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 Identity: S ± (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.
(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 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)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-…-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 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 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},
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
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}
(a+b)2 - (a-b)2 = 4ab
IV. Some Identities of Squares
Euler-Aida Ammei Identity
Brahmagupta-Fibonacci Two-Square Identity
Euler Four-Square Identity
Degen-Graves-Cayley Eight-Squares Identity
V. Arnold’s Perfect Forms
Lagrange’s Identity
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,