**10. Form: a**^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} (k = 2,4)

*Bastien’s Theorem* also given below states that the above eqn is trivial if simultaneously true for k = 1,2,3,4. However, this does have solns when k = 2,4; k = 1,2,4, or even k = 2,3,4 and, of course, when only for single exponents. When valid for both k = 2,4, this *quadratic-quartic system *will be called throughout this work as *Q*_{1}, and this form is significant because it may appear as a side condition for higher powers such as for the 5th, 6th, 7th, and 8th. First, we have these general theorems,

T. Sinha

*Theorem*: If,

a^{2}+b^{2}+c^{2} = d^{2}+e^{2}+f^{2},

a^{4}+b^{4}+c^{4} = d^{4}+e^{4}+f^{4},

then (a^{2}-f^{2})(b^{2}-f^{2}) = (d^{2}-c^{2})(e^{2}-c^{2}).

The eqn (a^{2}-f^{2})(b^{2}-f^{2}) = (d^{2}-c^{2})(e^{2}-c^{2}) is quite easy to satisfy and can be reduced to solving an eqn of form x^{2}+ny^{2} = z^{2}. However, it is not enough to imply *Q*_{1}. More generally, any two of the equations in the system implies the third, but it is not the case that one implies the other two.

Tarry, Escott

*Theorem*: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4, then

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

k = 1,2,3,4,5 for *any* *t*.

This is a special case of a theorem of the *Prouhet-Tarry-Escott Problem* which I’m assuming is due to Tarry or Escott since they established related theorems for this field. Note that appropriate pairs have the common sum *2t* and is known as a *symmetric solution*. Any multi-grade k = 1-5 of this form is then reducible to Q_{1}. By *Bastien's Theorem*, the system,

x_{1}^{k}+x_{2}^{k}+…+ x_{m}^{k} = y_{1}^{k}+y_{2}^{k}+…+ y_{m}^{k}

for k = 1,2,…*n* has non-trivial solns *only if m>n*. Thus, one side must have at least *n+1* terms. For the case k = 1,2,3,4,5, this implies an eqn with 6 terms on either side. However, it is possible some terms are equal to zero. For example, given the special case of Q_{1},

a^{k}+b^{k}+c^{k} = 2d^{k}+e^{k}, k = 2,4, then,

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

which has only 4 terms on the RHS. A broad class of solutions to,

a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4

also obey *a+b+c = d+e+f = 0* such as,

(p-q)^{k}+(p+q)^{k}+(-2p)^{k} = r^{k} + (-r)^{k}, where 3p^{2}+q^{2 }= r^{2}

which yields the smallest as {3,5,-8; 7,-7}. Another also has **f = 0**, has *a+b = d+e* instead,

(p-u)^{k}+(q+u)^{k}+(4x)^{k} = (p+u)^{k} + (q-u)^{k}

where {p,q,u} = y+1, y-1, 2x^{2}, and {x,y} solve 3y^{2 }= -4x^{4}+8x^{2}-1. This is birationally equivalent to an *elliptic curve* with rational point *x* = 7/19 and an infinite more. Others with *a+b = d+e *are,

Chernick

*Theorem*: If (a+d)^{k} + (-a+d)^{k} + b^{k} = (c+d)^{k} + (-c+d)^{k} + d^{k}, k = 2,4, then

(b-d)^{k} + (-b-d)^{k} + (2d)^{k} = (a+2d)^{k} + (-a+2d)^{k} + (-c-2d)^{k} + (c-2d)^{k}, k = 1,2,3,5

and the solution *a,b,c,d* given explicitly as one of two identities. Note that if the terms of Q_{1} are expressed as x_{i} and y_{i}, this also satisfies x_{1}+x_{2} = y_{1}+y_{2}. More generally,

Piezas

*Theorem*: If (p-q+r)^{k} + (p+q-r)^{k} + x^{k} = (p+q+r)^{k} + (p-q-r)^{k} + y^{k}, k = 2,4, then,

(2p+q+r)^{k} + (2p-q-r)^{k} + (-2p-q+r)^{k} + (-2p+q-r)^{k} = (p+x)^{k} + (p-x)^{k} + (-p+y)^{k} + (-p-y)^{k}, k = 1,2,3,5,

and where one of the terms of the latter eqn can be set to zero to get either of Chernick’s two solns.

Chernick, Escott

*Theorem*: If 1^{k} + 2^{k} + 3^{k} = u^{k} + v^{k} + (u+v)^{k}, k = 2,4, then,

(u-7)^{k} + (u-2v+1)^{k} + (3u+1)^{k} + (3u+2v+1)^{k} = (u+7)^{k} + (u-2v-1)^{k} + (3u-1)^{k} + (3u+2v-1)^{k}, k = 2,4,6.

Sinha

*Theorem*: If (a+3c)^{k} + (b+3c)^{k} + (a+b-2c)^{k} = (c+d)^{k} + (c+e)^{k} + (-2c+d+e)^{k}, k = 2,4, then,

a^{k} + b^{k} + (a+2c)^{k} + (b+2c)^{k} + (-c+d+e)^{k} = (a+b-c)^{k} + (a+b+c)^{k} + d^{k} + e^{k} + (3c)^{k}, k = 1,3,5,7

excluding the trivial case *c* = 0. This was derived using the next theorem.

Birck, Sinha

*Theorem*: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4 where a+b ≠ c; d+e ≠ f, then,

(a+b+c)^{k} + (-a+b+c)^{k} + (a-b+c)^{k} + (a+b-c)^{k} + (2d)^{k} + (2e)^{k} + (2f)^{k} =

(d+e+f)^{k} + (-d+e+f)^{k} + (d-e+f)^{k} + (d+e-f)^{k} + (2a)^{k} + (2b)^{k} + (2c)^{k}

for k = 1,2,4,6,8.

Piezas

*Theorem*: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 2,4. Define {x,y} = {a^{2}+b^{2}+c^{2}, a^{4}+b^{4}+c^{4}}. Then,

a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6} = 3(a^{2}-d^{2})(b^{2}-d^{2})(c^{2}-d^{2}) = 3(a^{2}b^{2}c^{2}-d^{2}e^{2}f^{2})

3(a^{8}+b^{8}+c^{8}-d^{8}-e^{8}-f^{8}) = 4(a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6})(x)

6(a^{10}+b^{10}+c^{10}-d^{10}-e^{10}-f^{10}) = 5(a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6})(x^{2}+y)

If we define *F*_{k}:= a^{k}+b^{k}+c^{k}-d^{k}-e^{k}-f^{k}, then for even k > 4, *F*_{k} is of the form *F*_{6}*Poly1* and also a highly composite number. More significantly, the identities can be combined and for a,b,c,d,e,f with appropriate properties, these can be woven together into a very neat form. For example, define *n* as,

a^{4}+b^{4}+c^{4} = n(a^{2}+b^{2}+c^{2})^{2}

and Q_{1} implies the general identity,

32(a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6})(a^{10}+b^{10}+c^{10}-d^{10}-e^{10}-f^{10}) = 15(n+1)(a^{8}+b^{8}+c^{8}-d^{8}-e^{8}-f^{8})^{2}

For the special case when a+b = ±c, since,

a^{4}+b^{4}+(a+b)^{4} = (1/2)(a^{2}+b^{2}+(a+b)^{2})^{2}

then *n* = 1/2, and the above becomes,

64(a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6})(a^{10}+b^{10}+c^{10}-d^{10}-e^{10}-f^{10}) = 45(a^{8}+b^{8}+c^{8}-d^{8}-e^{8}-f^{8})^{2}

which is a concise version of *Ramanujan’s 6-10-8 Identity* discussed in the next section. (In the general case, this turns out to be just the first in a family of identities.) (Update, 7/31/09): This can be extended to the "*solution chain*" a_{1}^{k}+a_{2}^{k}+a_{3}^{k} = a_{4}^{k}+a_{5}^{k}+a_{6}^{k} = a_{7}^{k}+a_{8}^{k}+a_{9}^{k} = a_{10}^{k}+a_{11}^{k}+a_{12}^{k} = ..., k = 2,4, for a chain divisible into pairs. For the special case when {a_{1}+a_{2}, a_{4}+a_{5}, a_{7}+a_{8}, a_{10}+a_{11}} = {a_{3}, a_{6}, a_{9}, a_{12}}, then,

64(a_{1}^{6}+a_{2}^{6}+a_{3}^{6}+a_{4}^{6}+a_{5}^{6}+a_{6}^{6}-a_{7}^{6}-a_{8}^{6}-a_{9}^{6}-a_{10}^{6}-a_{11}^{6}-a_{12}^{6})(a_{1}^{10}+a_{2}^{10}+a_{3}^{10}+a_{4}^{10}+a_{5}^{10}+a_{6}^{10}-a_{7}^{10}-a_{8}^{10}-a_{9}^{10}-a_{10}^{10}-a_{11}^{10}-a_{12}^{10})

= 45(a_{1}^{8}+a_{2}^{8}+a_{3}^{8}+a_{4}^{8}+a_{5}^{8}+a_{6}^{8}-a_{7}^{8}-a_{8}^{8}-a_{9}^{8}-a_{10}^{8}-a_{11}^{8}-a_{12}^{8})^{2},

One can test this by the soln chain [28, 175, 203] = [77, 140, 217] = [107, 113, 220] = [5, 188, 193] given here, found by Gloden in the 1940's. (*End update*.) It seems then the system Q_{1} is quite important and worth a second look. To solve,

a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}

for k = 2,4, one trick is *to reduce the degree* of the two eqns by using the substitution {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} to get,

pq+rs+tu = 0

p^{3}q+pq^{3}+r^{3}s+rs^{3}+t^{3}u+tu^{3} = 0

By eliminating *u* between them, one ends up with a quartic in *t*,

(pq+rs)t^{4}-(p^{3}q+pq^{3}+r^{3}s+rs^{3})t^{2}+(pq+rs)^{3} = 0

or a cubic in either *p,q,r,s*. However, it is possible to use another substitution that can reduce the degree *even* further.

*Theorem*: "The system a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k }for k = 2,4, call it *Q*_{1}, can be reduced to solving a quadratic. Its discriminant *D* is a quartic polynomial in one variable and the problem of making D a square can be treated as an elliptic curve."

(In fact, as was already mentioned, similar results can be given for the three systems k = 1,4, or k = 1,5, or k = 1,2,6. Another substitution will be used for these which is slightly more efficient since the first system ends up with a discriminant D that is only a quadratic polynomial.)

*Proof*: We simply use a different substitution, call this form *F*_{1},

{a,b,c,d,e,f} = {p+qu, r+su, t+u, t-u, r-su, p-qu} to get,

pq+rs+t = 0

p^{3}q+r^{3}s+t^{3}+(pq^{3}+rs^{3}+t)u^{2} = 0

where the degree of the variable *u* has been reduced to merely a quadratic. Eliminating *t* between them yields,

(Poly1)u^{2}-(Poly2) = 0

where,

Poly1:= p(q-q^{3})+r(s-s^{3})

Poly2:= p^{3}(q-q^{3})-3pqrs(pq+rs)+r^{3}(s-s^{3})

and the problem is reduced to making its discriminant a square,

y^{2} = (Poly1)(Poly2)

which is just a quartic in *p* (or *r*), thus proving the theorem. This quartic in fact has a square leading and constant term so is easily made a square and from that initial point, other rational ones may be computed. To prove that this is the *complete* characterization of non-trivial solns, solve for *p,q*… in terms of the original variables to get,

{p,q,r,s,t,u} = {(a+f)/2, (a-f)/(c-d), (b+e)/2, (b-e)/(c-d), (c+d)/2, (c-d)/2}

and even if there is division by a variable, when *c = d* the system a^{k}+b^{k} = e^{k}+f^{k} for k = 2,4 has only trivial solns hence this exceptional case is of no interest and does not affect the generality of the theorem. While it has been proven that Q_{1} can be reduced to solving a quadratic, a *third* constraint may be useful to explore various solns. This constraint is an appropriate linear or quadratic relation between the terms such that the final eqn remains a quadratic and involve a simpler elliptic curve. This author found *nine* relations, six linear and three quadratic,

10.1 a+b = nc; d+e = nf

10.2 a+b ≠ c; d+e ≠ f

10.3 a+b±c = n(d+e±f)

10.4 na+b+c = d+e+nf

10.5 na+b = e+nf

10.6 a+d = n(c+f)

10.7 (a^{2}-f^{2})c^{2} = -(b^{2}-e^{2})d^{2 }

10.8 a^{2}+nab+b^{2} = d^{2}+nde+e^{2}

10.9 a^{2}+nad+d^{2} = b^{2}+nbe+e^{2}

for *any n*. All of these have a *complete* soln in terms of an *elliptic curve* though may simplify into quadratic forms for particular values of *n* or be trivial when *n* = 0. The linear relations will be discussed in the next section.

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