001a: Assorted Identities, Part 1 (Various Forms)

(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 F_k = x^k+y^k+z^k. 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 F₁F_n+2 - F₃F_n with odd n > 3, though it will be odd if it stops. (End update)

6 (x₁²+x₂²+x₃²)² = ∑ (x_i ± x_j)⁴ + 2 ∑ x_i⁴

6 (x₁²+x₂²+x₃²+x₄²)² = ∑ (x_i ± x_j)⁴ + 0 ∑ x_i⁴

6 (x₁²+x₂²+ ... + x₅²)² = ∑ (x_i ± x_j)⁴ - 2 ∑ x_i⁴

and so on. For 3 variables taken at a time,

12 (x₁²+x₂²+x₃²)² = ∑ (x_i ± x_j ± x_k)⁴ + 8 ∑ x_i⁴

24 (x₁²+x₂²+x₃²+x₄²)² = ∑ (x_i ± x_j ± x_k)⁴ + 12 ∑ x_i⁴

36 (x₁²+x₂²+ ... + x₅²)² = ∑ (x_i ± x_j ± x_k)⁴ + 12 ∑ x_i⁴

48 (x₁²+x₂²+ ... + x₆²)² = ∑ (x_i ± x_j ± x_k)⁴ + 8 ∑ x_i⁴

60 (x₁²+x₂²+ ... + x₇²)² = ∑ (x_i ± x_j ± x_k)⁴ + 0 ∑ x_i⁴

72 (x₁²+x₂²+ ... + x₈²)² = ∑ (x_i ± x_j ± x_k)⁴ - 12 ∑ x_i⁴

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 = ∑ (x_i ± x_j ± x_k ± x_m)⁴ + 2⁴ ∑ x_i⁴

72 (x₁²+x₂²+ ... + x₅²)² = ∑ (x_i ± x_j ± x_k ± x_m)⁴ + 40 ∑ x_i⁴

144 (x₁²+x₂²+ ... + x₆²)² = ∑ (x_i ± x_j ± x_k ± x_m)⁴ + 64 ∑ x_i⁴

240 (x₁²+x₂²+ ... + x₇²)² = ∑ (x_i ± x_j ± x_k ± x_m)⁴ + 80 ∑ x_i⁴

360 (x₁²+x₂²+ ... + x₈²)² = ∑ (x_i ± x_j ± x_k ± x_m)⁴ + 80 ∑ x_i⁴

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 = ∑ (x_a ± x_b)⁴

60 (x₁²+x₂²+ ... + x₇²)² = ∑ (x_a ± x_b ± x_c)⁴

672 (x₁²+x₂²+ ... + x₁₀²)² = ∑ (x_a ± x_b ± x_c ± x_d)⁴

7920 (x₁²+x₂²+ ... + x₁₃²)² = ∑ (x_a ± x_b ± x_c ± x_d ± x_e)⁴

96096 (x₁²+x₂²+ ... + x₁₆²)² = ∑ (x_a ± x_b ± x_c ± x_d ± x_e ± x_f)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 2^m-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 = x_a⁴, 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 = ∑ (x_i ± x_j ± x_k)⁴ + 6 ∑ (x_i ± x_j)⁴

60 (x₁²+x₂²+ ...+ x₆²)² = ∑ (x_i ± x_j ± x_k)⁴ + 2 ∑ (x_i ± x_j)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*2^n-1 (x₁²+x₂²+ ... + x_n²)² = ∑ (x_a ± x_b ± ... ± x_n)⁴ + 2ⁿ ∑ x_n⁴ (Id.1)

Going higher, for 3rd powers,

60 (x₁²+x₂²+x₃²+x₄²)³ = ∑ (x_i ± x_j ± x_k)⁶ + 2 ∑ (x_i ± x_j)⁶ + 36 ∑ x_i⁶

120 (x₁²+x₂²+x₃²+x₄²)³ = ∑ (x_i ± x_j ± x_k ± x_m)⁶ + 6 ∑ (x_i ± x_j)⁶ + ∑ (2x_i)⁶

60 (x₁²+x₂²+x₃²+x₄²+x₅²)³ = ∑ (x_i ± x_j ± x_k)⁶ + 36 ∑ x_i⁶

For 4th powers,

5040 (x₁²+x₂²+x₃²+x₄²)⁴ = 6 ∑ (x_i ± x_j ± x_k ± x_m)⁸ + ∑ (2x_i ± x_j ± x_k)⁸ + 60 ∑ (x_i ± x_j)⁸ + 6 ∑ (2x_i)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+ ... + x_n²)² = ∑ (x_a ± x_b ± ... ± x_m)⁴ + q ∑ x_n⁴

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+ ... + x_n²)³ = ∑ (x_a ± x_b ± ... ± x_m)⁶ + q ∑ x_n⁶

Define r₂ = Binomial[3(m-2), m-2] and n = 3m-4, then,

p = (5/2) 2^m r₂

q = 2^m 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 ±…± x_k)^k = k! 2^k-1x₁x₂…x_k

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 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 (x₁²+x₂²+...+x_k²). 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

( ∑ a_k²) ( ∑ b_k²) = ( ∑ a_kb_k)² + ∑ (a_kb_j-a_jb_k)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 = ∑ (x_i ± x_j)⁴ + 4 ∑ x_i⁴