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005: Sums of four or five squares

 
 
III. Sums of four or five squares

 

  1. a2+b2+c2+d2 = ek
  2. a2+b2+c2+d2 = e2+f2
  3. a2+b2+c2+d2 = e2+f2+g2
  4. a2+b2+c2+d2 = e2+f2+g2+h2
  5. a2+b2+c2+d2+e2 = f2

  

1. Form: a2+b2+c2+d2 = ek

 

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.

 

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)

 

{x1, x2, x3, x4} = {ap-mbq-ncr+mnds,  aq+bp-ncs-ndr,  ar+mbs+cp+mdq,  as-br+cq-dp}

  

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 (Piezas):  For any 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)}

 

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.

 

 

2. Form: a2+b2+c2+d2 = e2+f2

 

Q: Any simple soln not a derivation of the forms in the subsequent section?

 

 

3. Form: a2+b2+c2+d2 = e2+f2+g2

 

Euler:

 

a2 + b2 + c2 + (a+b+c)2 = (a+b)2 + (a+c)2 + (b+c)2

 

E. Catalan

 

(a-d)2 + (b-d)2 + (c-d)2 + d2 = a2 + b2 + c2,  if a+b+c = 2d

 

This can be expressed alternatively as,

 

Birck

 

(a+b+c)2 + (-a+b+c)2 + (a-b+c)2 + (a+b-c)2 = (2a)2 + (2b)2 + (2c)2

 

The above is useful for eighth powers 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

 

(a+b+c-d)k + (-a+b+c+d)k + (a-b+c+d)k  + (a+b-c+d)k  = (2a)k + (2b)k + (2c)k + (2d)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

 

(4a+1)2 + (4b+1)2 + (4c+1)2 + (4d+1)2 = 4(a+b+c+d+1)2 + 4(a-b-c+d)2 + 4(a+b-c-d)2 + 4(a-b+c-d)2

 

or, Hirschhorn's Odd-Even Identity, proving that the sum of four distinct odd squares is the sum of four distinct even ones.

 

 

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?

 
 
 
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