(Update, 12/21/09): IIIb. Lamé-Type Identities
(x+y+z)3 - (x3+y3+z3) = 3(x+y)(x+z)(y+z)
(x+y+z)5 - (x5+y5+z5) = 5(x+y)(x+z)(y+z)(x2+y2+z2+xy+xz+yz)
(x+y+z)7 - (x7+y7+z7) = 7(x+y)(x+z)(y+z)((x2+y2+z2+xy+xz+yz)2+xyz(x+y+z))
The last was used by Gabriel Lamé in his proof for Fermat’s Last Theorem for k = 7. (I do not know a simple form for k = 11 and higher primes.) The form is simpler when z = 0. Define {a,b,c} = {xy, x+y, x2+xy+y2}. Then,
(x+y)3 - x3 - y3 = 3ab
(x+y)5 - x5 - y5 = 5abc
(x+y)7 - x7 - y7 = 7abc2
(x+y)11 - x11 - y11 = 11abc(a2b2+c3)
(x+y)13 - x13 - y13 = 13abc2(2a2b2+c3)
(x+y)17 - x17 - y17 = 17abc(a4b4+5a2b2c3+c6)
(x+y)19 - x19 - y19 = 19abc2(3a2b2+7a2b2c3+c6)
and so on. Note the slight difference between primes of form 6n-1 and 6n+1. (End update.)
(Update, 12/23/09): A.Verghote, Piezas
Define Fk = xk+yk+zk. Then,
F1F7 - F3F5 = (x+y)(x+z)(y+z) (2F5-F1F4+xyz(5F2-F12)/2)
F1F9 - F3F7 = (x+y)(x+z)(y+z) (2F7-F1F6+xyz(5F4-2F1F3+F22)/2)
F1F11 - F3F9 = (x+y)(x+z)(y+z) (2F9-F1F8+xyz(5F6-2F1F5+2F2F4-F32)/2)
F1F13 - F3F11 = (x+y)(x+z)(y+z) (2F11-F1F10+xyz(5F8-2F1F7+2F2F6-2F3F5+F42)/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 F1Fn+2 - F3Fn with odd n > 3, though it will be odd if it stops. (End update)
6 (x12+x22+x32)2 = ∑ (xi ± xj)4 + 2 ∑ xi4
6 (x12+x22+x32+x42)2 = ∑ (xi ± xj)4 + 0 ∑ xi4
6 (x12+x22+ ... + x52)2 = ∑ (xi ± xj)4 - 2 ∑ xi4
and so on. For 3 variables taken at a time,
12 (x12+x22+x32)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi4
24 (x12+x22+x32+x42)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi4
36 (x12+x22+ ... + x52)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi4
48 (x12+x22+ ... + x62)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi4
60 (x12+x22+ ... + x72)2 = ∑ (xi ± xj ± xk)4 + 0 ∑ xi4
72 (x12+x22+ ... + x82)2 = ∑ (xi ± xj ± xk)4 - 12 ∑ xi4
Note how for seven squares, the second summation has a zero coefficient like for the Lucas-Liouville identity. For 4 variables taken at a time,
24 (x12+x22+x32+x42)2 = ∑ (xi ± xj ± xk ± xm)4 + 24 ∑ xi4
72 (x12+x22+ ... + x52)2 = ∑ (xi ± xj ± xk ± xm)4 + 40 ∑ xi4
144 (x12+x22+ ... + x62)2 = ∑ (xi ± xj ± xk ± xm)4 + 64 ∑ xi4
240 (x12+x22+ ... + x72)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi4
360 (x12+x22+ ... + x82)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi4
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. (Anyone can check if this conjecture is true, as well as provide the identities for the next level of 5 variables taken at a time?) See the complete soln by Gerry Martens at the 9/30 update below.
Update 9/29/09: I guessed wrong, it should be 10 squares, and the infinite family is given by:
6 (x12+x22+x32+x42)2 = ∑ (xa ± xb)4
60 (x12+x22+ ... + x72)2 = ∑ (xa ± xb ± xc)4
672 (x12+x22+ ... + x102)2 = ∑ (xa ± xb ± xc ± xd)4
7920 (x12+x22+ ... + x132)2 = ∑ (xa ± xb ± xc ± xd ± xe)4
96096 (x12+x22+ ... + x162)2 = ∑ (xa ± xb ± xc ± xd ± xe ± xf)4
and so on. My thanks to Daniel Lichtblau of Wolfram Research (makers of Mathematica), and Renzo Benedetti, Martin Rubey, and Gerry Martens from sci.math.symbolic who answered my questions. One can see that the number of squares {4, 7, 10, 13, ...} is just in arithmetic progression and Benedetti observed that the numerical factor is generated by 2m-1 Binomial[3(m-1), m-1] where m is the number of objects taken at a time per term in the RHS. Including the leading coefficient of the trivial identity 1(x12)2 = xa4, this sequence {1, 6, 60, 672, 7920,...} is, as of this date, not yet in the Online Encyclopedia of Integer Sequences, but is there without the powers of 2 as sequence {1, 3, 15, 84, 495,...}. (End update, 9/29)
For a combination of 3 and 2 variables, these may be derived by combining the identities given previously and we get:
72 (x12+x22+ ...+ x52)2 = ∑ (xi ± xj ± xk)4 + 6 ∑ (xi ± xj)4
60 (x12+x22+ ...+ x62)2 = ∑ (xi ± xj ± xk)4 + 2 ∑ (xi ± xj)4
For seven squares, the second summation would have a zero coefficient. A general identity when the first summation involves choosing n objects from n squares is given by,
3*2n-1 (x12+x22+ ... + xn2)2 = ∑ (xa ± xb ± ... ± xn)4 + 2n ∑ xn4 (Id.1)
Going higher, for 3rd powers,
60 (x12+x22+x32+x42)3 = ∑ (xi ± xj ± xk)6 + 2 ∑ (xi ± xj)6 + 36 ∑ xi6
120 (x12+x22+x32+x42)3 = ∑ (xi ± xj ± xk ± xm)6 + 6 ∑ (xi ± xj)6 + ∑ (2xi)6
60 (x12+x22+x32+x42+x52)3 = ∑ (xi ± xj ± xk)6 + 36 ∑ xi6
For 4th powers,
5040 (x12+x22+x32+x42)4 = 6 ∑ (xi ± xj ± xk ± xm)8 + ∑ (2xi ± xj ± xk)8 + 60 ∑ (xi ± xj)8 + 6 ∑ (2xi)8
and so on for other k powers. (End update, 9/26.)
Update 9/30/09: Gerry Martens gave the complete solution to,
p (x12+x22+ ... + xn2)2 = ∑ (xa ± xb ± ... ± xm)4 + q ∑ xn4
where we choose m out of n objects. First, define r1 = Pochhammer[2-n, m-2] / (m-2)!, then,
p = (3/2)(-2)m r1
q = (1/2)(-2)m r1 (-n+3m-2)/(m-1)
Mathematica's Pochhammer[a, n] gives the rising factorial (a)n = a(a+1)...(a+n-1). Note that by choosing some constant m, we can set q = 0 by letting n = 3m-2, which then yields the infinite family starting with the Lucas-Liouville identity at m = 2. If m = n, this is the same as (Id.1). For 6th powers, Renzo Benedetti solved,
p (x12+x22+ ... + xn2)3 = ∑ (xa ± xb ± ... ± xm)6 + q ∑ xn6
Define r2 = Binomial[3(m-2), m-2] and n = 3m-4, then,
p = (5/2) 2m r2
q = 2m r2 m/(m-1)
Note that, unlike for 4th powers, n depends on m and one also can't set q = 0 except for the trivial case m = 0. Starting with m = 2 to prevent division by zero, we get,
10 (x12+x22)3 = (x1+x2)6 + (x1-x2)6 + 8(x16 +x26)
then the identity involving 5 squares given above, and so on. (End update, 9/30.)
IIb. Boutin’s Identity
S ± (x1 ± x2 ±…± xk)k = k! 2k-1x1x2…xk
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 k:
(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 2k-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 (x12+x22+...+xk2). For the first few k,
2[a2+b2][(a+b)2 - (a-b)2] = (a+b)4 - (a-b)4
10[a2+b2+c2][(a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3] = 3[(a+b+c)5 - (a-b+c)5 - (a+b-c)5 + (a-b-c)5]
5[a2+b2+c2+d2][(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =
(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6
and so on, with the last example tweaked a bit and also discussed in Section 8.4. (End update.)
IIIa. Lagrange’s Polynomial Identity
( ∑ ak2) ( ∑ bk2) = ( ∑ akbk)2 + ∑ (akbj-ajbk)2
where the first three summations are from k = 1 to n, while the fourth is 1≤k<j≤n. For n = 2,3,4 these are,
(a12+a22)(b12+b22) = (a1b1+a2b2)2 + (a1b2-a2b1)2
(a12+a22+a32)(b12+b22+b32) = (a1b1+a2b2+a3b3)2 + (a1b2-a2b1)2 + (a1b3-a3b1)2 + (a2b3-a3b2)2
(a12+a22+a32+a42)(b12+b22+b32+b42) = (a1b1+a2b2+a3b3+a4b4)2 + (a1b2-a2b1)2 + (a1b3-a3b1)2 + (a1b4-a4b1)2 + (a2b3-a3b2)2 + (a2b4-a4b2)2 +(a3b4-a4b3)2
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.
6 (x12+x22)2 = ∑ (xi ± xj)4 + 4 ∑ xi4
Back to Index Go to Updates Page
1. Various Forms
I. Maillet’s Identity
IIa. Lucas-Liouville Polynomial Identity
IIb. Boutin’s Identity
IIIa. Lagrange’s Polynomial Identity
IIIb. Lamé-Type Identities
IV. Waring-like Problems
I. Maillet's Identity (E. Maillet)
(a+b)3+(a+c)3+(a+d)3+(a-b)3+(a-c)3+(a-d)3 = 6a(a2+b2+c2+d2)
IIa. Lucas-Liouville Polynomial Identity
(a+b)k+(a+c)k+(a+d)k+(b+c)k+(b+d)k+(c+d)k+(a-b)k+(a-c)k+(a-d)k+(b-c)k+(b-d)k+(c-d)k = 6(a2+b2+c2+d2)k/2, for k = 2,4
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 (x12+x22+ ... +xn2)k can be written as a rational combination of (2k)th powers of linear forms in the xi. (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 (x12+x22+x32+x42)2 = ∑ (xi ± xj)4
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 Mathematica as Binomial[4, 2] = 6.) Since there are 2 sign changes, this gives a total of 6 x 2 = 12 terms on one side of the equation, which can be seen in the explicit identity. This in fact belongs to an infinite family where the addends involve 2 variables (xi ± xj) taken at a time and the xi taken singly: