12.1 Form:  a+b = nc, d+e = nf

 

The complete soln for the case n = 1 is given by,

 

a^k + b^k + (a+b)^k = c^k + d^k + (c+d)^k,  k = 2,4

 

where a²+ab+b² = c²+cd+d².  By modifying it slightly to the form,

 

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

 

makes it valid for k = 1 as well.  One soln is, define {m,n} = {a-3b, a+3b}, then for k = 1,2,4,

 

(2a-cm)^k + (2ac-n)^k + (-m-cn)^k = (2a-cn)^k + (2ac-m)^k + (-n-cm)^k

 

for arbitrary a,b,c. 

 

Ramanujan

 

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

 

if ad = bc, for k = 2,4.

 

This form appears in the beautiful 6-10-8 and 3-7-5 Identities,

 

Ramanujan

 

64[(a+b+c)⁶+(b+c+d)⁶+(a-d)⁶-(c+d+a)⁶-(d+a+b)⁶-(b-c)⁶][(a+b+c)¹⁰+(b+c+d)¹⁰+(a-d)¹⁰-(c+d+a)¹⁰-(d+a+b)¹⁰-(b-c)¹⁰] = 45[(a+b+c)⁸+(b+c+d)⁸+(a-d)⁸-(c+d+a)⁸-(d+a+b)⁸-(b-c)⁸]²

 

M. Hirschhorn

 

25[(a+b+c)³-(b+c+d)³-(a-d)³+(c+d+a)³-(d+a+b)³+(b-c)³][(a+b+c)⁷-(b+c+d)⁷-(a-d)⁷+ (c+d+a)⁷-(d+a+b)⁷+(b-c)⁷] = 21[(a+b+c)⁵-(b+c+d)⁵-(a-d)⁵+(c+d+a)⁵-(d+a+b)⁵+(b-c)⁵]²

 

where for both, ad = bc, with the discovery of the latter inspired by the former.  These can be generalized as,

 

Ramanujan-Hirshhorn

 

Theorem 1: If a^k+b^k+c^k = d^k+e^k+f^k, k = 2,4 where a+b+c = d+e+f = 0, then,

 

64(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 45(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸)²

 

and,

 

25(a³+b³+c³-d³-e³-f³)(a⁷+b⁷+c⁷-d⁷-e⁷-f⁷) = 21(a⁵+b⁵+c⁵-d⁵-e⁵-f⁵)²

 

Proof:  By doing the substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, all three equations have the common factor p²+pq+q²-(r²+rs+s²) = 0 so the problem is reduced to finding expressions {p,q,r,s} that satisfy this.  It has a complete soln.  Let {p,q,r,s} = {-u+v, u+v, -x+y, x+y}, and the condition becomes,

 

u²+3v² = x²+3y²

 

with the complete soln to u²+nv² = x²+ny² as,

 

(ac+nbd)² + n(bc-ad)² = (ac-nbd)² + n(bc+ad)²

 

Analogously, there is one dependent on a system with odd powers,

 

Piezas

 

Theorem 2: If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k,  k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then,

 

7(a⁴+b⁴+c⁴+d⁴-e⁴-f⁴-g⁴-h⁴)(a⁹+b⁹+c⁹+d⁹-e⁹-f⁹-g⁹-h⁹) = 12(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶)(a⁷+b⁷+c⁷+d⁷-e⁷-f⁷-g⁷-h⁷)

 

which again has a complete soln.  For higher powers,

 

Theorem 3: If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k,  k = 2,4,6, define n as,

 

a⁴+b⁴+c⁴+d⁴ = n(a²+b²+c²+d²)²

 

then,

 

25(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)(a¹²+b¹²+c¹²+d¹²-e¹²-f¹²-g¹²-h¹²) = 12(n+1)(a¹⁰+b¹⁰+c¹⁰+d¹⁰-e¹⁰-f¹⁰-g¹⁰-h¹⁰)²

 

Theorem 4: If a^k+b^k+c^k+d^k+e^k = f^k+g^k+h^k+i^k+j^k,  k = 2,4,6,8 define n as,

 

a⁴+b⁴+c⁴+d⁴+e⁴ = n(a²+b²+c²+d²+e²)²

 

then,

 

72(a¹⁰+b¹⁰+c¹⁰+d¹⁰+e¹⁰-f¹⁰-g¹⁰-h¹⁰-i¹⁰-j¹⁰)(a¹⁴+b¹⁴+c¹⁴+d¹⁴+e¹⁴-f¹⁴-g¹⁴-h¹⁴-i¹⁴-j¹⁴) = 35(n+1)(a¹²+b¹²+c¹²+d¹²+e¹²-f¹²-g¹²-h¹²-i¹²-j¹²)²

 

and so on based on higher systems k = 2,4,…2m.  Also, buried in the musty pages of an old journal, there is a similarly beautiful identity, Sinha's 2-4-6 Identity,

 

Sinha

 

5[a²+b²+c²]*[(a+b+c)⁴ + (a+b-c)⁴ + (a-b+c)⁴ + (-a+b+c)⁴ - (2a)⁴ - (2b)⁴ - (2c)⁴] = [(a+b+c)⁶ + (a+b-c)⁶ + (a-b+c)⁶ + (-a+b+c)⁶ - (2a)⁶ - (2b)⁶ - (2c)⁶]

 

which this author realized could be extended (and no further) as,

 

5[a²+b²+c²+d²]*

[(a+b+c-d)⁴ + (a+b-c+d)⁴ + (a-b+c+d)⁴ + (-a+b+c+d)⁴ - (2a)⁴ - (2b)⁴ - (2c)⁴ - (2d)⁴] =

[(a+b+c-d)⁶ + (a+b-c+d)⁶ + (a-b+c+d)⁶ + (-a+b+c+d)⁶ - (2a)⁶ - (2b)⁶ - (2c)⁶ - (2d)⁶]

 

One can see that Sinha’s is just the case d=0.  How this was derived will be discussed in the section on Eighth Powers.

 

V. Tariste

 

(23x-57y)^k + (40x+6y)^k + (17x+63y)^k = (23x+57y)^k + (40x-6y)^k + (17x-63y)^k

 

The identity is in fact valid for k =2,4.  There is nothing really special about these particular values as these can be generalized as,

 

Piezas

 

Let {u₁, u₂, u₃, c} = {a+2b, 2a+b, a-b, a+b},  then for k = 2,4,

 

(ax+u₁y)^k + (bx-u₂y)^k + (cx-u₃y)^k = (ax-u₁y)^k + (bx+u₂y)^k + (cx+u₃y)^k

 

(ax+u₂y)^k + (bx-u₃y)^k + (cx+u₁y)^k = (ax+u₃y)^k + (bx+u₁y)^k + (cx+u₂y)^k

 

And as sums of one side,

 

(ax+u₁y)^k + (bx-u₂y)^k + (cx-u₃y)^k = (a^k+b^k+c^k)(x²+3y²)^k/2

 

(ax+u₂y)^k + (bx-u₃y)^k + (cx+u₁y)^k = (a^k+b^k+c^k)(x²+3xy+3y²)^k/2

 

Furthermore,

 

(ax²+2u₁xy-3ay²)^k + (bx²-2u₂xy-3by²)^k + (cx²-2u₃xy-3cy²)^k = (a^k+b^k+c^k)(x²+3y²)^k

 

where the last is a template identity that can generate quadratic parametrizations using an initial soln, as long as it satisfies the condition a+b = c, an example of which to be given in a later section.  Note that the expressions {a+2b, 2a+b, a-b, a+b} are intimately connected to the form x²+3y² since,

 

4(a²+ab+b²) = (a+2b)² + 3a² = (2a+b)² + 3b² = (a-b)² + 3(a+b)²

 

However, for general n, given the system,

 

a+b = nc;  d+e = nf

a²+b²+c² = d²+e²+f²

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴

 

and the substitution {a,b,c,d,e,f} = {p+q,r+s, t+u, t-u, r-s, p-q}, where set t = 1 without loss of generality, solve for {p,s,u} and substitute these into the last eqn.  This will have a quadratic factor in r with discriminant D,

 

D:= (1+n²)²q⁴ + 8n(2+n²)q³ - 2(-17-6n²+n⁴)q² - 8n(2+n²)q + (1+n²)²

 

which must be made a square.  Since this quartic polynomial has a square leading and constant term, this is easily done.  Two small solns are,

 

q₁ = -2/n,   q₂ = (n-1)/(n+1)

 

Though these lead to a trivial soln, they can be used to compute other rational points on the curve.  A non-trivial point found using Fermat’s method is,

 

q₃ = (n⁶+2n⁴+n²-4)/(4n(n²+1)²)

 

 

12.2 Form: a+b ≠ c; d+e ≠ f

 

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 + (2e)^k + (2f)^k + (2g)^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. 

 

This author found an extension of this to tenth powers and will be discussed later.  The proof for this particular theorem of Birck will be discussed on the section on Eighth Powers.  From this, Sinha derived a version applicable to odd exponents,

 

T. Sinha

 

Theorem: If a^k+b^k+c^k = d^k+e^k+f^k,  k = 2,4  (or Q₁)

 

where a+b-c = 2(d+e-f) and a+b ≠ c, d+e ≠ f, then,

 

(2a-3h)^k + (2a-h)^k + (2b-3h)^k + (2b-h)^k + (2f+h)^k = (2c+h)^k + (2c+3h)^k + (2d-h)^k + (2e-h)^k + (3h)^k

 

for k = 1,3,5,7, where 2h = d+e-f. 

 

To solve Q₁ and the side conditions, Sinha gave a system of linear relations (modified by this author to be consistent with the others), call it S,

 

a+b-c = 2(d+e-f)

(a+b)-(d-e) = 0; 

13a+13b+11c+4f = 0

 

These can then define three of the variables.  For example, solving for {a,c,f}, and substituting into Q₁ for either k = 2 or 4, this forces the remaining three variables, say they are {x,y,z}, into a quadratic eqn of form,

 

a₁x² + a₂y² + a₃z² + a₄xy + a₅xz + a₆yz = 0

 

One can solve this by using an identity of Desboves’ (found in the section on Second Powers) that needs an initial soln {x,y,z}.  Another approach, producing more aesthetic results, is that since this is a quadratic in any of the variables, say z, for rational solns its discriminant must be made a square,

 

b₁x²+b₂xy+b₃y² = t²

 

where the b_i are in terms of the a_i.  Thus, if one can find a system of useful linear relations between the {a,b,c,d,e,f}, the problem will be reduced to solving this eqn.  As this author found out, there are at least four systems S,

 

1. a+b-c = 2(d+e-f);  (a+b) ± (d-e) = 0;  13a+13b+11c+4f = 0
2. a+b-c = 2(d+e-f);  (a-b) ± (d+e) = 0;  4c+7d+7e-f = 0

counting positive and negative cases of the second eqn as distinct.  These then yield the polynomial solns to Sinha's problem, 

 

1. (-3x+2y+z)^k + (-3x+2y-z)^k + (2x-4y)^k = (8x-y)^k + (2x+3y)^k + (14x-2y)^k

 

where 121x²-10xy-5y² = z²,  D = 70. 

 

2. (5x+2y+z)^k + (5x+2y-z)^k + (-14x-4y)^k = (14x+3y)^k + (4x-y)^k + (6x-2y)^k

 

where x²-50xy-5y² = z²,  D = 70. (Sinha’s)

 

3. (6x+3y)^k + (4x+9y)^k + (2x-12y)^k = (-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k

 

where x²+10y² = z²,  D = -10.

 

4. (2x+3y)^k + (-4x+y)^k + (-10x-4y)^k = (3x+y+z)^k + (3x+y-z)^k + (2x-2y)^k

 

where 49x²+40xy+10y² = z²,  D = -10.

 

where D is the discriminant of the conditional eqn and some manipulation was done to attain these simple forms.  Since all have a square leading coefficient, these can be transformed to the form u²+dv² = w² which has a complete soln.

 

Note 1: Any other linear relations or S that is not merely a permutation of variables and change of signs?  For ex, the first pair (a+b)±(d-e) = 0 is equivalent to either of the ff: (a+b)±(d+f), (a-c)±(d-e), (a-c)±(d+f), while the second pair (a-b) ± (d+e) = 0 is the same as (a-b)±(d-f), (a+c)±(d+e), (a+c)±(d-f).  I do not know if the above four solns are the only ones in terms of binary quadratic forms.

 

Note 2:  Sinha’s theorem can be expressed more concisely as given,

 

(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k, 

 

for 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

 

for k = 1,3,5,7, (excluding the trivial case c = 0). 

 

Piezas

 

There can be more solns to Q₁ with a+b-c = 2(d+e-f) though now an elliptic curve has to be solved.  The linear relation used is either case of (a+b) = ±(d+e).  The general form is,

 

(-y+t)^k + (y+t)^k + p^k = (-z±t)^k + (z±t)^k + q^k, 

 

where t is chosen so the eqn satisfies a+b-c = 2(d+e-f).  For the positive case,

 

(-y+t)^k + (y+t)^k + (48-2t)^k = (-12x+t)^k + (12x+t)^k + (24)^k,  where t = 13-x²,  and y² = -2x⁴+100x²+46

 

This elliptic curve has trivial solns x = {1,3,5,7} but from these we can find non-trivial ones such as 4307x = 2367, and so on.  For the negative case,

 

(-y+t)^k + (y+t)^k + (2u+6v)^k = (-z-t)^k + (z-t)^k + (-2u)^k,  where t = u+v,

 

and u,v must satisfy the simultaneous eqns u²+6v² = y² and u²+12uv+24v² = z².  Solving the first as {u,v} = {p²-6, 2p}, the second becomes,

 

p⁴+24p³+84p²-144p⁴+36 = z²

 

with integral solns p = {-6,-4,-2,0,1,3,5} from which rational ones can be derived though this has to be checked for triviality. 

 

Ramanujan

 

3⁴ + (2v⁴-1)⁴ + (4v⁵+v)⁴ = (4v⁴+1)⁴ + (6v⁴-3)⁴ + (4v⁵-5v)⁴

 

This has the nice form of involving a constant term and can be generalized as,

 

Piezas

 

For any constant n,

 

(2n)^k + (n+p)^k + (n-p)^k = (2r)^k + (q+r)^k + (q-r)^k,  k = 2,4

 

{p,q,r} = {mx²+nx+3m,  mx²+nx-3m,  2mx+n},  

 

and three free variables {m,n,x}.  Example, let n = 1, x = 2, so,

 

2^k + (7m+3)^k + (7m+1)^k = (3m-1)^k + (5m+3)^k + (8m+2)^k

 

though for any {m,n,x} this family belongs to the case a+b = c, d+e = f so can’t be used for Sinha’s problem.

 

R. Norrie

 

(a⁴-2)⁴b⁴ + a⁴(b⁴+2)⁴  + (2a³b)⁴ = (a⁴+2)⁴b⁴ + a⁴(b⁴-2)⁴ + (2ab³)⁴