formula for Pythagorean triples separately on {u,v} and {x,y}. The procedure can be extrapolated to higher systems.
The difficulty with the system,
a
valid only for odd powers is that terms can be transposed or moved around to the form,
c
and it is more challenging to find relations beyond k = 2n+1. The trick then is to “fix the variables in place” by finding a balanced partition such that,
a) a b) eq.1 is valid for k = 2 as well
and hence terms can no longer be transposed arbitrarily. (For the system k = 1,3,5, the two conditions are equivalent.)
a) Condition: a
7(a
More succinctly, "Define ^{k}+b^{k}+c^{k}+d^{k} -(e^{k}+f^{k}+g^{k}+h^{k}). If F = 0 for k = {1,2,3,5}, then _{k}7F."_{4}F_{9} = 12F_{6}F_{7}
Note: As these two systems have complete solns, given here (Form 13) and here (Form 5.3b), respectively, one can substitute those into the given identities and see that they are true. For higher than fifth powers, there are only isolated results with this condition so it harder to come up with general statements.
b) Condition: Valid for k = 2 as well.
As
Define F
Define F
Define F
Note:
(Update, 12/21/09): Jarek Wroblewski observed that the third conjecture can be generalized as (13F |