13. Form: a^k+b^k+c^k = d^k+e^k+f^k,  k =1,3

 

Just like for k = 1,2, the eqn

 

a^k+b^k+c^k = d^k+e^k+f^k,                (eq.1)

 

also has a complete soln for k = 1,3.  We can solve this in two ways.  First, using the form by Lander,

 

(u+x)^k + z^k + (v+y)^k = (v+x)^k + y^k + (u+z)^k

 

call this L₀, a complete soln for k = 1 which he used for k = 1,5.  Expanding L₀ for k = 3,

 

u²x-v²x+ux²-vx²+v²y+vy²-u²z-uz² = 0

 

As a quadratic in z, this has discriminant D,

 

D:= -4(x-y)uv²-4(x²-y²)uv+u²(u+2x)²

 

which must be made a square.  Since this discriminant as a polynomial in v is a quadratic with a square constant term, it is easily made a square using Fermat’s method.  A special case is when one of the terms of L₀ is zero, say letting y = 0, since this gives the complete symmetric ideal soln of degree four,

 

(t+a)^k+(t+b)^k+(t+c)^k+(t+d)^k+(t+e)^k = (t-a)^k+(t-b)^k+(t-c)^k+(t-d)^k+(t-e)^k

 

for k = 1,2,3,4 which is true for any t iff a^k+b^k+c^k+d^k+e^k = 0 for k = 1,3.  Another approach to solve the system k = 1,3 which turns out to be simpler is to use another form,

 

(a+bp+q)^k + (b-bp+q)^k + (c+ap+q)^k = (a+cp+q)^k + (b+ap+q)^k + (c-cp+q)^k

 

call this L₁ since this author found it while studying Lander’s work on L₀ for k = 1,5.  This is a subset of L₀, yet still complete for multi-grades with one k >1 (the easy proof will be given in Fifth Powers).  For k = 1,3, expanding L₁ at k = 3 gives just a linear condition in q,

 

-a²+b²+c²+ab+ac+bc+2(b+c)q = 0

 

Solving for q then gives the complete soln of eq.1 in four variables {a,b,c,p}.  (And just like in the previous approach, any of the terms of L₁ can be set equal to zero, say a+bp+q = 0, solve for q, and the linear condition will now be in the variable p.)   One can see the approach can easily be generalized for k = 1,4 and higher.  So L₁ for k = 1,4, as a polynomial in p after removing the trivial factor p(p-1), is,

 

(Poly10)p²+(Poly20)p+(Poly30) = 0

 

with discriminant D that is a quadratic polynomial in q, while for k = 1,5,

 

(Poly11)p²+(Poly21)p+(Poly31) = 0

 

has a D that is a quartic in q, and for k = 1,2,6,

 

(Poly12)p²+(Poly22)p+(Poly32) = 0

 

with D that is again a quartic in q, so the problem can be treated as finding rational points on the curve D = y².  Thus, L₁ is a very versatile form because when expanded for higher powers, the variable p remains of relatively low degree and for either of the systems k = 1,5 or k = 1,2,6, the complete soln involves solving only a quadratic eqn and an elliptic curve.  These will be discussed in more detail later.

 

Another special case of eq.1 also has abc = def.  One way to solve this is to use the form,

 

(ap)^k + (bq)^k + (cr)^k =  (bp)^k + (cq)^k + (ar)^k

 

which already satisfies the condition.  Expanding for k = 1,3 and solving the two eqns yields,

 

{p,q,r} = {ab-c²,  -a²+bc,  -b²+ac}

 

There is a second distinct form though,

 

(ap)^k + (bq)^k + (cr)^k =  (-bp)^k + (-cq)^k + (ar)^k

 

with {p,q,r} = {ab+c²,  -a²-bc,  b²-ac}.  In fact, it can easily be proven that if eq.1 has abc = def, then a+b+c = d+e+f = 0.

 

Proof:  The system a^k+b^k+c^k = d^k+e^k+f^k for k = 1,3 where abc = def, by eliminating {c,f} from this system depends on the final eqn,

 

ab(a+b) = de(d+e)

 

call this E₁. However, letting {c,f} = {-a-b, -d-e} and substituting into the system, this also satisfies k = 1,3 given E₁, proving the assertion. (End proof)

 

Theorem (Piezas):  If a^k+b^k+c^k = d^k+e^k+f^k for k = 1,3 where a+b+c = d+e+f = 0 (or equivalently, abc = def), then,

 

9abc(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶) = 2(a⁹+b⁹+c⁹-d⁹-e⁹-f⁹)

 

Proof:  Simply substitute {c,f} = {-a-b, -d-e} into the eqn and this has E₁ as a factor.

 

A. Gerardin

 

x³+y³+z³ = t³+u³+v³

 

{x,y,z} = {p²(p³+2q³),  -pq(2p³+q³),  q²(p³-q³)}

{t,u,v} = {pq(p³+2q³),  -q²(2p³+q³),  p²(p³-q³)}

 

This is just for k = 3 but satisfies xyz = tuv.

 

14. Form: x^k+y^k+z^k = t^k+u^k+v^k, k = 2,3

 

Identities that are good for squares can be extended for particular values to cubes as well, just as this one,

 

C. Goldbach

 

p² + q² + (p+q+3r)² = (p+2r)² + (q+2r)² + (p+q+r)²

 

 

 

15. Form: x³+y³+z³ = 3t³-t

 

M. Noble

 

Given a³+b³+c³ = 3d³-d, then,

 

{a,b,c,d} = {(4p-4r)/(5q),  (p+4r)/(5q),  (3p)/(5q),  (4p)/(5q)}

 

where {p,q,r} = {15u²+4v², 15u²+9uv-4v², 10uv+3v²}

 

This soln has the special property such that by defining the expressions,

 

{x,y,z} = {a³-d³, b³-d³, c³-d³}

 

then {x-(x+y+z)³,  y-(x+y+z)³,  z-(x+y+z)³} are perfect cubes.  (This is a particular case of a more general identity by Noble.)

 

 

16. Form: x³+y³+z³ = m(x+y+z)

 

L.Aubry

 

x³+y³+z³ ± (x+y+z) = 0

 

{x,y,z} = {u+v,  -u+v,  -2v(3n+1)},  if  u²-(1+12n+36n²+36n³)v² = ±n

 

with n some constant.  For example, for n=1 this entails solving either u²-85v² = ±1 depending on which sign is involved.  In general, if x,y,z need not be co-prime, then,

 

{x,y,z} = {n(u+v),  n(-u+v),  -2nv(3n²+1)},  if  u²-(1+12n+36n²+36n³)v² = ±1.

 

 

17. Form: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k,  k = 1,2,3

 

By Bastien’s Theorem, the system of eqns,

 

x₁^k+x₂^k+…+x_n^k = y₁^k+y₂^k+…+y_n^k,  k = 1,2,3

 

has a non-trivial soln only for n>3.  However, one can always set one term x₁ = 0.  A simple soln is,

 

J. Nicholson

 

a^k + b^k + (2a+4b)^k + (3a+3b)^k = (a+3b)^k + (2a+b)^k + (3a+4b)^k,  k = 1,2,3

 

 

a^k + b^k + c^k + d^k = e^k + f^k,

 

{d,e,f} = {(-abc)/x,  ((a+b)(a+c)(b+c)+y)/(2x),  ((a+b)(a+c)(b+c)-y)/(2x)}

 

where x = ab+ac+bc and a,b,c satisfy (a²-x)(b²-x)(c²-x) = y².  Two easy solns can be found.  First, let b = c, and,

 

{a,b,c,y} = {p+q, q, q, 2(p+q)qr},  where p² = 2q² + r².

 

Second, let a+b = c,

 

{a,b,c,y} = {-2p+q,  2p+q,  2q,  4(2p-q)(2p+q)r},  where p² = q² + r².  Any more?

 

 

18. Form: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k,  k = 1,3

 

(Update, 8/6/09):  T. Rizzo gave the soln,

 

(p+1)^k + (q+1)^k + (r-1)^k + (s-1)^k = (p-1)^k + (q-1)^k + (r+1)^k + (s+1)^k,  for k = 1,3,

 

where p²+q² = r²+s² and solved the condition as,

 

 

19. Form: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k,  k = 1,2,3

 

The complete soln to this system is already known.  For multigrades valid for the consecutive powers k = 1,2,…m by Frolov’s Theorem it is always possible to add a constant to each term such that,

 

x₁+x₂+… +x_n = y₁+y₂+…+y_n = 0         (eq.1)

 

so we can make use of this property.  One approach to completely solving k = 1,2,3 is then to use the Chernick-Lander form,

 

(a+b+c)^k + (a-b-c)^k + (-a-b+c)^k + (-a+b-c)^k = (d+e+f)^k + (d-e-f)^k + (-d-e+f)^k + (-d+e-f)^k

 

which satisfies eq.1.  The solution to this, which is in binary quadratic forms, also solves k = 5 (by Gloden’s Theorem) so will be discussed in the section on Fifth Powers.  Necessarily some of the terms will be negative so this form does not parametricize the even system,

 

x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k

 

for k = 2,4,6.  To include this, we have to solve k = 1,2,3 without appealing to eq.1.  Use the general form F₂,

 

(a+bh)^k + (c+dh)^k + (e+fh)^k + (g+h)^k = (a-bh)^k + (c-dh)^k + (e-fh)^k + (g-h)^k

 

To satisfy k = 1,2, set {g,f} = {-(ab+cd+ef),  -(1+b+d)}.  Expanding for k = 3, we get a quadratic in h,

 

(1+b)(1+d)(b+d)h²-(Poly1) = 0

 

where Poly1 is a quadratic in the variables a,b,c,d,e.  To find rational h, one must make its discriminant a square,

 

(1+b)(1+d)(b+d)(Poly1) = y²               (eq.2)

 

which is easily done since the expression is just a quadratic in {a,c,e}.  It is well-known that given an initial soln to F(x) = y² where F(x) is a quadratic polynomial, then one can find its complete soln. (One way is to use Fermat’s method.)  As a polynomial in e, an initial soln to eq.2 is,

 

e = (a+ab-c+cd)/(b+d)

 

from which one can then find the complete soln.  Using this, one can prove that,

 

Theorem:  Given a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k for k = 1,2,3.  Define n as a²+b²+c²+d² = n(a+b+c+d)², then,

 

25(a⁴+b⁴+c⁴+d⁴-e⁴-f⁴-g⁴-h⁴)(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶) = 12(n+1)(a⁵+b⁵+c⁵+d⁵-e⁵-f⁵-g⁵-h⁵)²

 

This 4-6-5 Identity, by squaring all variables, can become a 8-12-10 Identity good for the even system k = 2,4,6 and analogous to Ramanujan’s 6-10-8 which depends on a k = 2,4.  This will be discussed more in Fourth Powers.

 

 

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