Theorem 1: (Piezas) "The system x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}, for k = 1,2,3,5,7, with,
x_{1}y_{1} = (x_{3}y_{3}) = (x_{4}y_{4})
has a complete soln in terms of making two quadratic polynomials into squares." This is given by,
(a+bj)^{k }+ (c+dj)^{k }+ (e+fj)^{k }+ (g+hj)^{k }+ (i+j)^{k} = (abj)^{k }+ (cdj)^{k }+ (efj)^{k }+ (ghj)^{k }+ (ij)^{k}
where,
{a, i, d, f, h} = {1, (b+cd+ef+gh), b1, b, b}
e = ((b2c+bc)+u) j/v g = ((b2c+bc)u) j/v j = v/(3+2b)
and {u,v} are expressed in the free variables {b,c} as,
(99b+27b^{2}+21b^{3}+2b^{4})  2(1+2b)(69b6b^{2}+b^{3})c + (3264bb^{2}+33b^{3}+2b^{4})c^{2} = nu^{2}
(915b+2b^{2}) + 2(1+2b)(3+5b)c + (2323b+18b^{2})c^{2} = nv^{2}
where n = (1+b)(1+2b). Hence, two quadratic polynomials in the variable c must be made squares. One polynomial soln is given by,
c = 2(2b^{2}+18b9)/(14b^{2}59b+27)
This, and a special case, will be explained more below. Also, since the theorem involves an even power, then we can give this relevant conjecture,
Conjecture: "Define F_{k}:= x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} – (y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}). If F_{k} = 0 for k = 1,2,3,5,7, then (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2})(11F_{9}) = 9F_{11}."
The conjecture is the generalization of the proven theorem for k = 1,2,3,5 and, using the soln given above, is easily shown to be true for this special case. However, I do not know how to prove this in general.
Proof (for theorem): Using F_{3}, the more general version of F_{2},
(a+bj)^{k }+ (c+dj)^{k }+ (e+fj)^{k }+ (g+hj)^{k }+ (i+j)^{k} = (abj)^{k }+ (cdj)^{k }+ (efj)^{k }+ (ghj)^{k }+ (ij)^{k}
where set a = 1 without loss of generality. For k = 1,2, let,
i = (ab+cd+ef+gh), d = (1+b+f+h)
To satisfy the two constraints, let,
(a+bj)  (abj) = (e+fj) + (efj) (a+bj)  (abj) = (g+hj) + (ghj)
so f = b, h = b. Expanding F_{3} for k = 3,5,7, we get,
(Poly01)j^{2} + (Poly02) = 0 (eq.1) (Poly11)j^{4} + (Poly12)j^{2} + (Poly13) = 0 (eq.2) (Poly21)j^{6} + (Poly22)j^{4} + (Poly23)j^{2} + (Poly24) = 0 (eq.3)
Eliminate j between eq.1 and 2, then between eq.1 and 3 (set j = √v for ease of computation) to get two eqs, call these the auxiliary resultants,
(Poly01)g + (Poly02) = 0 (eq.4) (Poly11)g^{3} + (Poly12)g^{2} + (Poly13)g + Poly(14) = 0 (eq.5)
where the first fortunately is only linear in the variable g. (It can get as high as sextic in the general case, with the second resultant even higher.) Eliminate g between the two and we get the final resultant which has a nontrivial factor only a quadratic in e, and some trivial linear ones. This has discriminant D that is only a quadratic in c so is low enough to be handled by standard algebraic methods. One can then finally express {e,g,j} in terms of free variables {b,c}, the explicit expressions given earlier. Note that {e,g} are the two roots of the same quadratic.
The problem then is reduced to making two quadratic polynomials in c as squares, with the first as the discriminant D of the final resultant. Two special cases were found by this author: one is when b = 3/5 which makes D a square and yields the identity dependent on x^{2}+6y^{2} = 1 given previously; the second is based on a numerical result by Gloden,
[39, 65, 1, 57, 53] = [33, 11, 73, 15, 71]
rearranged by Shuwen to make it valid for k = 1,2,3,5,7. This author noticed it had another side condition x_{3}+x_{4} = y_{4}+y_{5} so using this constraint found a polynomial for c given at the start as,
c = 2(2b^{2}+18b9)/(14b^{2}59b+27)
which makes both quadratics as squares and should yield terms as quartic polynomials, with Gloden’s soln as b = 4. In fact, there is an infinite family of polynomial c though what is desirable are those of small degree. In general, it’s long been known that given an initial soln to a quadratic polynomial that is to be made a square, P_{1}(x) = y^{2}, then the complete soln can be given as a binary quadratic form, x = p_{1}m^{2}+p_{2}mn+p_{3}n^{2}, where we can suppress n for conciseness. If there is a second quadratic also in x which is also to be made a square, P_{2}(x) = z^{2}, by substituting this value for x into it, one gets a quartic expression of form,
q_{1}m^{4}+q_{2}m^{3}+q_{3}m^{2}+q_{4}m+q_{5} = z^{2}
Using an initial polynomial soln m, one can then compute an infinite sequence for the system k = 1,2,3,5,7. (In our case, one initial value c, which incidentally makes both quadratics as squares, though trivial, is c = b/(1b). Appropriate application of what was discussed can then generate an infinite sequence of polynomial solns.) Note: This approach can also be used for k = 1,2,4,6,8 with the same constraints. Unfortunately, neither of the auxiliary resultants is linear in some variable but instead are of quite high degree. However, using a simple algebraic trick on the variables {e,g} discussed in the next section, these can be reduced and the final resultant turns out to be a quartic in one variable with odd exponents so, in general, needs the cube roots of unity. I do not know if this is nontrivially solvable.
Even more generally, the odd system k = 1,3,5,7,
(a+bj)^{k }+ (c+dj)^{k }+ (ebj)^{k }+ (gbj)^{k }+ (i+j)^{k} = (abj)^{k }+ (cdj)^{k }+ (e+bj)^{k }+ (g+bj)^{k }+ (ij)^{k}
where a=1 for convenience and which has x_{1}y_{1} = (x_{3}y_{3}) = (x_{4}y_{4}), other than one exceptional case b(c±e) – c = ±i, then this has a complete soln where d = b1, j is a root of a quadratic, and {e,g} are two roots of a single quartic, both only in even exponents with coefficients in the remaining variables {b,c,i}. In other words, the radical soln of this system does not employ the cube roots of unity. (Of course, for certain variables this radical soln becomes rational.) Using the same approach as the one above, we find the final resultant of this system factors as the four cases of b(c±e) – c = ±i and an irreducible quartic factor, call this Q_{4}, with only even exponents. Two of the roots of Q_{4} turn out to define {e,g}, though unfortunately is too tedious to write down here. To find j, use the eqn above at k = 3 to get,
bc^{2}+bc^{2}b(e^{2}+g^{2})+i^{2}+(bb^{2})j^{2} = 0
and using the appropriate two roots of Q_{4} as {e,g}, it transforms to,
c^{2}+i^{2}b(3+c^{2}+2i^{2}) + 3b(1+b)(1+2b)j^{2} = 0
so we have all the unknowns {e,g,j} in terms of {b,c,i}. If the exceptional case b(c±e) – c = ±i (or equivalently, b(c±g) – c = ±i) is used to define i, two of the variables {e,g,j} are roots of two different quartics while the third variable is a root of a quadratic. This system can have a nontrivial radical soln but I don’t know if it can be rational. However, if an appropriate relationship can be found between {b,c,i}, such as,
(3b3c+5bc) + (3+2b)i = 0
then Q_{4} can factor into two quadratics as in the explicit identity for k = 1,2,3,5,7 given earlier where {e,g} are the two roots of one quadratic factor. To find a second relationship, using one soln by Gloden,
[51, 19, 9, 43, 55] = [3, 53, 45, 11, 37]
in addition to the aforementioned constraints, this also has x_{1}+x_{3} = (y_{2}+y_{4}) and we can have a second theorem,
Theorem 2: (Piezas) "The odd system x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}, for k = 1,3,5,7 where,
x_{1}y_{1} = (x_{3}y_{3}) = (x_{4}y_{4}), as well as x_{1}+x_{3} = (y_{2}+y_{4}),
other than one exceptional case, has a complete soln in terms of making two quadratic polynomials into squares." (See update below.)
Proof: Given F_{3} again,
(a+bj)^{k }+ (c+dj)^{k }+ (e+fj)^{k }+ (g+hj)^{k }+ (i+j)^{k} = (abj)^{k }+ (cdj)^{k }+ (efj)^{k }+ (ghj)^{k }+ (ij)^{k}
The case to be avoided is the same b(c±e) – c = ±i. Set a = 1 and {d,f,h} = {b1, b, b} to satisfy the usual conditions and k = 1. The new constraint then entails making a+c+e+g+j = 0. To satisfy this and k = 3,5,7, let,
e = ((1+b+bc)+u) i/v g = ((1+b+bc)u) i/v i = v/(3+2b) j = (1+3c)/(3+2b)
where {u,v} are,
(8+24b+15b^{2}+2b^{3}) – 2(1+2b)(62b+b^{2})c + (36+12b+23b^{2}+2b^{3})c^{2} = (2b1)u^{2}
(24b+39b^{2}+18b^{3}) + 18b(1+b)(1+2b)c + (1+b)(939b+50b^{2})c^{2} = (2b1)v^{2}
so the problem is reduced to making these two quadratic polynomials in c as squares. By completely solving one, say the second, using a binary quadratic form (an initial soln for both is c = 1) then solving the other one entails making a quartic polynomial (with a square leading term) a square, just like for Theorem 1. By using Fermat’s method an infinite sequence of polynomial solns can then be easily found. This also has a special case at b = 1/6 where the two become,
(4/81)(119+406c+1009c^{2}) = u^{2} (4/81)(9+5c)^{2} = v^{2}
so one merely has to solve the first quadratic condition. Explicitly, the identity is,
(17+3p)^{k} + (7+37p)^{k} + (62pq)^{k} + (62p+q)^{k} + (1513p)^{k} =
(153p)^{k} + (75p)^{k} + (4+4pq)^{k} + (4+4p+q)^{k} + (3+23p)^{k}
for k = 1,3,5,7, if 119+406p+1009p^{2} = q^{2} (which has discriminant D = 70).
Note 1: As one can see, in Theorem 2 the relationship between {b,c,i} is quadratic while that of {b,c,j} is linear. (In Theorem 1, this is reversed.)
Note 2: Sometimes it is a matter of choosing the right form to solve a particular system of eqns. For ex, suppose one chose the more obvious form,
(a+b)^{k }+ (c+d)^{k }+ (e+f)^{k }+ (g+h)^{k }+ (i+j)^{k} = (ab)^{k }+ (cd)^{k }+ (ef)^{k }+ (gh)^{k }+ (ij)^{k}
To satisfy the three constraints and k = 1, set f = b, h = b, g = (a+b+cd+e), j = bd, where a = 1 for convenience. Expanding for k = 3,5,7, by eliminating {i,c} one gets a final resultant that is a quadratic in e with a quartic discriminant in {b,d}. Recovering i, this is also a quadratic with a different quartic discriminant in {b,d}. Thus, one ends up with two quartic polynomials to be made as squares which is hard to solve when in fact using the first approach, the polynomials can be reduced to just two quadratics. It makes one wonder whether in other situations where one ends up with two quartic polynomials, if there is a right approach to simplify the problem.
Note 3: Authors who have written on the seventh degree are Sinha, Letac, Gloden, Moessner, etc: A. Letac, Gazeta Matematica, 48 (1942), p. 6869. A. Gloden, Mehrgradige Gleichungen, Groningen, 1944; Two theorems on multidegree equalities, Amer. Math. Monthly, 55 (1948), p. 8688. A. Moessner, On equal sums of powers, Math. Student, 15 (1947), p. 8388. I’ll add the information in these sources once I can access them (or if someone is nice enough to send me a pdf copy), though I presume Gloden’s and Moessner’s are just special cases of the computerassisted general theorems in this section. On the other hand, Letac’s has one term x_{i} = 0 with different constraints so he must have used another method.
(Update, 7/26/09): I just realized that Sinha's multigrade system has an extra sidecondition such that it is essentially equivalent to Theorem 2. With a small linear change of variables, it is quite easy to find the complete soln. Let,
(a+4e)^{k }+ (b+4e)^{k }+ (a+b)^{k } = (c+2e)^{k }+ (d+2e)^{k }+ (c+d)^{k}, for k = 2,4
then,
(abe)^{k }+ (ce)^{k }+ (b+e)^{k }+ (a+e)^{k }+ (c+d+e)^{k} = (3e)^{k }+ (a+b+3e)^{k }+ (a3e)^{k }+ (b3e)^{k }+ (d+e)^{k}, for k = 1,3,5,7 (eq.1) Rearranged this way, note that eq.1 also satisfies,
x_{1}y_{1} = (x_{3}y_{3}) = (x_{4}y_{4}), as well as x_{1}+x_{3} = (y_{2}+y_{4}),
just like in Theorem 2. The complete soln for this simple version is also much simpler given by,
b = ((a+4e)r+u)/(2r)
c = ((d+2e)r+v)/(2r) r = 2ad
where {u,v} are in the free variables {a,d,e} as,
2a^{3}+3a^{2}d4d^{3 }+ 8(2ad)(a+d)e + 32(a+d)e^{2 } = (1/r)u^{2}
8a^{3}6ad^{2}d^{3 }+ 4(2ad)(4a+d)e + 4(26ad)e^{2 } = (1/r)v^{2}
hence, naturally enough, involves making two quadratics as squares, though,
ad(a+d)(ad)(a+2e)(a2e)(d+4e)(d4e) = 0
and a few other relations are trivial. Small nontrivial polynomial solns which makes the two conditions squares are,
e = (3a^{2}4ad5d^{2})/(6a+22d) e = (5a^{2}2ad+21d^{2})/(18a+66d)
from which an infinite number of other polynomial solns can be derived. One can wonder if Theorem 1 has a simpler version as well. It can also be shown that there is no permutation of the {x_{i}, y_{i}} of Sinha's system such that it is valid for k = 2 for all {a,d,e}, except for special cases, so it is not equivalent to Theorem 1. (End update.)
Note: By a suitable change of variables, two cubic polynomials to be made squares may be reduced to just two quadratics, and hence amenable to being treated as an elliptic curve. For example, given,
2b^{3}+3b^{2}c4c^{3}2(7b^{2}+8bc8c^{2})x+32(b+c)x^{2}96x^{3} = u^{2}
8b^{3}6b^{2}cc^{3}2(26b^{2}20bcc^{2})x+8(19b8c)x^{2}288x^{3} = v^{2}
Let {b,c} = {a+mx, d+nx}, and we get,
P_{3}x^{3}+P_{2}x^{2}+P_{1}x+P_{0} = u^{2} Q_{3}x^{3}+Q_{2}x^{2}+Q_{1}x+Q_{0} = v^{2}
where the P_{i} and Q_{i} are polynomials in {a,d,m,n}. The objective is to eliminate the cubic term by setting P_{3} = Q_{3} = 0. Since these are only in the {m,n}, one can get their resultant and, for these pair, it turns out the suitable values are {m,n} = {4,2}. (Of course, not all cubic pairs will fortuitously have rational {m,n}, but it will be worth checking.)
