016: Sums of three Fourth Powers (Part 3)

 

 

(The identities discussed here are in blue.)

 

  1. a4+b4 = c4+d4
  2. pq(p2+q2) = rs(r2+s2)
  3. pq(p2-q2) = rs(r2-s2)
  4. pq(p2+hq2) = rs(r2+hs2)
  5. x4+y4 = z4+nt2
  6. x4+y4 = z4+nt4
  7. u4+nv4 = (p4+nq4)w2
  8. u4+nv4 = x4+y4+nz4
  9. u4+v4 = x4+y4+nz4
  10. ak+bk+ck = dk+ek+fk,  k = 2,4
  11. ak+bk+ck = 2dk+ek,  k = 2,4
  12. ak+bk+ck = dk+ek+fk,  k = 2,3,4
  13. x4+y4+z4 = 2(x2y2+x2z2+y2z2)-t2
  14. x4+y4+z4 = t4
  15. x4+y4+z4 = ntk
  16. v4+x4+y4+z4 = ntk
  17. vk+xk+yk+zk = ak+bk+ck+dk,  k = 2,4
  18. 2(v4+x4+y4+z4) = (v2+x2+y2+z2)2
  19. x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k, k = 1,2,3,4
  20. x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k, k = 1,2,3,4
  21. x14+x24+…xn4, n > 4

 

11. Form: ak+bk+ck = 2dk+ek  (k = 2,4)

 

The case e = 0 was studied by Ramanujan.  To recall, he gave the formulas for k = 2,4,

 

ak + bk + ck = 2(ab+ac+bc)k/2

 

ak(b-c)k + bk(a-c)k + ck(a-b)k = 2(ab+ac+bc)2k/2

 

(a2b+b2c+c2a)k + (ab2+bc2+ca2)k + (3abc)k = 2(ab+ac+bc)3k/2

 

(and so on) with a+b+c = 0.  As was already mentioned, this form gives rise to a 5th power multi-grade with just 10 terms,

 

(a+d)k + (-a+d)k + (b+d)k + (-b+d)k + (c+d)k + (-c+d)k = (2d)k + (2d)k + (e+d)k + (-e+d)k,  k = 1,2,3,4,5. 

 

For the complete radical soln to the system x1k+x2k+x3k = 2x4k+ x5k, k = 2,4, express it in the form,

 

(2a)k + (2b)k + (p+q)k = 2(2c)k + (p-q)k

 

which gives,

 

a2+b2-2c2+pq = 0

2a4+2b4-4c4+p3q+pq3 = 0

 

a useful trick to reduce the degree.  Combining the two by eliminating q then gives the beautifully simple condition,

 

(a2+b2-2c2)p4 - 2(a4+b4-2c4)p2 + (a2+b2-2c2)3 = 0

 

Unfortunately, the discriminant of this is a sextic in a,b (and an octic in c) so cannot be treated as an elliptic curve.  However, it does have solns one of which, implied in Ramanujan’s case when last term is p-q = 0, is

 

{a,b,c,p} = {u2-v2,  2uv+v2,  u2+uv+v2,  u2+2uv}

 

which makes the discriminant zero.  (There are other solns as well.)

 

Piezas

 

Parametric solns to ak+bk+ck = 2dk+ek have been found in two cases: 

 

I.  When a+b = c

  1. If e = 0, then a2+ab+b2 = d2.  (Ramanujan’s)
  2. If e = 2d, then a2+ab+b2 = 3d2.

II.  When a+b = 2d. 

 

Let, (p+r)k + (-p+r)k + qk = 2rk + sk,  then p2+2q2 = 6r2, and 2p2+q2 = s2. 

 

The first eqn can be solved as {p,q,r} = {2u2-4uv-4v2,  u2+4uv-2v2,  u2+2v2} and substituting this into the second gives the elliptic curve,

 

(3u2-4uv-6v2)2 + 32u2v2 = s2

 

with one non-trivial soln being {u,v} = {12, 5}, and so on.  Note: Any other case?

 

 

12. Form: ak+bk+ck = dk+ek+fk  (k = 2,3,4)

 

A. Choudhry

 

(ay+b)k + (cy+ad)k + (-ay+b+cd)k = (ay-b)k + (cy-ad)k + (-ay-b-cd)k

 

This is already identically true when k = 2.  But let,

 

{a,b,c,d} = {x2+3,  (x+3)3(x-1),  -2(x2-3),  2(x2+2x+3)}, 

 

where x satisfies the elliptic curve,

 

y2 = (x2-4x+3)2-48x2

 

and this also becomes valid for k = 3,4.  Some integer points on this curve are given by x = {-3, -1, 0, 15} while one non-trivial soln is x = 1/5, and so on.

 

Piezas

 

Rational solns to ak+bk+ck = dk+ek+fk,  k = 2,3,4 imply a rational soln t to the quadratic eqn,

 

(a2+b2-e2-f2)t2 - 2(a3+b3-e3-f3)t + (a4+b4-e4-f4) = 0

 

Proof: The above is equivalent to the identity,

 

(c2-d2)t2 - 2(c3-d3)t + (c4-d4) = 0,  where t = c+d.

 

 

13. Form: x4+y4+z4 = 2(x2y2+x2z2+y2z2)-t2

 

This is equivalent to (x+y-z)(x-y+z)(-x+y+z)(x+y+z) = t2 hence is essentially the problem of finding a triangle with rational area using Heron’s formula.  One approach is to reduce it to the equation p2+q2+r2 = s2 of which the complete soln by Desboves is known.  Equate the first three factors as,

 

{(x+y-z), (x-y+z), (-x+y+z)} = {p2, q2, r2},

 

and by solving for x,y,z, and substituting into the last factor as x+y+z = s2, one then gets the afore-mentioned condition.  Other ways are,

 

1. Brahmagupta

 

{x,y,z} = {a2/b+b,  a2/c+c,  a2/b+a2/c-b-c}

 

2. Kurushima (Euler would later give a similar identity.)

 

{x,y,z} = {ac(b2+d2),  bd(a2+c2),  (bc+ad)(ab-cd) }

 

3. Gauss

 

{x,y,z} = {abcd(a2+b2),  ab(c+d)(a2c-b2d),  ab(a2c2+b2d2) }

 

4. A. Martin

 

Let {m,n} = {a2+b2, a2-b2}, then,

 

{x,y,z} = {(c2-d2)m,  2cdm+2d2n,  (c2+d2)m+2cdn}

 

 

 

14. Form: x4+y4+z4 = t4

 

This equation is significant since Euler made the rather reasonable conjecture that it would take at least k kth powers to sum up to a kth power which turned out to be erroneous.  The first counter-example was found by Lander, Parkin as,

 

275 + 845 + 1105 + 1335 = 1445

 

via computer search but there is no systematic method to find other primitive solns for fifth powers.  For fourth powers, a counter-example was later found as well as a method to generate further solns, albeit they can quickly become enormous.

 

  

(Update, 2/8/10)Piezas

 

If p4+q4+r4 = 1, then (p+q)4 - r4 + 1 = 2(p2+pq+q2)2  also holds.

 

  

(Update, 2/8/10):  Demjanenko, Elkies

 

The complete soln to

 

p4+q4+r4 = 1                          (eq.0)

 

can be given in the form,

 

(x+y)4 + (x-y)4 + z4 = 1

 

where,

 

ay2 = (8mn-3a)x2-2bx-2mn       (eq.1)

±az2 = 4bx2+8mnx-b                 (eq.2)

 

and {a,b} = {2m2+n2, 2m2-n2}, for some constants {m,n} hence is a problem of making two quadratics in x as squares and, for appropriate {m,n}, reduces to solving an elliptic curve.  One can solve {y,z} in the radicals to see that it holds and it is easily proven that this is the complete soln. 

 

Proof:  Let {x+y, x-y, z} = {p,q,r}.  As eq.1 and eq.2 are also quadratics in {m,n}, solve for m, assume eq.0 as true to rationalize the discriminant, and take the positive sign of the square root to get,

 

m1 = n((p+q)2+r2-1) / (2(p2+pq+q2+p+q))

m2 = n(p2+pq+q2-p-q) / ((p+q)2-r2-1)

 

(One can use the identity given by the author to rationalize the disciminant of m2.)  It will then be seen that m1 = m2 if eq.0 is true, hence one can always find rational {m,n} in terms of {p,q,r}.  For example, using the smallest {p,q,r} = {217519/s, 414560/s, 95800/s} and s = 422481, we get {m,n} = {6007, 30080}.  (End proof

 

However, one can also start with {m,n} as it is possible they are relatively small.  Elkies’ first soln used only {m,n} = {8,-5}, and (eq.1) becomes,

 

153y2 = -779x2-206x+80

 

which has initial soln {x1,y1} = {3/14, 1/42}.  This yields a parametrization for x as,

 

x = (51p2-34p-5221)/(238p2+10906)

 

which, when substituted into (eq.2), becomes the problem of finding a value such that the quartic polynomial in p is a square.  Clever transformations may then reduce the size of the coefficients.  This can be treated as an elliptic curve and since x1 = 3/14 has p = -3779/17, from this initial point one can calculate more, proving that eq.0 has an infinite number of distinct solns.  (End update.)

 

Elkies

 

(85v2+484v-313)4 + (68v2-586v+10)4 + (2u)4 = (357v2-204v+363)4

 

if u2 = 22030+28849v-56158v2+36941v3-31790v4,

 

a soln of which is v = -31/467.  This gave the first rational soln to x4+y4+z4 = t4 (eq.0) and is just the case {m,n} = {8,-5} of the previous identity.  This elliptic curve in fact has an infinite number of rational points, thus providing an infinite number of solns to eq.0.

 

 

15. Form: x4+y4+z4 = ntk

 

E. Fauquembergue

 

(ab)4 + (ac)4 + (bc)4 = (a4+a2b2+b4)2,  if a2+b2 = c2

 

This beautifully simple soln has a counterpart p4+q4+r4+s4 = t2 (eq.1) which also depends on Pythagorean triples and again found by Fauquembergue,

 

(2a2bc3)4 + (2ab2c3)4 + (2ab(a4+b4))4 + ((a2-b2)c4)4 = (4a2b2(a4+b4)2-c12)2,   if a2+b2 = c2

 

However, since eq.1 has a lot of small solns, there is probably a parametrization where {p,q,r,s} are binary quadratic forms, and t is a quartic form (which can then be specialized to become a square). 


(Update, 6/23/13):  Jean-Joël Delorme solved,


x4 y4 z4 t4  = (x2+y2+z2-t2)2 


If we let a2+b2 = c2 and,


{x, y, z, t} = {(a2-b2)c4,   2a2bc3,   2ab2c3,   2ab(a4+b4)}


then this is equivalent to Fauquembergue's. Delorme also gave a new solution as,


{x, y, z, t} = {2a2b2c(a4+b4),   a(a8-b8),   b(a8-b8),   abc5(a2-b2)}


and pointed out that if {x, y, z, t} is a solution, then so is {1/x, 1/y, 1/z, 1/t}. 

 

F. Proth (and others)

 

ak + bk + (a+b)k = 2(a2+ab+b2)k/2,  for k =2,4

 

As was already pointed out, this is quite a ubiquitous algebraic form.  Its first even kth powers can be expressed as a sum of three squares,

 

(ab)2 + (ab+b2)2 + (a2+ab)2 = (a2+ab+b2)2,

 

and a square-biquadratic sum by E. Escott,

 

(ab)4 + (ab+b2)4 + ((a2+ab)(a2+ab+2b2))2 = (a2+ab+b2)4,

 

These imply that the form (a2+ab+b2)2k is expressible as the sum of three squares using the method discussed below.

 

Ramanujan

 

Let a+b+c = 0, then,

 

a4 + b4 + c4 = 2(ab+ac+bc)2

 

a4(b-c)4 + b4(a-c)4 + c4(a-b)4 = 2(ab+ac+bc)4

 

(a2b+b2c+c2a)4 + (ab2+bc2+ca2)4 + (3abc)4 = 2(ab+ac+bc)6

 

In his Notebooks, Ramanujan gave these as well as one for k=8 and wrote “…and so on”.  The rest can be found by noting that the first is equivalent to Proth’s so what one has to do is to find expressions a,b such that a2+ab+b2 = (p2+pq+q2)k which can be done by factoring over a complex cube root of unity ω,

 

(a-bω) (a-bω2) = (p-qω)k (p-qω2)k

 

and by using the reliable method of equating factors, one can then easily solve for {a,b}.  It should be pointed out Ramanujan may have used another method to derive his more elegant expressions which are conditionally dependent on a+b+c = 0.  (Incidentally, this form has a generalization for third and fifth powers.)

 

S. Realis

 

x4+y4+z4= 3t2

 

{x,y,z} = {5a4+4a3+9a2+10a+5,  5a4+10a3+9a2+4a+5,  5a4+16a3+27a2+16a+5}

 

Any other solns to x4+y4+z4= ntk for n>2?

 

E. Fauquembergue

 

(2p2-2q2)4 + (p2-4q2)4 + (3pq)4 = (p2+2q2)4 + (4p4-11p2q2+16q4)2

 

 

 

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