005: Sums of four or more squares
(Update, 2/2/6/10): Alain Verghote gave a generalization as two (2.5.6) identities,
Form 1:
(-a+b+c+d+e)k + (a-b+c+d+e)k + (a+b-c+d+e)k + (a+b+c-d+e)k + (a+b+c+d-e)k = (2a)k + (2b)k + (2c)k + (2d)k + (2e)k + (a+b+c+d+e)k, for k = 1,2
Form 2:
(-a+b+c+d+e)2 + (a-b+c+d+e)2 + (a+b-c+d+e)2 + (a+b+c-d+e)2 + (a+b+c+d-e)2 = (-a+b+c+d)2 + (a-b+c+d)2 + (a+b-c+d)2 + (a+b+c-d)2 + (a+b+c+d+e)2 + (2e)2
Note how for e = 0, Form 1 reduces to Zehfuss', while Form 2 reverts into a tautology. These can be further generalized as,
(-a+b+c+d+e+f)k + (a-b+c+d+e+f)k + (a+b-c+d+e+f)k + (a+b+c-d+e+f)k + (a+b+c+d-e+f)k + (a+b+c+d+e-f)k = (2a)k + (2b)k + (2c)k + (2d)k + (2e)k + (2f)k + 2(a+b+c+d+e+f)k, for k = 1,2
and so on. Setting f = 0 reduces to Form 1. The significance of Birck's and Zehfuss' symmetric versions is that they can give rise to the multi-grade identity,
(-a+b+c+d)k + (a-b+c+d)k + (a+b-c+d)k + (a+b+c-d)k + (2e)k + (2f)k + (2g)k + (2h)k =
(-e+f+g+h)k + (e-f+g+h)k + (e+f-g+h)k + (e+f+g-h)k + (2a)k + (2b)k + (2c)k + (2d)k, for k = 1,2,4,6,8
if abcd = efgh and an+bn+cn+dn = en+fn+gn+hn, for n = 2,4. This has an infinite number of solns and one can also let d = h = 0 for a simpler version. See Eighth Powers for details. However, to create an analogous multi-grade using Verghote’s identity would entail a more complicated set of conditions on ten variables. (End update.)
5. Form: a2+b2+c2+d2+e2 = f2
The complete soln is given by the scaled Euler-Aida Ammei identity,
((a2-b2-c2-d2-e2)t)2 + (2abt)2 + (2act)2 + (2adt)2 + (2aet)2 = ((a2+b2+c2+d2+e2)t)2
Other solns are,
Fauquembergue n-Squares Identity
(a2+b2-c2-d2+x)2 + (2ac+2bd)2 + (2ad-2bc)2 + 4x(c2+d2) = (a2+b2+c2+d2+x)2
where it can be set as x = e2 (or any n squares). It can also be shown that the square of six squares is identically the sum of five squares in two ways:
Piezas
(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
(a2+b2-c2-d2+e2+f2)2 + 4(ac+bd)2 + 4(ad-bc)2 + 4(ce+df)2 + 4(cf-de)2 = (a2+b2+c2+d2+e2+f2)2
Q: Any other square of n distinct squares expressible as the sum of n-1 squares (or less) other than this and the Lebesgue Three-Square? (See update in next page.)
E. Barisien
(x6)2 + (4x2y4)2 + (4xy5)2 + (2y6)2 + (2xy(2x2+y2)(x2+2y2))2 = (x6+8x4y2+8x2y4+2y6)2
Q: Any other similar polynomial whose square is expressible as the sum of 5 or less squares?
6. Form: 2(a2+b2+c2+d2) = (a+b+c+d)2
(Update, 6/26/11) Contributed by Christoph Soland. A configuration of four mutually tangent circles is known as Soddy circles. Configurations come in dual pairs with the same tangency points. Let {e1, e2, e3, e4} be the curvatures (1/radius) of the four circles and {f1, f2, f3, f4} those of the dual circles. Then,
2(e12+e22+e32+e42) = (e1+e2+e3+e4)2 (eq.1)
and similarly for the fi. For solutions to eq.1, let x2+y2+z2 = t2. Then,
{e1, e2, e3, e4} = {t+x+y+z, t+x-y-z, t-x+y-z, t-x-y+z}
{f1, f2, f3, f4} = {t-x-y-z, t-x+y+z, t+x-y+z, t+x+y-z}
Ref.: Coxeter, H.S.M. Introduction to Geometry, 2ed.
Note: Parametric solutions to eq.1 where all ei are squares 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2 are also known. See Form 18 here.
(t+x+y+z)k + (t-x-y+z)k + (t-x+y-z)k + (t+x-y-z)k = (t-x-y-z)k + (t+x+y-z)k + (t+x-y+z)k + (t-x+y+z)k, for k = 1,2
where {x,y,z} in the LHS is just negated in the RHS.
These three are different forms of the same identity. Birck's form appears in the 8th power Birck-Sinha Identity as well as for Sinha’s 2-4-6 Identity to be discussed later.
4. Form: a2+b2+c2+d2 = e2+f2+g2+h2
J. Zehfuss
(2a)k + (2b)k + (2c)k + (2d)k = (-a+b+c+d)k + (a-b+c+d)k + (a+b-c+d)k + (a+b+c-d)k, for k = 1,2
Note that for d = 0, this reduces to Birck’s version. This can also be expressed, after minor changes in signs as,
M. Hirschhorn
(4p+1)2 + (4q+1)2 + (4r+1)2 + (4s+1)2 = 4(p+q+r+s+1)2 + 4(p-q-r+s)2 + 4(p+q-r-s)2 + 4(p-q+r-s)2
or, Hirschhorn's Odd-Even Identity, proving that the sum of four distinct odd squares is the sum of four distinct even ones. Proof: This can easily be shown true by equating the terms on the LHS of both forms, solving for {p,q,r,s}, and substituting into the RHS of the second. A third equivalent form, more symmetrical in one sense, is,
J. Wroblewski
(-a+b+c)k + (a-b+c)k + (a+b-c)k + (a+b+c)k = (2a)k + (2b)k + (2c)k, for k = 1,2
If a = b = z, and c = d = t, then we get the special case,
e2+f2 = 2(z2+t2) (eq.2)
which can be given a complete soln. See also Form 8 of "Sums of Two Squares". Q: Any simple identity for eq.1 not a derivation of the others in the subsequent section?
3. Form: a2+b2+c2+d2 = e2+f2+g2
Euler:
ak + bk + ck + (a+b+c)k = (a+b)k + (a+c)k + (b+c)k, for k = 1,2
E. Catalan
(-a+d)k + (-b+d)k + (-c+d)k + dk = ak + bk + ck, if a+b+c = 2d
Birck
The smallest positive primitive solns are,
12 + 12 + 32 + 32 = 22 + 42 = 20
22 + 22 + 32 + 32 = 12 + 52 = 26
12 + 12 + 42 + 42 = 32 + 52 = 34
22 + 22 + 22 + 52 = 12 + 62 = 37
For k > 2, just like the other sums of n squares, the equation a2+b2+c2+d2 = ek has an identically true soln for all positive integer k. In fact, by Lagrange’s Four-Square Theorem, all positive integers can be expressed as the sum of at least four squares. Other results are,
Euler
{ap+bq+cr+ds, ar-bs-cp+dq, -as-br+cq+dp, aq-bp+cs-dr}
{-aq+bp+cs-dr, as+br+cq+dp, ar-bs+cp-dq, ap+bq-cr-ds}
{ar+bs-cp-dq, -ap+bq-cr+ds, aq+bp+cs+dr, as-br-cq+dp}
{-as+br-cq+dp, -aq-bp+cs+dr, -ap+bq+cr-ds, ar+bs+cp+dq}
The sum of the squares of each row or column is (a2+b2+c2+d2)(p2+q2+r2+s2). Note that each term has two negative signs other than those on the main diagonal slanting towards the right which have none. By letting {p,q,r,s} = {b,c,d,a} the common sum becomes (a2+b2+c2+d2)2 thus giving a 4x4 semi-magic square of squares. (If the two main diagonals also have the same sum, then it is known as a full magic square.) Another form is {p,q,r,s} = {d,b,a,c}. However, let {p,q,r,s} = {c,d,a,b},
{2ac+2bd, a2-b2-c2+d2, -2ab+2cd, 0}
{2bc-2ad, 2ab+2cd, a2-b2+c2-d2, 0}
{a2+b2-c2-d2, -2ac+2bd, 2bc+2ad, 0}
{0, 0, 0, (a2+b2+c2+d2)}
or a 3x3 semi-magic square of squares and which gives equivalent forms of the Lebesgue Three-Square Identity,
(2ac+2bd)2 + (a2-b2-c2+d2)2 + (2ab-2cd)2 = (a2+b2+c2+d2)2
(There is probably an octonion 8x8 version of Euler’s 4x4 square though this author has not been able to find it yet.) A more general version was given by Lagrange as,
x12+mx22+nx32+mnx42 = (a2+mb2+nc2+mnd2) (p2+mq2+nr2+mns2)
where {x1, x2, x3, x4} = {ap-mbq-ncr+mnds, aq+bp-ncs-ndr, ar+mbs+cp+mdq, as-br+cq-dp}
2. Form: a2+b2+c2+d2 = e2+f2
Theorem: “The complete soln to a2+b2+c2+d2 = e2 is given by the scaled Euler-Aida Ammei identity,
((p2-q2-r2-s2)t)2 + (2pqt)2 + (2prt)2 + (2pst)2= ((p2+q2+r2+s2)t)2
where t is a scaling factor.”
Proof: For any given soln {a,b,c,d}, one can always find rational {p,q,r,s,t} using the formulas,
{p,q,r,s,t} = {a+e, b, c, d, (e-a)/(2(b2+c2+d2)} (End proof.)
Other solns are:
J. Neuberg:
a2(a+b+c)2 + b2(a+b+c)2 + c2(a+b+c)2 + (ab+ac+bc)2 = (a2+b2+c2+ab+ac+bc)2
E. Catalan:
(p2-q2)2 + ((p+q)(p+r))2 + ((p+q)(p-r))2 + (2pq-q2-r2)2 = (p2+2q2+r2)2
A. Martin
(4pr+s)2 + (4pr-s)2 + (4qr+s)2 + (4qr-s)2 = (4p2+4q2+2r2)2, if s = 2p2+2q2-r2
G. Metrod
x2 + (x+2pq)2 + (x+4pq)2 + (x+6pq)2 = (p2+5q2)2, if x = (p2-6pq-5q2)/2
The last is a special case of (x+ay)2 + (x+by)2 + (x+cy)2 + (x+dy)2 = z2. This can be solved as a quadratic in x and making its discriminant a square entails solving an expression of the form ry2+16z2 = t2 which can easily be solved.
III. Sums of four or more squares
a2+b2+c2+d2 = ek
a2+b2+c2+d2 = e2+f2
a2+b2+c2+d2 = e2+f2+g2
a2+b2+c2+d2 = e2+f2+g2+h2
a2+b2+c2+d2+e2 = f2
2(a2+b2+c2+d2) = (a+b+c+d)2
1. Form: a2+b2+c2+d2 = ek