I. IIa. IIb. IIIa. IIIb. IV.
(a+b) IIa. Lucas-Liouville Polynomial Identity
(a+b)
Note: It seems E. Lucas knew this identity before Liouville. Anything else for k > 4?Update 9/26/09: William Ellison pointed out that David Hilbert has proven (x
_{1}^{2}+x_{2}^{2}+ ... +x_{n}^{2})^{k} can be written as a rational combination of (2k)th powers of linear forms in the x_{i}. (You can read his paper on Waring's Problem and how Hilbert solved it in the Attachments section at the bottom of this page.) For example, the Lucas-Liouville Polynomial Identity above can be concisely written as,6 (x
To determine the number of terms in the summation, if we are to choose 2 objects from 4, this gives 6 ways. (This can be calculated in
6 (x 6 (x 6 (x 6 (x
and so on. For 3 variables taken at a time,
12 (x 24 (x 36 (x 48 (x 60 (x 72 (x
Note how for
24 (x 72 (x 144 (x 240 (x 360 (x
I was not yet able to compute when the second summation has a zero coefficient but, as it seems to be reversing just like for 3 variables, I'm guessing it will be for 12 squares. (
Update 9/29/09: I guessed wrong, it should be
6 (x 60 (x 672 (x 7920 (x 96096 (x
and so on. My thanks to Daniel Lichtblau of
For a combination of 3 and 2 variables, these may be derived by combining the identities given previously and we get:
72 (x 60 (x
For seven squares, the second summation would have a zero coefficient. A general identity when the first summation involves choosing
3*2
Going higher, for 3rd powers,
60 (x 120 (x 60 (x
For 4th powers,
5040 (x
and so on for other k powers. (
Update 9/30/09: Gerry Martens gave the complete solution to,
p (x
where we choose
p = (3/2)(-2) q = (1/2)(-2)
p (x
Define
p = (5/2) 2 q = 2
Note that, unlike for 4th powers,
10 (x
then the identity involving 5 squares given above, and so on. (
S ± (x
where the exterior sign is the product of the interior signs. (In other words, the term is negative if there is an odd number of negative interior signs; positive if even.) For the first few
(a+b)
^{2} - (a-b)^{2} = 4ab(a+b+c)
^{3} - (a-b+c)^{3} - (a+b-c)^{3} + (a-b-c)^{3} = 24abc(a+b+c+d)
^{4} - (a-b+c+d)^{4} - (a+b-c+d)^{4} - (a+b+c-d)^{4} + (a-b-c+d)^{4} + (a-b+c-d)^{4} + (a+b-c-d)^{4} - (a-b-c-d)^{4} = 192abcd
and so on for other
kth powers. This then generalizes the difference of two squares to a sum and difference of 2^{k-1 }kth powers.(Update, 11/18/09): I realized that
Boutin's Identity is behind some unusual but elegant identities as the ratio between exponents k and k+2 is simply a rational multiple of (x_{1}^{2}+x_{2}^{2}+...+x_{k}^{2}). For the first few k,2[a
^{2}+b^{2}][(a+b)^{2} - (a-b)^{2}] = (a+b)^{4} - (a-b)^{4}10[a
5[a (a+b+c-d)
and so on, with the last example tweaked a bit and also discussed in Section 8.4. ( IIIa. Lagrange’s Polynomial Identity
( ∑ a
where the first three summations are from k = 1 to
(a
(a
(a and so on. Note that for n = 4, while it is the sum of seven squares by Lagrange's Identity, it is just the sum of
four squares by Euler's Four-Square Identity. The case n = 3 shows that the product of two sums of three squares is likewise the sum of four squares.(Update, 12/21/09):
(x+y+z) (x+y+z) (x+y+z)
The last was used by Gabriel Lamé in his proof for
(x+y) (x+y) (x+y) (x+y) (x+y) (x+y) (x+y)
and so on. Note the slight difference between primes of form 6n-1 and 6n+1. (
(Update, 12/23/09): A.Verghote, Piezas
Define F
_{k} = x^{k}+y^{k}+z^{k}. Then,F F F F
_{1}F_{13} - F_{3}F_{11} = (x+y)(x+z)(y+z) (2F_{11}-F_{1}F_{10}+xyz(5F_{8}-2F_{1}F_{7}+2F_{2}F_{6}-2F_{3}F_{5}+F_{4}^{2})/2)and so on. When z = 0, the form is much simpler as the long expression on the RHS vanishes. Verghote provided the first identity, and I used
Mathematica to find the higher ones. After finding the next two, the pattern was easily seen, though I have no proof it goes on for all F_{1}F_{n+2} - F_{3}F_{n} with odd n > 3, though it will be odd if it stops. (End update) |