015: Sums of three Fourth Powers (Part 2)

-3p²+q²+3r² = u²

-3p²+3q²+r² = v²

and so on for others involving a+b = c. Note that the expressions {a+2b, 2a+b, a-b, a+b} are intimately connected to the form x²+3y² since,

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

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)

10.2. Constraint: a+b ≠ c; d+e ≠ f

Birck, Sinha

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, 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 2: 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 two 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

These then yield polynomial solns to Sinha's problem as,

1. (-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = (9x-y)^k + (-x+3y)^k + (16x-2y)^k

where 126x²-5y² = z², D = 9(70). (Sinha’s)

2. (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.

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, k = 2,4

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

10.3. Constraint: a+b±c = n(d+e±f)

Piezas

It turns out the system,

a+b±c = n(d+e±f) (eq.1)

a²+b²+c² = d²+e²+f² (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴ (eq.3)

also has a soln for any n of which Sinha’s problem was just the special case n = 2. The general problem is equivalent to finding rational points on a certain elliptic curve (though this curve may be degenerate for some n). To show this, first express the variables in the form,

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

reducing the degrees of eq.2 and 3, a trick which will be also useful when dealing with sixth powers. As the explicit expressions are a bit messy, only the steps will be given:

1. Solve for p,u using eqs.1,2.

2. Substituting these into eq.3, it has a quadratic factor in r with discriminant D.

3. D is a quartic in s. Hence this must be made a square with two small solns for any n as s = q±t. By treating D = y² as an elliptic curve, from an initial soln subsequent ones can then be found.

Example: Using Sinha’s problem n = 2, and the negative case a+b-c = 2(d+e-f), doing the substitutions, we get,

3p-q-r+3q-3t+u = 0 (eq.1)

pq+rs+tu = 0 (eq.2)

p³q+pq³+r³s+rs³+t³u+tu³ = 0 (eq.3)

Solve for {p,u} and substitute into eq.3. One factor will be a quadratic in r. The square-free discriminant D of this is of form,

D: = c₁s⁴+c₂s³+c₃s³+c₄s+c₅² = y²

with a square constant term and where the c_i are polynomials in {q,t}. Four small solns are: s = {q, q-t, q+t, (q-t)/7}.

These may then be used to generate subsequent ones. For general n, using s = q+t with t = 1, a quadratic form soln can be found for the + case of eq.1 as,

{a,b,c} = {1+2nq+r, n(1+2q), -1+n-r}

{d,e,f} = {n+q+nq+r, 1-n+q-nq-r, 1+2q}

where r = q+(n+1)q², for arbitrary {n,q}. This also satisfies {a-b+c, d+e-f} = {0,0} though one can also find parametric ones such that this is not the case.

10.4. Constraint: na+b+c = d+e+nf

Piezas

Just like the previous, this has a soln for any n. The inspiration for this form is the smallest polynomial soln to the sixth power multi-grade a^k+b^k+c^k = d^k+e^k+f^k, k = 2,6 which, among others, surprisingly obeys the side condition 3a+b+c = d+e+3f. (Other n for sixth powers obeying na+b+c = d+e+nf are now known.) For the quartic version, for general n again this can be reduced to finding rational points on an elliptic curve. Let,

na+b+c = d+e+nf (eq.1)

a²+b²+c² = d²+e²+f² (eq.2)

a⁴+b⁴+c⁴ = d⁴+e⁴+f⁴ (eq.3)

and likewise {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q},

1. Solve for r,u in eq.1,2. When substituted into eq.3, this becomes a quadratic in q.

2. Discriminant D of eq.3 is a quartic polynomial in s (but a sextic in p,t).

3. For general n, four small solns are s = {p-nt, p-t, p+t, p}. Using these and others, subsequent ones can then be found. Thus, any soln to eq.1,2,3 must solve D = y².

Example: Using s = p+t with t = 1 yields,

{a,b,c} = {2n-3p+np, -n-n²+3p+np, -n(-1+n-2p)}

{d,e,f} = {n(1+n-2p), -n+n²+3p-np, -2n+3p+np}

for arbitrary {n,p}, with this particular soln also satisfying {a+b-c, d-e+f} = {0,0}.

10.5. Constraint: na+b = e+nf

Piezas

For n = 1, this can be seen as the case n = 0 of a+b+nc = nd+e+f of the system above, and can be completely solved as polynomials thus providing a four-parameter soln to a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,4. Interestingly, this ultimately involves solving the simple equation x²+my² = z². Furthermore, any soln implies a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k for k = 1,2,3,5.

Theorem: If (p+u)^k + (p-u)^k + (2q)^k = (p+v)^k + (p-v)^k + (2r)^k, for k = 2,4, then,

(p+2q)^k + (p-2q)^k + (-p+2r)^k + (-p-2r)^k = (-2p+u)^k + (-2p-u)^k + (2p+v)^k + (2p-v)^k, for k = 1,2,3,5

The complete soln is given by the simultaneous eqns,

(50x²+300xy-150y²)⁴ + (50x²-300xy-150y²)⁴ + (100x²-300y²)⁴ + (4x²+12y²)⁴ + (15x²+45y²)⁴ = 103⁴(x²+3y²)⁴

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. For example, the smallest soln, 2⁴+2⁴+4⁴+3⁴+4⁴ = 5⁴, was used by Ramanujan. Another, found by this author, is 50⁴+50⁴+100⁴+4⁴+15⁴ = 103⁴ which yields,

Proof: Doing the same substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, the two equations have the common factor pq(p+q)-rs(r+s) = 0. (End proof) This is also discussed in Form 13 of Third Powers. Another uses k = 1,3,5,