005b: Sums of n squares

(a²+b²+c²+d²+e²+f²+g²+h²)² - (a²+b²+c²+d²-e²-f²-g²-h²)² = 4(a²+b²+c²+d²)(e²+f²+g²+h²)

5. V. Arnold’s Perfect Forms

Let {a,b,c} be in the integers. Given the equation (au²+buv+cv²)(ax²+bxy+cy²) = az₁²+bz₁z₂+cz₂². If for any integral integral {u,v,x,y} one can always find integral {z₁, z₂}, then the binary quadratic form F(a,b,c) is defined as a perfect form.

Theorem 1: “The product of three binary quadratic forms F(a,b,c) is of like form.”

Proof: (an₁²+bn₁n₂+cn₂²)(au²+buv+cv²)(ax²+bxy+cy²) = az₁²+bz₁z₂+cz₂²

where,

z₁ = u(n₃x+cn₂y)+cv(n₂x-n₁y)

z₂ = v(an₁x+n₄y)-au(n₂x-n₁y)

and {n₃, n₄ } = {an₁+bn₂, bn₁+cn₂} from which immediately follows,

Corollary:“If there are integers {n₁, n₂} such that F(a,b,c) = 1, then F(a,b,c) is a perfect form.”

Note that if F(a,b,c) is monic, the soln {x,y} = {1,0} immediately implies this form is perfect. But by dividing z₁, z₂ with c, a, respectively, and modifying the expressions for {n₃, n₄} will result in a second theorem,

Theorem 2: “If there are integers {n₁, n₂, n₃, n₄} such that an₁²+bn₁n₂+cn₂² = ac, n₃ = (an₁+bn₂)/c, n₄ = (bn₁+cn₂)/a, then F(a,b,c) is a perfect form.”

Proof: (an₁²+bn₁n₂+cn₂²)(au²+buv+cv²)(ax²+bxy+cy²) = (az₁²+bz₁z₂+cz₂²)(ac)

where,

z₁ = v(n₁x+n₄y)-u(n₂x-n₁y)

z₂ = u(n₃x+n₂y)+v(n₂x-n₁y)

and {n₃, n₄ } = {(an₁+bn₂)/c, (bn₁+cn₂)/a}.

The expressions are essentially the same as in Theorem 1 but have been divided by c,a. This second class is relevant to quadratic discriminants d with class number h(d) = 3m. For imaginary fields with h(-d) = 3, there are sixteen fundamental d, all of which have its associated F(a,b,c) as perfect forms. For brevity, only the first three will be given and in the format {a,b,c}, {n₁, n₂, n₃, n₄}:

d = 23; {2,1,3}, {1,1,1,2}

d = 31; {2,1,4}, {2,0,1,1}

d = 59; {3,1,5}, {-2,1,-1,1}

For real fields with h(d) = 3, there are forty-two d in the Online Encyclopedia of Integer Sequences, all of which also have F(a,b,c) as perfect. However, while most of the n_i for negative d with h(d) = 3,6 were only single digits, for positive d these can get quite large. For ex,

d = 2857; {2,51,-32}, {3326866, -127404, -4879, 86873547}

Q: Any other theorems regarding the product of two or three binary quadratic forms? We can generalize this somewhat and go to diagonal n-nary quadratic forms (one without cross terms),

F(a₁,a₂,…a_n):= a₁x₁² + a₂x₂² +…+ a_nx_n²

If we consider the equation,

(a₁x₁² + a₂x₂² +…+ a_nx_n²)(a₁y₁² + a₂y₂² +…+ a_ny_n²) = a₁z₁² + a₂z₂² +…+ a_nz_n²

then for what constants {a₁, a₂,…a_n} is there such that the product of two diagonal n-nary quadratic forms is of like form? Most of the results have been limited to the special case of all a_i = 1 and n = 2,4,8, namely the Brahmagupta-Fibonacci, Euler, and Degen-Graves identities discussed above. The first can be generalized to the form {1,p}, the second to {1, p, q, pq} by Lagrange, and the third to {1,1, p,p, q,q, pq, pq}. Ramanujan in turn generalized Lagrange’s Four-Square Theorem and found 54 {a,b,c,d} such that ax₁²+bx₂²+cx₃²+dx₄² can represent all positive integers, namely,

{1,1,1,v}; v = 1-7

{1,1,2,v}; v = 2-14

{1,1,3,v}; v = 3-6

{1,2,2,v}; v = 2-7

{1,2,3,v}; v = 3-10

{1,2,4,v}; v = 4-14

{1,2,5,v}; v = 6-10

which is the complete list. (Note: Incidentally, it would have been expected that the last would be for v = 5-10. What is the smallest positive integer not expressible by the form {1,2,5,5}?) All 54 are then perfect quaternary forms since, needless to say, the product of two positive integers is always a positive integer. For the first case {1,1,1,1} this is just Euler’s four-square identity and the z_i have a bilinear expression in terms of the x_i and y_i. It might be interesting to know if the other 53 {a,b,c,d} have similar formulas for their z_i. In general, for a_i not all equal to unity, what other results are there for the product of two n-nary quadratic forms, especially for n not a power of two?

Update, 9/21/09: Turns out the form {1,2,5,5} cannot express the number 15. See "The 15 and 290 Theorems" by Conway, Schneeberger, and Bhargava.

6. Lagrange’s Identity

A faintly similar identity to the sum-product of n squares given previously is,

(x₁y₁ + … + x_ny_n )² + S (x_ky_j – x_jy_k)² = (x₁² + … + x_n²) (y₁² + … + y_n² ) for 1£ k<j£n.

For n=3, this has 4 addends,

(x₁y₁+x₂y₂+x₃y₃ )² + (x₁y₂ – x₂y₁ )² +(x₁y₃ – x₃y₁ )² + (x₂y₃ – x₃y₂ )² = (x₁²+x₂²+x₃²) (y₁²+y₂²+y₃²)

while n=4 already involves 7 addends, and is an alternative way to express Euler's Four-Square Identity. A general identity was found by A. Cauchy and J. Young. The case n=3 was also rediscovered by T.Weddle while studying the semi-axes of an ellipsoid.

7. Difference of Two Squares Identity

A difference-product identity on the other hand is given by,

(x₁²+ … +x_n² + (y₁²+…+y_n))² – (x₁²+ … +x_n² – (y₁²+…+y_n))² = 4(x₁²+ … +x_n²)(y₁²+ … +y_n² )

Proof: Let a = x₁²+ … +x_n², b = y₁²+ … +y_n², then this is just the basic identity, (a+b)²-(a-b)² = 4ab. (End proof.) If the right hand side of the identity is non-trivially expressible as the sum of n squares, as is the case for n = 2,4,8, this automatically implies a square of 2n squares expressible as the sum of n+1 squares, thus explaining the Three, Five, Nine-Square Identities above. (The case n = 1 just gives the formula for Pythagorean triples.) Let,

(a²+b²+c²+d²+e²+f²)² - (a²+b²+c²-d²-e²-f²)² = 4(a²+b²+c²)(d²+e²+f²)

(a²+ ub²+c²+ud²+ve²+uvf²+vg²+uvh²) (m²+un²+o²+up²+vq²+uvr²+vs²+uvt²) = x₁²+ux₂²+x₃²+ux₄²+vx₅²+uvx₆²+vx₇²+uvx₈²

x₁ = am-bnu-co-dpu-eqv-fruv-gsv-htuv

x₂ = bm+an+do-cp+fqv-erv-hsv+gtv

x₃ = cm-dnu+ao+bpu+gqv+hruv-esv-ftuv

x₄ = dm+cn-bo+ap+hqv-grv+fsv-etv

x₅ = em-fnu-go-hpu+aq+bru+cs+dtu

x₆ = fm+en-ho+gp-bq+ar-ds+ct

x₇ = gm+hnu+eo-fpu-cq+dru+as-btu

x₈ = hm-gn+fo+ep-dq-cr+bs+at

(End update)

and so on. (End update.)

1. Euler-Aida Ammei Identity

Theorem: “The square of the sum of n squares is itself a sum of n squares.”

(x₁²-x₂²-…-x_n)² + å(2x₁x_n)² = (x₁²+x₂²+…+x_n²)²

Examples:

(a²-b²)² + (2ab)² = (a²+b²)²

(a²-b²-c²)² + (2ab)² + (2ac)² = (a²+b²+c²)²

(a²-b²-c²-d²)² + (2ab)² + (2ac)² + (2ad)² = (a²+b²+c²+d²)²

and so on. Note that these can be alternatively expressed as the basic identity,

(x₁²-x₀)² + (2x₁)²x₀ = (x₁²+x₀)²

for arbitrary x₁ and x₀, where for the theorem it was set x₀ = x₂²+x₃²+…+x_n². By letting x₀ = x₂², one can see this basic identity is essentially the formula for Pythagorean triples. A stronger result by M. Moureaux is that, “The kth power of the sum of n squares, for k a power of 2, is itself a sum of n squares.” After some time with Mathematica, I observed that there seems to be this beautifully consistent pattern in the identities based on a certain algebraic form. Let the summation å be from m = {2 to n}, then,

(a)² + å(2x₁x_n)² = (x₁²+x₂²+…+x_n²)²

(b)² + å(4ax₁x_n)² = (x₁²+x₂²+…x_n²)⁴,

(c)² + å(8abx₁x_n)² = (x₁²+x₂²+…x_n²)⁸,

(d)² + å(16abcx₁x_n)² = (x₁²+x₂²+…x_n²)¹⁶,

(e)² + å(32abcdx₁x_n)² = (x₁²+x₂²+…x_n²)³²,

and so on, where a = -(x₁²-x₂²-…-x_n) and b = (4x₁⁴+4ax₁²-a²), c = (4a⁴+4a²b-b²), d = (4b⁴+4b²c-c²), e = (4c⁴+4c²d-d²), etc. It seems there is this “recurrence relation” involving the algebraic form 4u⁴+4u²v-v². I have no proof this in fact goes on, though it would be odd if the pattern stops. (And the algebraic form factors over √2, which is only appropriate since we are dealing with powers of two.)

To generalize the Euler-Aida Ammei identity: For what k is the kth power of n diagonal quadratic forms identically a sum of like form? Or,

c₁w₁² + c₂w₂² + …+ c_nw_n² = (c₁x₁² + c₂x₂² + …+ c_nx_n²)^k

It turns out for the monic case, or when c₁ = 1, the answer is for all positive integer k. In a previous section, it was proven that, “the kth power of the sum of n squares (x₁²+x₂²+… x_n²)^k is itself the sum of n squares”. It takes only a very small modification of the proof to generalize this.

Theorem 1. “The expression (c₁x₁² + c₂x₂² + …+ c_nx_n²)^k, for c₁ = 1, is identically a sum of n squares of like form for all positive integer n and k.”

Proof: (Piezas) To recall, let the expansion of the complex number (a±bi)^k be,

U+Vi = (a+bi)^k; U-Vi = (a-bi)^k

where U,V are expressions in the arbitrary a,b. Their product, or norm, is,

U²+V² = (a²+b²)^k

Since b can be factored out in V, or V = V₁b, if we let {a,b} = {p₁,√p₀}, then,

U² + p₀V₁² = (p₁²+p₀)^k

All these are familiar. The slightly different step is that since p₀ is arbitrary, one can choose it to be the sum of squares of form p₀ = c₂p₂²+c₃p₃²+…+c_np_n² and distributing terms, we get,

U² + c₂p₂²V₁² + … + c_np_n²V₁² = (p₁² + c₂p₂² +…+ c_np_n²)^k

thus proving the kth power of the right hand side of the eqn is a sum of squares of like form. (End proof.) For k = 2 and all c_i = 1, this is of course the Euler-Aida Ammei identity and the inspiration for the proof. For k = 3, this is,

(p₁³-3p₀p₁)² + p₀(3p₁²-p₀)² = (p₁²+p₀)³

for p₀ = c₂p₂² +…+ c_np_n² and so on. For the more general non-monic case (c₁y₁² + …+ c_ny_n²)^k, some particular identities are also known for the case k=3,

J. Neuberg

ax₁²+bx₂²+cx₃² = (ap²+bq²+cr²)³

{x₁, x₂, x₃} = {p(y-2z), q(y-2z), ry}, if y = 4(ap²+bq²)-z, z = ap²+bq²+cr²

G. de Longchamps

ax₁²+bx₂²+cx₃²+dx₄² = (ap²+bq²+cr²+ds²)³

{x₁, x₂, x₃, x₄} = {p(y-2z), q(y-2z), ry, sy}, if y = 4(ap²+bq²)-z, z = ap²+bq²+cr²+ds²

This author observed that this can be generalized as,

ax₁²+bx₂²+cx₃²+dx₄²+ex₅² = (ap²+bq²+cr²+ds²+et²)³

{x₁, x₂, x₃, x₄, x₅} = {p(y-2z), q(y-2z), ry, sy, ty}, if y = 4(ap²+bq²)-z, z = ap²+bq²+cr²+ds²+et²

and so on for n variables by simply modifying z. A more systematic approach is given below.

Theorem 2. “The non-monic expression (c₁ x₁² + c₂x₂² + …+ c_nx_n²)^k is identically a sum of n squares of like form for all positive integer n and odd k.”

Proof: The proof is a variation of the one above. One simply solves the equation,

U²+c₁V² = (x²+c₁y²)^k,

by equating its linear factors,

U+V√-c₁ = (x+y√-c₁)^k, U-V√-c₁ = (x-y√-c₁)^k,

and easily solving for U,V as expressions in x,y. Since, for odd k, x can be factored out in U, or U = U₁x, if we let {x,y} = {√p₀, p₁} then,

p₀U₁² + c₁V² = (p₀+c₁p₁²)^k,

where p₀ can then be set as the sum of squares p₀ = c₂p₂² +…+ c_np_n², giving,

c₁V² + (c₂p₂² +… + c_np_n²)U₁² = (c₁p₁² + c₂p₂² +…+ c_np_n²)^k

proving that, for odd k, the right hand side is identically the sum of squares of like form. (End of proof.) For k = 3, this is,

c₁(c₁p₁³-3p₀p₁)² + p₀(3c₁p₁²-p₀)² = (c₁p₁²+p₀)³

for p₀ = c₂p₂² +…+ c_np_n², and so on for all odd k.

2. Brahmagupta-Fibonacci Two-Square Identity

(ac+bd)² + (ad-bc)² = (a²+b²)(c²+d²)

This can be generalized as,

(ac+nbd)² + n(ad-bc)² = (a²+nb²)(c²+nd²)

From the Two-Square we can derive the Euler-Lebesgue Three-Square,

(a²+b²-c²-d²)² + (2ac+2bd)² + (2ad-2bc)² = (a²+b²+c²+d²)²

This can be generalized by the Fauquembergue n-Squares Identity. It is a bit difficult to convey with limited notation but in one form can be seen as,

(a²+b²-c²-d²+x)² + (2ac+2bd)² + (2ad-2bc)² + 4x(c²+d²) = (a²+b²+c²+d²+x)²

where x is arbitrary and can be chosen as any sum of n squares. Note that for x = 0 this reduces to the Euler-Lebesgue. For the case x as a single square this gives, after minor changes in variables,

(a²+b²+c²-d²-e²)² + (2ad+2ce)² + (2ae-2cd)² + (2bd)² + (2be)² = (a²+b²+c²+d²+e²)²

distinct from the Euler-Aida Ammei identity for n = 5 which is given by,

(a²-b²-c²-d²-e²)² + (2ab)² + (2ac)² + (2ad)² + (2ae)² = (a²+b²+c²+d²+e²)²

For x = e²+f², it results in seven squares whose sum is the square of six squares:

(a²+b²+c²+d²-e²-f²)² + (2ae+2df)² + (2af-2de)² + (2be)² + (2bf)² + (2ce)² + (2cf)² = (a²+b²+c²+d²+e²+f²)²

and so on for other x.

3. Euler Four-Square Identity

(a²+b²+c²+d²) (e²+f²+g²+h²) = u₁² + u₂² + u₃² + u₄²

u₁ = ae-bf-cg-dh

u₂ = af+be+ch-dg

u₃ = ag-bh+ce+df

u₄ = ah+bg-cf+de

Note that a cubic version, in fact, is possible, (x₁³+x₂³+x₃³+x₄³) (y₁³+y₂³+y₃³+y₄³) = z₁³+ z₂³+z₃³+ z₄³, to be discussed below. Also, by Lagrange's Identity, the product can be expressed as the sum of seven squares,

(a²+b²+c²+d²) (e²+f²+g²+h²) = (ae+bf+cg+dh)² + (af-be)² + (ag-ce)² + (ah-de)² + (bg-cf)² + (bh-df)² + (ch-dg)²

A more general version for squares was also given by Lagrange as,

(a²+mb²+nc²+mnd²) (p²+mq²+nr²+mns²) = x₁²+mx₂²+nx₃²+mnx₄²

x₁ = ap-mbq-ncr+mnds,

x₂ = aq+bp-ncs-ndr,

x₃ = ar+mbs+cp+mdq,

x₄ = as-br+cq-dp

In analogy to the Three-Square, we can also find a Five-Square (by yours truly),

(a²+b²+c²+d²-e²-f²-g²-h²)² + (2u₁)² + (2u₂)² + (2u₃)² + (2u₄)² = (a²+b²+c²+d²+e²+f²+g²+h²)²

with the u_i as defined above. Hence this is another case of a square of n squares expressed in less than n squares. And, in analogy to Fauquembergue’s n squares, another kind of n squares identity can be derived from the Five-Square as,

(a²+b²+c²+d²-e²-f²-g²-h²+x)² + (2u₁)² + (2u₂)² + (2u₃)² + (2u₄)² + 4x(e²+f²+g²+h²) = (a²+b²+c²+d²+e²+f²+g²+h²+x)²

where x again can be any number of squares. For x a square, this can give a 9-square identity.

4. Degen-Graves-Cayley Eight-Squares Identity (DGC)

(a²+b²+c²+d²+e²+f²+g²+h²) (m²+n²+o²+p²+q²+r²+s²+t²) = v₁²+v₂²+v₃²+v₄²+v₅²+v₆²+v₇²+v₈²

v₁ = am-bn-co-dp-eq-fr-gs-ht

v₂ = bm+an+do-cp+fq-er-hs+gt

v₃ = cm-dn+ao+bp+gq+hr-es-ft

v₄ = dm+cn-bo+ap+hq-gr+fs-et

v₅ = em-fn-go-hp+aq+br+cs+dt

v₆ = fm+en-ho+gp-bq+ar-ds+ct

v₇ = gm+hn+eo-fp-cq+dr+as-bt

v₈ = hm-gn+fo+ep-dq-cr+bs+at

For convenience, let {a, b,…t} = {a₁, a₂,…a₁₆}. This can also give a Nine-Square Identity (distinct from the one in the previous section) as,

(a₁²+…+ a₈²- a₉²-…- a₁₆²)² + (2v₁)² + …+ (2v₈)² = (a₁²+ a₂² +… + a₁₆²)²

and the DGC n-Squares identity,

(a₁²+…+ a₈²- a₉²-…- a₁₆² + x)² + (2v₁)² + …+ (2v₈)² + 4x(a₉² +… + a₁₆²) = (a₁²+ a₂² +… + a₁₆² + x)²

Since the DGC is the last bilinear n-squares identity, these two should also be the last of their kind.

(Update, 10/26/09): Just like the Two-Square and Four-Square, the Eight-Square Identity can be generalized. For arbitrary {u, v},

b) {m,n} = {4v-3, 4v}

c) {m,n} = {8v-7, 8v}

III. For m < n, with n odd (v > 1):

a) For n = 4v-1, then m = 4v-3.

b) For n = {8v-1, 8v-3, 8v-5}, then m = 8v-7.

Thus, for example, (y₁²+y₂²+… +y₁₆²)² can be identically expressed as m squares for m = {1, 9, 13, 15, 16}. (Q. Is this the most number of m when m ≤ n?) However, when m < n, and after eliminating the possibility of (x₁²+x₂²+x₃²)² expressed as two non-zero squares, two cases are still unresolved: n = 5 and n = 8v-7. Whether or not there are identities for these remains to be seen.

Proof: One simply uses the difference of two squares identity:

(a+b)² + (a-b)² = 4ab (eq.1)

Let a = (p₁²+p₂²+p₃²+p₄²+ ... +p_2u²). Disregarding the square numerical factor, the RHS of eq.1 becomes (by distributing in pairs) as,

b(p₁²+p₂²) + b(p₃²+p₄²) + ...

Let b = (q₁²+q₂²). Since by the Bramagupta-Fibonacci Identity (x₁²+x₂²)(y₁²+y₂²) = z₁²+z₂², then the RHS can be expressed as 2u squares. Thus, {m,n} = {2u+1, 2u+2} or, equivalently, {2v-1, 2v}. However, if the identity is not used on one pair which is instead expressed as four squares, then the RHS has 2u+2 squares. Thus {m,n} = {2u+3, 2u+2} or, {2v+1, 2v}, proving part (Ia) of Theorem 3.

Similarly, let a = (p₁²+p₂²+p₃²+p₄²+ ... +p_4u²). Disregarding again the square numerical factor, the RHS (by distributing fourwise) is,

b(p₁²+p₂²+p₃²+p₄²) + b(p₅²+p₆²+p₇²+p₈²) + ...

Let b = (q₁²+q₂²+q₃²+q₄²). Since by the Euler Four-Square Identity (x₁²+x₂²+x₃²+x₄²)(y₁²+y₂²+y₃²+y₄²) = z₁²+z₂²+z₃²+z₄², then the RHS can be expressed as 4u squares. Thus, {m,n} = {4u+1, 4u+4}, or {4v-3, 4v}, proving part (Ib). The last part is proven using the last such identity (by the Hurwitz Theorem) namely, the Degen Eight-Square Identity. Finally, part II of the Theorem can be proven by simply setting the appropriate number of variables y_i as equal to zero. For example, we can reduce the {5,8} Identity to a {5,7} by letting one of the eight squares as zero to get,

(a²+b²+c²+d²-e²-f²-g²)² + 4(ae-bf-cg)² + 4(af+be-dg)² + 4(ag+ce+df)² + 4(bg-cf+de)² = (a²+b²+c²+d²+e²+f²+g²)²

a special case of Boutin's Theorem also given at the end of this section which generalizes it to a sum and difference of kth powers. Since (p₁²+p₂²)(q₁²+q₂²) = r₁²+r₂² by the Bramagupta-Fibonacci Two-Square Identity, then the RHS of eq.1 can be expressed as the sum of two squares, which explains the Lebesgue Identity. But there are the higher Euler Four-Square Identity and Degen Eight-Square Identity. (See also the article, Pfister's 16-Square Identity.) Thus, let {a,b} = {p₁²+p₂²+p₃²+p₄², q₁²+q₂²+q₃²+q₄²} and we get,

(p₁²+p₂²+p₃²+p₄²+q₁²+q₂²+q₃²+q₄²)² - (p₁²+p₂²+p₃²+p₄²-q₁²-q₂²-q₃²-q₄²)² = 4(p₁²+p₂²+p₃²+p₄²)(q₁²+q₂²+q₃²+q₄²)

Since the RHS can be expressed as four squares, this gives an identity for the square of 8 squares as the sum of five squares. Using the Degen Eight-Square, this gives one for the square of 16 squares as nine squares. In general, we can have a Theorem 3 to complement the first two in Sums of Three Squares.

Theorem 3: Given x₁²+x₂²+… +x_m² = (y₁²+y₂²+… +y_n²)², where m ≤ n, one can identically express the square of n squares as m squares for:

I. All m = n.

II. For m < n, with n even (v > 1):

a) {m,n} = {2v±1, 2v}

(a+b)² - (a-b)² = 4ab