005b: Sums of n squares

 

 

IV. Some Identities of Squares

 

  1. Euler-Aida Ammei Identity
  2. Brahmagupta-Fibonacci Two-Square Identity
  3. Euler Four-Square Identity
  4. Degen-Graves-Cayley Eight-Squares Identity
  5. V. Arnold’s Perfect Forms
  6. Lagrange’s Identity
  7. Difference of Two Squares Identity

(Update, 10/26/09):  The Lebesgue Polynomial Identity is given by,

 

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

 

We can generalize this to the form,

 

x12+x22+… +xm2 = (y12+y22+… +yn2)2 

 

It can be proven there are polynomial identities with integer coefficients for all m = n (discussed in Sums of Three Squares).  But what about when m < n?  It turns out we can generalize the one by Lebesgue by looking at its underlying structure.  Note that this can be expressed as,

 

(p12+p22+q12+q22)2 - (p12+p22-q12-q22)2 = 4(p12+p22)(q12+q22)        (eq.1)

 

Let {a,b} = {p12+p22, q12+q22} and this reduces to the basic Difference of Two Squares Identity,

 

(a+b)2 - (a-b)2 = 4ab

 

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+… +xm2 = (y12+y22+… +yn2)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}
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+ ... +p2u2).  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+ ... +p4u2).  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 yi 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
 
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-…-xn)2 + å(2x1xn)2 = (x12+x22+…+xn2)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+…+xn2.  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(2x1xn)2 = (x12+x22+…+xn2)2

 

(b)2 + å(4ax1xn)2 = (x12+x22+…xn2)4,

 

(c)2 + å(8abx1xn)2 = (x12+x22+…xn2)8,

 

(d)2 + å(16abcx1xn)2 = (x12+x22+…xn2)16,

 

(e)2 + å(32abcdx1xn)2 = (x12+x22+…xn2)32,

 

and so on, where a = -(x12-x22-…-xn) 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 + …+ cnwn2 = (c1x12 + c2x22 + …+ cnxn2)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+… xn2)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 + …+ cnxn2)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+…+cnpn2 and distributing terms, we get,

 

U2 + c2p22V12 + … + cnpn2V12 = (p12 + c2p22 +…+ cnpn2)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 ci = 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 +…+ cnpn2 and so on.  For the more general non-monic case (c1y12 + …+ cnyn2)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 + …+ cnxn2)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 +…+ cnpn2, giving,

 

c1V2 + (c2p22 +… + cnpn2)U12 = (c1p12 + c2p22 +…+ cnpn2)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 +…+ cnpn2, 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 later.  Also, by Lagrange's Identity discussed below, 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 ui 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},
 

(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)

 

 

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, n} = {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, n} = {(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 ni 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,…an):= a1x12 + a2x22 +…+ anxn2

 

If we consider the equation,

 

(a1x12 + a2x22 +…+ anxn2)(a1y12 + a2y22 +…+ anyn2) = a1z12 + a2z22 +…+ anzn2

 

then for what constants {a1, a2,…an} 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 ai = 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 zi have a bilinear expression in terms of the xi and yi.  It might be interesting to know if the other 53 {a,b,c,d} have similar formulas for their zi.  In general, for ai 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 + … + xnyn )2 + S (xkyj – xjyk)2 = (x12 + … + xn2) (y12 + … + yn2 )  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+ … +xn2 + (y12+…+yn))2 – (x12+ … +xn2 – (y12+…+yn))2 = 4(x12+ … +xn2)(y12+ … +yn2 )

 

Proof:  Let a = x12+ … +xn2, b = y12+ … +yn2, 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) = 4(a2+b2+c2)(d2+e2+f2)
(a2+b2+c2+d2+e2+f2+g2+h2)2 - (a2+b2+c2+d2-e2-f2-g2-h2) = 4(a2+b2+c2+d2)(e2+f2+g2+h2)
 
and together with Lagrange’s Identity for n = 3,4 applied on the RHS respectively, this can be used to prove that the square of 6 squares can be expressed as the sum of five squares,

 

(a2+b2+c2-d2-e2-f2)2 + 4(ad+be+cf)2 + 4(ae-bd)2 + 4(af-cd)2 + 4(bf-ce)2 = (a2+b2+c2+d2+e2+f2)2

 
and provide an alternative soln to expressing the square of 8 squares as a sum of 8 squares,

 

(a2+b2+c2+d2-e2-f2-g2-h2)2 + 4((ae+bf+cg+dh)2 + (af-be)2 + (ag-ce)2 + (ah-de)2 + (bg-cf)2 + (bh-df)2 + (ch-dg)2) = (a2+b2+c2+d2+e2+f2+g2+h2)2

 
since the Euler-Aida Ammei gives it as,

 

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

 

Note 1:  By not using the Euler-Aida Ammei identity, are there always alternative solns to the square of n squares as the sum of n squares?

Note 2:  For what n is the square of n squares identically the sum of less than n squares?  (It can be for n = 4, 8, 16 (as 3, 5, 9 squares, respectively).  In fact, by setting appropriate variables equal to zero, it is the case for all n ≤ 16 other than n = 3, 5, 9.)  (See update at start of this section.)

 

A. Boutin

 

Boutin’s IdentityS ± (x1 ± x2 ±± xk)k = k! 2k-1x1x2…xk

 

where the exterior sign is the product of the interior signs.  (Or, the term is negative if there is an odd number of negative interior signs.)  The case k=2 gives the well-known,

 

(a+b)2 - (a-b)2 = 4ab

 

while for k = 3,4,

 

(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.  The case k=3 then implies that,

 

Piezas

 

(x13+x23+…+xn3) (y13+y23+…+yn3) = z13+ z23+z33+ z43

 

or, “The product of two sums of n cubes is the sum of four cubes.”

 

Proof:  Simply let a = x13+x23+…+xn3, b = y13+y23+…+yn3, and c = 9 in Boutin's Identity for k = 3.

 

 

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