PART 9. Sum / Sums of Sixth Powers
6.1 Four terms
As was already discussed in a previous section, the equation (√p+√q)^{6} + (√p-√q)^{6} = (√r+√s)^{6} + (√r-√s)^{6} has a polynomial soln. It is conjectured that,
x_{1}^{6} + x_{2}^{6} = y_{1}^{6} + y_{2}^{6}
is only trivially solvable. However, if we relax it a bit, there is an infinite number of solutions to,
x_{1}^{6} + x_{2}^{3} = y_{1}^{6} + y_{2}^{3} (1)
Proof:
We use the form,
(c-x)^{6} + (x+y)^{3} = (c+x)^{6} + (-x+y)^{3}
Expanding this, one must solve,
6cx^{4 }+ (20c^{3}-1)x^{2} + 6c^{5} = 3y^{2}
which, for an appropriate constant c, can be treated as an elliptic curve. For example, for c = 7/2, one gets {x,y} = {1/2, 67/2} which yields,
3^{6} + 34^{3} = 4^{6} + 33^{3}
From this initial point {x,y}, one can then easily get subsequent ones, ad infinitum. Surprisingly, there is also a small solution to, x_{1}^{6} + x_{2}^{6} = y_{1}^{6} + y_{2}^{3 } (2)
as, 15^{6} + 18^{6} = 19^{6} + (-118)^{3} It can also be shown there is an infinite number of primitive solutions to (2) using the well-known identity,
(3t^{2})^{6} + (3t-9t^{4})^{3} = 1 + (9t^{3}-1)^{3}
by making the second term a square,
3t-9t^{4} = y^{2}
This can be treated as an elliptic curve. One such point is t = 3/13 which yields,
27^{6} + 138^{6} = 13^{12} + (-25402)^{3}
Another is t = 208098704151/634343459737 (though presumably there may be points of smaller height), and so on. A different approach is used to solve the analogously relaxed,
x_{1}^{8} + x_{2}^{8} + x_{3}^{4} = y_{1}^{8} + y_{2}^{8} + y_{3}^{4}
discussed in Eighth Powers.
6.2 Six Terms: Equal sum of three sixth powers, x_{1}^{6}+x_{2}^{6}+x_{3}^{6} = x_{4}^{6}+x_{5}^{6}+x_{6}^{6}
Duncan Moore has done an exhaustive search within a radius of 17,800. Out of about 400 solns, it turns out that 92% (!) are good for k = 2,6. Why that is the case, no one knows, though it may have to do with the algebraic form x^{2}+xy-y^{2}. (See Bremner and Kuwata's work below.) Data results are here. In fact, the smallest (6.3.3), (6.3.4), (6.4.4) are multi-grade for k = 2,6, [1] 23^{k} + 10^{k }+ 15^{k }= 3^{k }+ 19^{k }+ 22^{k} ^{} [2] 73^{k} + 58^{k }+ 41^{k }= 70^{k }+ 65^{k }+ 32^{k }+ 15^{k} ^{} [3] 10^{k} + 6^{k }+ 5^{k }+ 3^{k }= 9^{k }+ 9^{k }+ 2^{k }+ 2^{k} The situation can be contrasted to the smallest known (8.4.4), (8.4.5), (8.5.5), (8.5.6), (8.6.6), none of which is multi-grade. [1] and [3] belong to parameterizations as, (ad+b)^{k} + (c-2d)^{k} + (de+f)^{k} = (de-f)^{k} + (c+2d)^{k} + (ad-b)^{k},
where {a,b,c,d,e,f} = {-4x+11y, y, 2-5(2x-5y)y, x-y, 2x-3y, 2x-5y}, and {x,y} satisfies x^{2}-6y^{2} = 1, with {x,y} = {7/5, -2/5} yielding [1] while, (a+d)^{k} + (-a+d)^{k} + (b+c)^{k} + (b-c)^{k }= 2(c+d)^{k} + 2(-c+d)^{k} where 5a^{2}+2b^{2} = 7c^{2}, and 2a^{2}+5b^{2} = 7d^{2}, with {a,b,c,d}= {13, 1, 11, 7} giving [3], after removing common factors for both examples. It is not known whether [2] belongs to a family. For 6th powers, identities, whether in terms of a polynomial or an elliptic curve, are known only for multi-grades k = 2,6, or 1,2,6:
1. x_{1}^{k}+x_{2}^{k}+x_{3}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}, for k = 2,6, and x_{1}y_{1}(x_{1}^{2}-y_{1}^{2})+x_{2}y_{2}(x_{2}^{2}-y_{2}^{2})+x_{3}y_{3}(x_{3}^{2}-y_{3}^{2}) = 0. Additional side conditions include, a) 3x_{1}+x_{2}+x_{3} = y_{1}+y_{2}+3y_{3} b) 2x_{2} = y_{2}+y_{3}, 2x_{3} = y_{2}-y_{3}
2. x_{1}^{k}+x_{2}^{k}+x_{3}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}, for k = 1,2,6. Additional conditions include, a) nx_{1}+x_{2}-x_{3} = ny_{1}+y_{2}-y_{3}, for n = 12, 15, 21, etc.
If you know of another class, pls send it. The second condition of the first class can be expressed as a set of three simultaneous eqns given below. As was mentioned, small solns to (k.3.3) for k = 4 or 6 also turn out to be valid for k = 2. The smallest for k = 6 is,
23^{k} + 10^{k }+ 15^{k }= 3^{k }+ 19^{k }+ 22^{k}
which, in fact, is for k = 2,6. This deceptively simple-looking eqn has a lot of structure. If expressed as,
{-23, 10, 15} = {3, 19, -22}
labelled {a,b,c,d,e,f} respectively, then,
a^{2}+ad-d^{2} = -(b^{2}-be-e^{2}) b^{2}+be-e^{2} = -(c^{2}-cf-f^{2}) c^{2}+cf-f^{2} = -(a^{2}-ad-d^{2}) 3a+b+c = d+e+3f
Not surprisingly, this is just the smallest instance of a parametric soln. Cubic and 4th powers have already been discussed and it was seen that a lot of identities involved equivalent forms of F_{3}:= x^{2}+xy+y^{2}. For 5th and 6th powers, it seems now it is the form F_{5}:= x^{2}+xy-y^{2} that is implicit in some identities. For example, recall that,
(√p+√q)^{5} + (√p-√q)^{5} = (√r+√s)^{5} + (√r-√s)^{5}
{p,q,r,s} = {5vw^{2}, -1+uw^{2}, 5v, -(u+10v)+w^{3}}
where w = u^{2}+10uv+5v^{2} and it takes only a small change of variables {u,v} = {x-2y, x+2y} to transform this to the form x^{2}+xy-y^{2}. Though F_{3} and F_{5} are similar-looking, they are quite different since the former has discriminant d = -3 and hence factors over the imaginary field √-3, while the latter has d = 5 and factors over the real field √5. It has been shown that solns to a^{2}+ab+b^{2} = c^{2}+cd+d^{2} also solve a^{k}+b^{k}+(a+b)^{k} = c^{k}+d^{k}+(c+d)^{k} for k = 2,4, and vice versa. For k = 2,6, one now needs a system of three equations,
Theorem 1: (Bremner, Kuwata) If there are {a,b,c,d,e,f} such that, call this system S_{1},
a^{2}+ad-d^{2} + (b^{2}-be-e^{2}) = 0 b^{2}+be-e^{2} + (c^{2}-cf-f^{2}) = 0 c^{2}+cf -f^{2} + (a^{2}-ad-d^{2}) = 0
is true, then so is system S_{2},
a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,6
Proof: Simply add the expressions together to get the identically true statement,
(a^{2}+ad-d^{2})^{k} + (b^{2}-be-e^{2})^{k} + (b^{2}+be-e^{2})^{k} + (c^{2}-cf-f^{2})^{k} + (c^{2}+cf-f^{2})^{k} + (a^{2}-ad-d^{2})^{k} = 2(a^{2k}+b^{2k}+c^{2k}-d^{2k}-e^{2k}-f^{2k}), for k = 1,3
If the LHS is zero, then so is the RHS, which proves the theorem. (Masato Kuwata, Equal Sums of Sixth Powers and Quadratic Line Complexes, 2007. End proof.) (Note: This probably contributes to the reason why more than 90% of solns are multi-grade for k = 2,6 since being valid for one power means being valid for the other. Unfortunately, no similar expressions are yet known for 8th powers!)
Corollary 1: (Piezas) Solutions to S_{1} imply the ff sums are squares,
4(a^{2}+ad-d^{2}) + 5b^{2} = (b+2e)^{2}, 4(b^{2}+be-e^{2}) + 5c^{2} = (c+2f)^{2}, 4(c^{2}+cf-f^{2}) + 5a^{2} = (a+2d)^{2}
4(a^{2}-ad-d^{2}) + 5c^{2} = (c-2f)^{2}, 4(b^{2}-be-e^{2}) + 5a^{2} = (a-2d)^{2}, 4(c^{2}-cf-f^{2}) + 5b^{2} = (b-2e)^{2}
In Unsolved Problems in Number Theory, R. Guy asked if the reverse was true: does S_{2} imply S_{1}? This was answered in the negative by Delorme who gave one more condition, and Choudhry who gave a parametric example that solved S_{2} but not S_{1}.
Theorem 2: (Delorme) The system S_{2} implies S_{1} if it is also the case that,
ad(a^{2}-d^{2}) + be(b^{2}-e^{2}) + cf(c^{2}-f^{2}) = 0
S. Brudno
This is a simple example that solves both S_{1} and S_{2} ,
a^{k }+ b^{k }+ c^{k} = d^{k} + (b+c)^{k }+ (b-c)^{k}, k = 2,6
Note that for k = 2, one must solve a^{2} = b^{2}+c^{2}+d^{2}. The complete soln for k = 2,6 is then,
{c,d} = {ax/(a^{2}+9b^{2}), 3bx/(a^{2}+9b^{2})}
where {a,b,x} satisfies the elliptic curve,
(a^{2}-b^{2})(a^{2}+9b^{2}) = x^{2}
Labelled as x_{i}, y_{i}, the terms also satisfy x_{1}y_{1}(x_{1}^{2}-y_{1}^{2}) + x_{2}y_{2}(x_{2}^{2}-y_{2}^{2}) + x_{3}y_{3}(x_{3}^{2}-y_{3}^{2}) = 0. One small soln, among infinitely many, is {a,b} = {5/4, 1} which yields,
65^{k} + 52^{k} + 15^{k} = 36^{k} + 67^{k} + 37^{k}
Piezas
This is another simple example that solves both systems,
(ad+b)^{k} + (c-2d)^{k} + (de+f)^{k} = (de-f)^{k} + (c+2d)^{k} + (ad-b)^{k}, k = 2,6
where {a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y^{2}, x-y, 2x-3y, 2x-5y}, and {x,y} satisfies x^{2}-6y^{2} = 1.
It is quite interesting that this particular Pell equation appears in the context of 6th powers. However, since the eqn is homogeneous, one can just as well solve it in the rationals. This is also discussed several sections below.
A. Bremner, Piezas
It turns out the complete soln of S_{1}, S_{2} can be given in terms of a quartic polynomial that is to be made a square, hence can be treated as an elliptic curve. R. Guy in the same book mentions that Bremner gave a method that can find all parametric solns. I don’t have access to this paper yet but after some experimentation found a procedure. It suffices to use S_{2} and one of the eqns of S_{1}. Using the general form,
(p+q)^{k} + (r+s)^{k} + (t+u)^{k} = (t-u)^{k} + (r-s)^{k} + (p-q)^{k}
for k = 2,6 and one of the others, say c^{2}+cf-f^{2} = -(a^{2}-ad-d^{2}) with variables changed appropriately, the complete soln is then,
{q,u} = {-rs(p+2t)/w, rs(2p-t)/w}, where w = p^{2}+t^{2} and {p,r,s,t} satisfy,
((p^{2}-11pt-t^{2})s^{2} + w^{2})r^{2} = (p^{2}+pt-t^{2}+s^{2})w^{2}
(Perhaps not surprisingly, the form F_{5}:= x^{2}+xy-y^{2} appears again.) Thus, the problem is to find {p,t,s} such that the expression,
((p^{2}-11pt-t^{2})s^{2} + (p^{2}+t^{2})^{2})(p^{2}+pt-t^{2}+s^{2}) = y^{2}
which is only a quartic in the variable s is a square, though some {p,t,s} are trivial. So, given an initial soln, this can be treated as an elliptic curve to generate more. For ex, let,
{p,t,s} = {1, n, n-1}
This is trivial with respect to the original sextic eqn S_{2} but, using the same {p,t}, we can find more rational points s on the curve with the next one (using Fermat’s method) non-trivial of deg-18. There are also small non-trivial solns,
{p,t,s} = {-n-1, (n-1)(n-2), 5+n^{2}} {p,t,s} = {(-3+n)(2+n), 3+3n+2n^{2},^{ } 3+6n+n^{2}}
either one of which yields scaled versions of the 4th-deg Brudno-Delorme identity given below. Again this can be used to generate more rational points with the first one yielding a rational 18-deg. (Using the same {p,t}, there is another rational point s that is also only a quadratic, s = 3-2n+n^{2} for the first and s = 9+4n+n^{2} for the second, but these are trivial.) Other solns are,
{p,t,s} = {2(1+n)(-1-n+n^{2})(-1+n+n^{2}), (1+n)(1-n+n^{2})(1-5n+n^{2}), (-1+n)(1+2n+7n^{2}+2n^{3}+n^{4})} {p,t,s} = {2(3+3n+2n^{2})(3+2n+2n^{2}), (3+3n+2n^{2})(3-2n+4n^{2}), (1+n)(9+4n^{2})}
The first, after a small adjustment, yields Delorme’s 5th-deg identity given below, while the second yields a different 5th deg. Delorme in his paper gave identities of deg n for all 4 ≤ n ≤ 11, except n = 6,10. Q: Can anyone give an identity with polynomials of deg n = 6?
Note: It can be shown that the system a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,6 with,
a^{2}+nad-d^{2} = -(b^{2}-nbe-e^{2}) b^{2}+nbe-e^{2} = -(c^{2}-ncf-f^{2}) c^{2}+ncf-f^{2} = -(a^{2}-nad-d^{2})
has non-trivial solns only for n = ±1. Proof: Using the general form and the method described above, and elimination of appropriate terms, the final resultant eqn has either trivial factors, or a quadratic factor solvable over √(-1), or if (n+1)(n-1) = 0. Since the first two can be disregarded then only the last applies, proving the assertion.
S. Brudno, J. Delorme
The 4th-deg Brudno-Delorme identity also has the side condition 3a+b+c = d+e+3f, call this S_{3}. Let,
a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,6 where,
{a,b,c} = {-n^{4}-n^{3}-5n^{2}+8n+8, (n^{3}+7n-2)(n+2), 3(3n^{2}+2n+4)} {d,e,f} = {(n^{2}-n+3)(n+2)^{2}, -4n^{3}-5n^{2}-8n+8, -n^{4}+n^{2}+14n+4}
(modified to reduced the size of the coefficients), then these satisfy S_{1}, S_{2}, S_{3}. The smallest non-trivial example is n = -3 which, after removing common factors, yields {-23, 10, 15} = {3, 19, -22}, the example given earlier. Interestingly, it can be shown that this identity depends on the equation x^{2}-6y^{2} = z^{2}. This author found that some of the relations are enough to derive a version using: one eqn from S_{2} (at k = 2), two from S_{1}, and S_{3}. Let,
(ad+b)^{k} + (c-2d)^{k} + (de+f)^{k} = (de-f)^{k} + (c+2d)^{k} + (ad-b)^{k}
and using this substitution also on the side conditions, one can derive the variables as,
{a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y^{2}, x-y, 2x-3y, 2x-5y}, where x^{2}-6y^{2} = 1.
One can easily solve for {x,y} in the integers as a Pell equation or, since the eqn is homogeneous, in the rationals. Q: Is it possible to solve S_{2}, S_{3} without solving S_{1}?
J. Delorme
Delorme gave identities of deg n for all 4 ≤ n ≤ 11 , except n = 6,10, that solves S_{1}, S_{2}, one of which is the 5th deg,
{x_{1}, x_{2}, x_{3}} = {3n^{5}+8n^{4}+9n^{3}-4n^{2}-9n-2, -2n^{5}-n^{4}+12n^{3}+13n^{2}+4n-1, -n^{5}-9n^{4}-13n^{3}-7n^{2}-7n-3} {y_{1}, y_{2}, y_{3}} = {2n^{5}+9n^{4}+4n^{3}-9n^{2}-8n-3, -3n^{5}-7n^{4}-7n^{3}-13n^{2}-9n-1, -n^{5}+4n^{4}+13n^{3}+12n^{2}-n-2}
Q: Any linear relations between the x_{i} and y_{i}?
Update (7/14/09): Since the x_{i}, y_{i} are polynomials in n, it turns out the problem is equivalent to finding six unknown rational constants p_{i} such that p_{1}x_{1}+p_{2}x_{2}+p_{3}x_{3}+p_{4}y_{1}+p_{5}y_{2}+p_{6}y_{3} = m for some constant m, preferably m = 0, for all n. Expanding and collecting powers of n, set the coefficients, which are polynomials in the p_{i}, equal to zero. For this particular identity, we can't find p_{i} such that m = 0, but instead find m = 40 given by the linear relation,
5x_{1}-12x_{2}-7x_{3}-12y_{1}+5y_{2}+7y_{3} = 40 (End note)
Update (6/24/09): A. Choudhry
The fact that solving the systems S_{1}, S_{2} is reducible to an elliptic curve implies that one soln can lead to another. Choudhry ("On Equal Sums of Sixth Powers", 1994) has given an explicit construction for this. Given the system,
x^{k}+y^{k}+z^{k} = u^{k}+ v^{k}+ w^{k}, k = 2,6 with,
x^{2}+xu-u^{2} = w^{2}+wz-z^{2} y^{2}+yv-v^{2} = u^{2}+ux-x^{2} z^{2}+zw-w^{2} = v^{2}+vy-y^{2}
Let {x,y,z,u,v,w} = {x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}} be a soln, then a new one is given by, {x,y,z,u,v,w} = {ap+x_{1}q, bp+x_{2}q, cp+x_{3}q, dp+y_{1}q, ep+y_{2}q, y_{3}q}, where {p,q} and {a,b,c,d,e} are,
{p,q} = {-a(2x_{1}-y_{1})-b(2x_{2}+y_{2})+d(x_{1}+2y_{1})-e(x_{2}-2y_{2}), a^{2}+b^{2}-d^{2}-e^{2}-ad+be}
{a,b,c,d,e} = {(r_{1}^{2}+2r_{1}r_{2})s_{2}, (r_{3}^{2}-2r_{3}r_{4})s_{1}, s_{1}s_{2}, -(r_{1}^{2}+r_{2}^{2})s_{2}, (r_{3}^{2}+r_{4}^{2})s_{1}}
{s_{1}, s_{2}} = {r_{2}^{2}+r_{2}r_{1}-r_{1}^{2}, r_{3}^{2}+r_{3}r_{4}-r_{4}^{2}}, and {r_{1}, r_{2}, r_{3}, r_{4}} = {x_{1}-y_{3}, x_{3}-y_{1}, x_{2}-y_{3}, x_{3}-y_{2}}.
Example: The initial soln {x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}} = {3, 22, -19, 23, 15, 10} gives {x,y,z,u,v,w} = {4513, -104, 4693, -2273, -5099, 3352} after removing the common factor 3*39200^{2}. (End note.) A. Choudhry There is a 3-parameter soln for k = 1,2,6 which entails solving an elliptic curve. This is a class that solves S_{2}, but not S_{1}.
x_{1}^{k}+x_{2}^{k}+x_{3}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}
{x_{1}, x_{2}, x_{3}} = { 2(α+β)m+(α-β+t)n, -2αm+(α+β+t)n, -2βm-(α+β-t)n} {y_{1}, y_{2}, y_{3}} = {-2(α+β)m+(α-β+t)n, 2αm+(α+β+t)n, 2βm-(α+β-t)n}
for k = 1,2 is true for all variables, though is not a complete soln. We can do the substitution {α+β, α-β, t} = {a,b,c} to get the slightly more symmetric,
{x_{1}, x_{2}, x_{3}} = { 2am+(b+c)n, -(a+b)m+(a+c)n, -(a-b)m-(a-c)n} {y_{1}, y_{2}, y_{3}} = {-2am+(b+c)n, (a+b)m+(a+c)n, (a-b)m-(a-c)n}
When expanded for k = 6 this has the condition,
(Poly1)m^{2}+(Poly2)n^{2} = 0
where,
Poly1:= b(11a^{2}+b^{2})+5(3a^{2}+b^{2})c; Poly2:= (a^{2}+b^{2})(b+5c)+10(bc^{2}+c^{3})
linear and cubic in c, respectively. The problem then is: Given constants a,b, find c such that,
y^{2} = -(Poly1)(Poly2)
avoiding the cases c = -b/3 and y = Poly1 = 0 since these give trivial results. For a non-trivial ex., let {α,β} = {1, 7}, hence {a,b} = {8,-6} giving,
y^{2} = -6(-74+19c)(-60+50c-6c^{2}+c^{3})
one soln being c = 38/23 with this elliptic curve having an infinite number of rational points. (Choudhry gave more than twenty {a,b} with another as {11,-9}.)
Piezas
While Choudhry did not give the complete soln of k = 1,2,6, it can be shown that the general soln entails making a certain quartic polynomial into a square (just like for the case k = 2,6 plus S_{2}). We can show this in two ways.
First method: Using a small variant of the general form,
(p+q)^{k} + (r+s)^{k} + (t+u)^{k} = (-t+u)^{k} + (-r+s)^{k} + (-p+q)^{k}
and letting r = np for some rational constant n, one can completely solve k = 1,2 with,
{t, u} = {-p(n+1), (q+ns)/(n+1)}
With these values, k = 6 becomes a polynomial of form (Poly1)p^{2}+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {q,s}, so their product is a quartic and the problem is reduced to finding,
y^{2} = -(Poly1)(Poly2)
a situation similar to Choudhry’s, though the tricky part is finding appropriate rational n which may be the ratio of relatively large integers.
Second method: We simply use the old form L_{1} again. Expanding for k = 2, the non-trivial condition is b+c = 0, hence we modify it to,
(a+bp+q)^{k} + (b-bp+q)^{k} + (-b+ap+q)^{k} = (a-bp+q)^{k} + (b+ap+q)^{k} + (-b+bp+q)^{k}
which solves k = 1,2. Expanding for k = 6 results in a quadratic in b of form (Poly1)b^{2}+(Poly2) = 0, where Poly1 is linear and Poly2 is cubic in {p,q}. The objective then is to make its discriminant D a square,
y^{2} = -(Poly1)(Poly2)
similar to the first method, but now the expressions are simple enough to write down,
Poly1:= a(1+p)(1+p^{2}) + 5(1-p+p^{2})q Poly2:= a^{3}(1+p)(1+p^{2}) + 5a^{2}(1+p+p^{2})q + 10a(1+p)q^{2} + 10q^{3}
Thus any solution to k = 1,2,6 must satisfy D = y^{2}. For example, using a result of Choudhry’s,
43^{k} + (-372)^{k} + 371^{k} = 307^{k} + (-405)^{k} + 140^{k}
for k = 1,2,6, we equate its first four terms with that of the modified L_{1} to get,
{a,b,p,q} = {291, -388, 33/97, -116}
Using the values for a,p gives the elliptic curve in q,
y^{2} = -(818844+7297q)(7369596+123291q+780q^{2}+2q^{3})
where a square numerical factor has been removed. A rational point, of course, is Choudhry's q = -116 (with another one as q = -130 though this gives a trivial result). From these, more points q can then be computed though typically they are rational numbers with many digits. Note: Incidentally, if we are to add more constraints to this system, then there are no non-trivial rational solns to k = 1,2,6 plus any of the eqns of S_{1}, for ex. c^{2}+cf-f^{2} = -(a^{2}-ad-d^{2}). These four eqns are enough to linearly derive a soln. Using the general form,
(p+q)^{k} + (r+s)^{k} + (t+u)^{k} = (t-u)^{k} + (r-s)^{k} + (p-q)^{k}
then,
{q,r,u} = {(p+2t)s/v, -(p^{2}+t^{2})/v, -(2p-t)s/v}, where v = p-3t
and {p,s,t} satisfies p^{3}-2p^{2}t+pt^{2}+8t^{3}-(p-8t)s^{2} = 0. One must then solve the elliptic curve,
(p^{3}-2p^{2}t+pt^{2}+8t^{3})(p-8t) = y^{2}
and as far as I checked this has only trivial solns. However, the curve,
(p^{3}-2p^{2}t+pt^{2}+8t^{3})(p-8t) = -y^{2}
does have non-trivial ones, like {p,t,y} = {24, 13, 1280} and many more thus giving an imaginary value to s. So this system is solvable in terms of Gaussian rationals. |