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1. Various Forms I. Maillet's Identity (E. Maillet)
(a+b)3+(a+c)3+(a+d)3+(a-b)3+(a-c)3+(a-d)3 = 6a(a2+b2+c2+d2) II. Lucas-Liouville Polynomial Identity
(a+b)k+(a+c)k+(a+d)k+(b+c)k+(b+d)k+(c+d)k+(a-b)k+(a-c)k+(a-d)k+(b-c)k+(b-d)k+(c-d)k = 6(a2+b2+c2+d2)k/2, for k = 2,4
Note: It seems E. Lucas knew this identity before Liouville. Anything else for k > 4?
Update 9/26/09: William Ellison pointed out that David Hilbert has proven (x12+x22+ ... +xn2)k can be written as a rational combination of (2k)th powers of linear forms in the xi. (You can read his paper on Waring's Problem and how Hilbert solved it in the Attachments section at the bottom of this page.) For example, the Lucas-Liouville Polynomial Identity above can be concisely written as,
6 (x12+x22+x32+x42)2 = ∑ (xi ± xj)4
To determine the number of terms in the summation, if we are to choose 2 objects from 4, this gives 6 ways. (This can be calculated in Mathematica as Binomial[4, 2] = 6.) Since there are 2 sign changes, this gives a total of 6 x 2 = 12 terms on one side of the equation, which can be seen in the explicit identity. This in fact belongs to an infinite family where the addends involve the xi and 2 variables (xi ± xj) taken at a time:
6 (x12+x22)2 = ∑ (xi ± xj)4 + 4 ∑ xi4 6 (x12+x22+x32)2 = ∑ (xi ± xj)4 + 2 ∑ xi4 6 (x12+x22+x32+x42)2 = ∑ (xi ± xj)4 + 0 ∑ xi4 6 (x12+x22+ ... + x52)2 = ∑ (xi ± xj)4 - 2 ∑ xi4
and so on. For 3 variables taken at a time,
12 (x12+x22+x32)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi4 24 (x12+x22+x32+x42)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi4 36 (x12+x22+ ... + x52)2 = ∑ (xi ± xj ± xk)4 + 12 ∑ xi4 48 (x12+x22+ ... + x62)2 = ∑ (xi ± xj ± xk)4 + 8 ∑ xi4 60 (x12+x22+ ... + x72)2 = ∑ (xi ± xj ± xk)4 + 0 ∑ xi4 72 (x12+x22+ ... + x82)2 = ∑ (xi ± xj ± xk)4 - 12 ∑ xi4
Note how for seven squares, the second summation has a zero coefficient like for the Lucas-Liouville identity. For 4 variables taken at at time,
24 (x12+x22+x32+x42)2 = ∑ (xi ± xj ± xk ± xm)4 + 24 ∑ xi4 72 (x12+x22+ ... + x52)2 = ∑ (xi ± xj ± xk ± xm)4 + 40 ∑ xi4 144 (x12+x22+ ... + x62)2 = ∑ (xi ± xj ± xk ± xm)4 + 64 ∑ xi4 240 (x12+x22+ ... + x72)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi4 360 (x12+x22+ ... + x82)2 = ∑ (xi ± xj ± xk ± xm)4 + 80 ∑ xi4
I was not yet able to compute when the second summation has a zero coefficient but, as it seems to be reversing just like for 3 variables, I'm guessing it will be for 12 squares. (Anyone can check if this conjecture is true, as well as provide the identities for the next level of 5 variables taken at a time?) See the complete soln by Gerry Martens at the 9/30 update below. Update 9/29/09: I guessed wrong, it should be 10 squares, and the infinite family is given by:
6 (x12+x22+x32+x42)2 = ∑ (xa ± xb)4 60 (x12+x22+ ... + x72)2 = ∑ (xa ± xb ± xc)4 672 (x12+x22+ ... + x102)2 = ∑ (xa ± xb ± xc ± xd)4 7920 (x12+x22+ ... + x132)2 = ∑ (xa ± xb ± xc ± xd ± xe)4 96096 (x12+x22+ ... + x162)2 = ∑ (xa ± xb ± xc ± xd ± xe ± xf)4
and so on. My thanks to Daniel Lichtblau of Wolfram Research (makers of Mathematica), and Renzo Benedetti, Martin Rubey, and Gerry Martens from sci.math.symbolic who answered my questions. One can see that the number of squares is just in arithmetic progression and Benedetti observed that the numerical factor is generated by 2m-1 Binomial[3(m-1), m-1] where m is the number of objects taken at a time per term in the RHS. This sequence {6, 60, 672, 7920,...} is, as of this date, not yet in the Online Encyclopedia of Integer Sequences. (End update, 9/29.)
For a combination of 3 and 2 variables, these may be derived by combining the identities given previously and we get:
72 (x12+x22+ ...+ x52)2 = ∑ (xi ± xj ± xk)4 + 6 ∑ (xi ± xj)4 60 (x12+x22+ ...+ x62)2 = ∑ (xi ± xj ± xk)4 + 2 ∑ (xi ± xj)4
For seven squares, the second summation would have a zero coefficient. A general identity when the first summation involves choosing n objects from n squares is given by,
3*2n-1 (x12+x22+ ... + xn2)2 = ∑ (xa ± xb ± ... ± xn)4 + 2n ∑ xn4 (Id.1)
Going higher, for 3rd powers,
60 (x12+x22+x32+x42)3 = ∑ (xi ± xj ± xk)6 + 2 ∑ (xi ± xj)6 + 36 ∑ xi6 120 (x12+x22+x32+x42)3 = ∑ (xi ± xj ± xk ± xm)6 + 6 ∑ (xi ± xj)6 + ∑ (2xi)6 60 (x12+x22+x32+x42+x52)3 = ∑ (xi ± xj ± xk)6 + 36 ∑ xi6
For 4th powers,
5040 (x12+x22+x32+x42)4 = 6 ∑ (xi ± xj ± xk ± xm)8 + ∑ (2xi ± xj ± xk)8 + 60 ∑ (xi ± xj)8 + 6 ∑ (2xi)8
and so on for other k powers. (End update, 9/26.)
Update 9/30/09: Gerry Martens gave the complete solution to,
p (x12+x22+ ... + xn2)2 = ∑ (xa ± xb ± ... ± xm)4 + q ∑ xn4
where we choose m out of n objects. First, define r1 = Pochhammer[2-n, m-2] / (m-2)!, then,
p = (3/2)(-2)m r1 q = (1/2)(-2)m r1 (-n+3m-2)/(m-1)
Mathematica's Pochhammer[a, n] gives the rising factorial (a)n = a(a+1)...(a+n-1). Note that by choosing some constant m, we can set q = 0 by letting n = 3m-2, which then yields the infinite family starting with the Lucas-Liouville identity at m = 2. If m = n, this is the same as (Id.1).
For 6th powers, Renzo Benedetti solved,
p (x12+x22+ ... + xn2)3 = ∑ (xa ± xb ± ... ± xm)6 + q ∑ xn6
Define r2 = Binomial[3(m-2), m-2] and n = 3m-4, then,
p = (5/2) 2m r2 q = 2m r2 m/(m-1)
Note that, unlike for 4th powers, n depends on m and one also can't set q = 0 except for the trivial case m = 0. Starting with m = 2 to prevent division by zero, we get,
10 (x12+x22)3 = (x1+x2)6 + (x1-x2)6 + 8(x16 +x26)
then the identity involving 5 squares given above, and so on. (End update, 9/30.)
II. Boutin’s Identity
S ± (x1 ± x2 ±…± xk)k = k! 2k-1x1x2…xk
where the exterior sign is the product of the interior signs. (In other words, the term is negative if there is an odd number of negative interior signs; positive if even.) For the first few k:
(a+b)2 - (a-b)2 = 4ab
(a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3 = 24abc
(a+b+c+d)4 - (a-b+c+d)4 - (a+b-c+d)4 - (a+b+c-d)4 + (a-b-c+d)4 + (a-b+c-d)4 + (a+b-c-d)4 - (a-b-c-d)4 = 192abcd
and so on for other kth powers. (This generalizes the difference of two squares to a sum and difference of 2k-1 kth powers.)
(Update, 11/18/09): I realized that Boutin's Identity is behind some unusual but elegant identities as the ratio between exponents k and k+2 is simply a rational multiple of (x12+x22+...+xk2). For the first few k,
2[a2+b2][(a+b)2 - (a-b)2] = (a+b)4 - (a-b)4
10[a2+b2+c2][(a+b+c)3 - (a-b+c)3 - (a+b-c)3 + (a-b-c)3] = 3[(a+b+c)5 - (a-b+c)5 - (a+b-c)5 + (a-b-c)5]
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
and so on, with the last example tweaked a bit and also discussed in Section 8.4. (End update.) III. Lagrange’s Polynomial Identity
( ∑ ak2) ( ∑ bk2) = ( ∑ akbk)2 + ∑ (akbj-ajbk)2
where the first three summations are from k = 1 to n, while the fourth is 1≤k<j≤n. For n = 2,3,4 these are,
(a12+a22)(b12+b22) = (a1b1+a2b2)2 + (a1b2-a2b1)2
(a12+a22+a32)(b12+b22+b32) = (a1b1+a2b2+a3b3)2 + (a1b2-a2b1)2 + (a1b3-a3b1)2 + (a2b3-a3b2)2
(a12+a22+a32+a42)(b12+b22+b32+b42) = (a1b1+a2b2+a3b3+a4b4)2 + (a1b2-a2b1)2 + (a1b3-a3b1)2 + (a1b4-a4b1)2 + (a2b3-a3b2)2 + (a2b4-a4b2)2 +(a3b4-a4b3)2 and so on. Note that for n = 4, while it is the sum of seven squares by Lagrange's Identity, is just the sum of four squares by Euler's Four-Square Identity, and the case n = 3 shows that the product of two sums of three squares is likewise the sum of four squares. The following identities are good for Waring-like Problems since they involve rational, instead of integral, terms.
2. Form: a2+b2+c2+d2 = N
By Langrange's Four-Square Theorem, any positive integer N is the sum of four integral squares. If we extend this to rational numbers, then,
Theorem: "Any positive rational N is the sum of four rational squares in an infinite number of ways."
Proof: x12+x22+x32+x42 = (y12+y22+y32+y42)(z12+z22+z32+z42)2
where {x1, x2, x3, x4} = {uy1-vz1, uy2-vz2, uy3-vz3, uy4-vz4} and {u,v} = {z12+z22+z32+z42, 2(y1z1+y2z2+y3z3+y4z4)}
with initial soln yi and four arbitrary zi. Since there is always yi for any positive rational N, then by dividing the eqn with the square factor of the RHS, one can find an infinite number of xi. This is just a particular case of a general identity involving n sums of squares discussed more here.
Note: It is easily proven that the rational number p/q is the sum of four rational squares. Multiplying this by q/q, the denominator becomes a square and the problem is reduced to expressing pq as the sum of four squares. Also, one can't solve a2+b2+c2 = Ny2, where N = 8m+7. To see this, there are only two possibilities: either y = 2n+1 or 2n. If y is odd, expanding (8m+7)(2n+1)2, it can be seen this has the form 8p+7 and can't be expresssed as three squares. Likewise, (8m+7)(2n)2 = 4(8m+7)(n)2 = 4(8m+7)(2r+1)2 = 4(8q+7) hence can't be expressed as three squares either. Thus, Lagrange's Four-Square Theorem applies whether the variables are integral or rational.
3a. Form: x3+y3+z3 = N
Ryley's Theorem: “Any rational number N is the sum of three rational cubes in an infinite number of ways.” (S. Ryley, 1825)
Proof: (p3+qr)3 + (-p3+pr)3 + (-qr)3 = N (6Nvp2)3,
where {p,q,r}= {N2+3v3, N2-3v3, 36N2v3}, with v arbitrary.
Corollary: "Any positive rational N is the sum of three positive rational cubes in an infinite number of ways."
Since v is an arbitrary variable, one can choose it such that all the terms are positive. For small integral N, the ranges roughly are:
N = 1: v = {1.32 - 1.42}
N = 2: v = {1.20 - 1.40}
N = 3: v = {2.8 - 3.0}
N = 4: v = {3.4 - 3.7}
and so on. Q: Does anyone know how to calculate v and express it in terms of N such that all terms are positive?
(Update, 11/11/09): Laurent Bartholdi from sci.math.research gave the answer. Setting x = N2 and y = v3 for ease of notation, one gets the constraints,
x-3y < 0 x2-30xy+9y2 < 0 x3+45x2y-81xy2 +27y3 > 0
which define 6 lines in the {x,y} plane delimiting two regions. These regions are,
R1 = [a1 to a2]; where {a13, a23} = {(1/3)N2, u1N2}. R2 = [b1 to b2]; where {b13, b23} = {u2N2, w1N2}
and {u1, u2} = {0.768..., 2.2529...} are the positive roots of 27u3-81u2+45u+1 = 0, while w1 = 3.299... is the larger root of 9w2-30w+1 = 0. For ex., for N = 163, then approx R1 = [20.7 to 27.3] and R2 = [39.2 to 44.4], and one can choose an infinite number of rational v within those two ranges. (End update.)
3b. Form: (a3+b3)(c3+d3) = N
Theorem: “Any rational number N is the product of two sums of rational cubes."
Proof: ((1+18N-27N2)3 + (-1+18N+27N2)3) ((1+3N)3 + (1-3N)3) = 63N (1+27N2)3
where the RHS is then divided by its cubic factor.
Note: Who discovered this identity? Euler? Lebesgue? 4. Form: x4+y4-(z4+t4) = N
Norrie's Theorem: “Any rational number N is the sum and difference of four rational fourth powers in an infinite number of ways.” (R. Norrie)
((2a+b)c3d)4 + (2ac4-bd4)4 - (2ac4+bd4)4 - ((2a-b)c3d)4 = a(2bcd)4
where b = c8-d8, for arbitrary {c,d}.
5. Form: x15+x25+x35+x45+x55+x65 = N
(Update, 11/17/09): Choudhry's Fifth Powers Theorem: “Any rational number N is the sum of six rational fifth powers in an infinite number of non-trivial ways.”
Note: I don't have access yet to Choudhry's paper, "Representation of every rational number as an algebraic sum of fifth powers of rational numbers" so I came up with a method which may be what Choudhry used. As this update is after the one for Seventh Powers, this incorporates some insights from that section.
Proof (Piezas): Expand the system,
F(x): = (x+a)5+(x+b)5+(mx+mc)5-(x+d)5-(x+e)5-(mx+mf)5
and collecting powers of x, this resolves to,
F(x): = 5P1x4 + 10P2x3 + 10P3x2 + 5P4x + P5
where Pk = ak+bk+m5ck-dk-ek-m5fk, for k = {1,2,3,4,5}. The objective is to eliminate all terms of F(x) other than the linear term, or to find {a,b,c,d,e,f,m} such that,
ak+bk+m5ck = dk+ek+m5fk, for k = 1,2,3,5 (eq.1)
which is just an Equal Sums of Like Powers problem. Once found, since x is arbitrary, let x = 54(a4+b4+m5c4-d4-e4-m5f4)4 N. Thus,
F(x): = 55(a4+b4+m5c4-d4-e4-m5f4)5 N Dividing by the numerical factor, any rational N then is the sum of six rational fifth powers as claimed. The problem, of course, is if there are non-trivial values such that eq.1 can be solved. It turns out this system can be reduced to an elliptic curve, hence N in fact is the sum of six rational fifth powers in an infinite number of ways. To see this, let,
{a,b,c,d,e,f,m5} = {p+q, r+s, 1+t, -p+q, -r+s, -1+t, n}
where the symmetry of terms simplifies the system considerably. Then let,
{q, t, r} = {(-rs-nt)/p, (rs-h)/r, -(n+p)}
with s = (-5n2-4np+16p2+5n4+16n3p-4n2p2-60np3-60p4) / (60hn+120hp)
and the multi-grade system (eq.1) is solved if {n,p} satisfies the curve,
p(n+p)(n2+3np+3p2-1) = 3h2
Note that n must be the 5th power of a rational. One value I found is n = 25 = 32 which gives,
p(32+p)(1023+96p+3p2) = 3h2
an initial rational point is p = 50/9, from which others can then be calculated. This gives an initial set of 9-digit solns to eq.1 as,
{m5,a,b,c,d,e,f} = {32; -225,478,523; -172,632,729; 112,165,576; -291,996,273; -277,027,261; 100,192,381}
thus providing an algebraic identity proving Choudhry's Fifth Powers Theorem. (End proof.) The approach can be extended to higher odd powers, with a corresponding increase of the complexity of the system to be solved, but Choudhry found an elegant soln for Seventh Powers discussed in the next section.
Q: Can a theorem be found to reduce it to five rational fifth powers?
6. Form: u6+v6+w6-(x6+y6+z6) = N
(Update, 11/30/09): To prove that any N is the sum and difference of six 6th powers, analogous to Norrie's Theorem for 4th powers, one way is to use a similar method as in 5th powers. Expanding,
F(x): = (x+a)6+(mx+mb)6+(nx+nc)6-(x+d)6-(mx+me)6-(nx+nf)6 and collecting powers of x, this resolves to,
F(x): = 6P1x5 + 15P2x4 + 20P3x3 + 15P4x2 + 6P5x + P6
where Pk = ak+m6bk+n6ck-dk-m6ek-n6fk, for k = {1,2,3,4,5,6}. The objective is to eliminate all terms of F(x) other than the linear and constant term, or to find {a,b,c,d,e,f,m,n} such that,
ak+m6bk+n6ck = dk+m6ek+n6fk, for k = 1,2,3,4 (eq.1)
Thus, what will be left is,
F(x): = 6P5x + P6 = N
where x is then easily solved for any N. Do the substitution,
{a,b,c,d,e,f} = {p+s, r+s, q+s, -p+s, -q+s, -r+s}, where p = -(m6q+n6r)
and eq.1 is satisfied for arbitrary s if rational and non-trivial {m,n,q,r} can be found that satisfies the cubic,
m6(m12-1)q3 + 3m12n6q2r + 3m6n12qr2 + n6(n12-1)r3 = 0
though this author hasn't been able to found any. Even if {m,n,q,r} can't be found, another approach might prove (or disprove) the statement. (End update.)
(Update, 11/10/09): Choudhry's Seventh Powers Theorem: “Any rational number N is the sum of eight rational seventh powers in an infinite number of non-trivial ways.”
Proof: Expand the system,
F(x): = (x+a)7+(x-a)7+(mx+b)7+(mx-b)7-(x+c)7-(x-c)7-(mx+d)7-(mx-d)7
and collecting powers of x, this resolves to,
F(x): = 42(a2+m5b2-c2-m5d2)x5 + 70(a4+m3b4-c4-m3d4)x3 + 14(a6+mb6-c6-md6)x
To get rid of the x5 and x3 terms, find {a,b,c,d,m} such that,
a2+m5b2 = c2+m5d2
a4+m3b4 = c4+m3d4
and only the linear term is left. Since x is arbitrary, let x = 146(a6+mb6-c6-md6)6 N. Thus,
F(x): = 147(a6+mb6-c6-md6)7 N
Dividing by the numerical factor, N then is the sum of eight rational seventh powers as claimed. (End proof) Choudhry chose m = 2 and found {a,b,c,d} with 33-digits! I'm assuming there might be smaller solns, so I posted the problem in sci.math.symbolic to see if someone can find one, for any m > 1.
Note 1: It can be shown that the general system,
a2+mb2 = c2+md2 (eq.1) a4+nb4 = c4+nd4 (eq.2)
can be reduced to solving an elliptic curve. Euler’s complete soln to eq.1 (see Form 10 of Sums of Two Squares) is given by,
(pr+mqs)2 + m(ps-qr)2 = (pr-mqs)2 + m(ps+qr)2
Apply these values for {a,b,c,d} on eq.2, and it reduces to the eqn,
(mp2-nq2)r2 = (np2-m3q2)s2
which have rational solns if,
(mp2-nq2)(np2-m3q2) = y2
For the special case relevant to 7th powers when {m,n} = {h5, h3}, the curve is,
(h2p2-q2)(p2-h12q2) = y2 (eq.3)
For h = 2, Choudhry found the 16-and-17-digit soln {p,q} = {5911167604843137, 12317476831120126} which gives his 33-digit {a,b,c,d}. It might be interesting to know if eq.3 has small solns for some other rational h > 1.
Update (11/16/09) Note 2: D. Rusin gave a very thorough analysis here and, as a first step, simplified it to a problem of Equal Sums of Like Powers. By letting {b,d} = {Bm, Dm}, one is to solve,
ak+m7Bk = ck+m7Dk, for k = 2,4 (eq.4)
Then, by a series of clever transformations, he reduced this system to the rather simple elliptic curve,
V2 = U(U+1)(U+q2)
where q = (1-m7) / (1+m7) and using Maple's APECS and Magma, eventually gave an explicit soln {U,V}, apparently the smallest, and the same as what Choudhry found. From this initial point, an infinite more can be calculated proving that any rational N is the sum of eight rational 7th powers in an infinite number of ways. For other m such that eq.4 has non-trivial solns, because of its symmetry, it suffices to look within the interval m = {0 to 1}. In addition to m = 2 (or equivalently m = 1/2), Rusin found possible candidates m = {3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/7} though solns, if any, must be large. (End update.) Q: Can this be reduced to seven rational seventh powers?
8. Form: x111+x211+x311+x411+ ... +x1211 = N
Using Choudhry's and Rusin's approach for 7th powers, if there is a non-trivial soln {xi, yi} to the system S11,
ax1k+bx2k+cx3k = ay1k+by2k+cy3k, for k = 2,4,6,8
for some rational {a,b,c} = {p11, q11, r11}, then any rational N is the sum of 12 rational 11th powers. Multi-grade systems valid for k = {2,4,6,8} or even k = {2,4,6,8,10} are known (see the section on Tenth Powers) so it might be possible, with certain assumptions, that S11 can be reduced to an elliptic curve as well. As a numerical experiment, one may test this by using small {p,q,r} < 11, like Choudhry's m = 2, excluding the case p = q = r, and testing values {xi, yi} below some reasonable bound. In general, the conjecture would be that any rational N is the sum of at most 4m rational (4m-1)th powers (with Ryley's Theorem reducing the cubic case to just three cubes), with the problem reducible to a system of equal sums of like powers.
9. Form: u2x+x2y+y2z+z2u = 0
Q. In what other form can this be expressed, or in what context does it appear? It looks familiar.
1. C. Hermite
{u,x,y,z} = {a2c-b3, a2b-ac2, a3-b2c bc2-ab2}
2. S.Realis
{u,x,y,z} = {ab+b2+c, 3(ab-b2+c), ab-3b2-c 3(ab-b2-c)}, if c = a2-2ab+2b2
3. M. Weill
{u,x,y,z} = {c, a2d, -abc, ad}, where c = a3b-1, d = ab2+1
(Update, 8/7/09):
10. Equal Sums of Like Powers
Piezas
The results in this section are by the author.
I. Even Powers (Part 1)
Theorem 1: Define Ek:= ak+bk+ck-(dk+ek+fk). Let {u,v} = {a2+b2+c2, a4+b4+c4}. If Ek = 0 for k = 2,4 then,
a6+b6+c6-d6-e6-f6 = ± 3(a2-d2)(b2-d2)(c2-d2) 3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6) (a2+b2+c2) 6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6) ((a2+b2+c2)2+a4+b4+c4)
Note: The d in the RHS of the first eqn can be either {d,e,f}. And the ± sign depends on what side of the eq is chosen to be {a,b,c} or {d,e,f}. Small soln: [6, 23, 25]k = [10, 19, 27]k, for k = 2,4.
Corollary: These relations can be woven together. Define n as a4+b4+c4 = n(a2+b2+c2)2 then,
32(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 15(n+1)(a8+b8+c8-d8-e8-f8)2
Proof: We can concisely express the last two of the eqns above as,
3 E8 = 4 E6(u) (eq.1) 6 E10 = 5 E6(u2+v) (eq.2)
Square eq.1, multiply eq.2 by E6, and then eliminate E62 between eq.1 and eq.2 to get,
32 E6E10 = 15 E82 (1+v/u2), or, 32 E6E10 = 15 E82 (1+n)
as stated. (End of proof). For the special case when a+b = ± c, since it is true that,
a4+b4+(a+b)4 = (1/2)(a2+b2+(a+b)2)2
then n = 1/2 and the above becomes,
64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2
which is a concise version of the Ramanujan 6-10-8 Identity. The general case is in fact the first in an infinite family to be given in the following theorems.
Theorem 2: Define Fk:= ak+bk+ck+dk-(ek+fk+gk+hk), and qk:= ak+bk+ck+dk. Let {u,v,w} = {q2, q4, q6}. If Fk = 0 for k = 2,4,6 then,
F8 = ± 4(a2-e2)(b2-e2)(c2-e2)(d2-e2) 4 F10 = 5 F8 (u) 8 F12 = 6 F8 (u2+v) 24 F14 = 7 F8 (u3+3uv+2w)
Corollary: 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
or, equivalently, 25 F8F12 = 12(n+1)(F10)2. Small soln: [2, 16, 21, 25]k = [5, 14, 23, 24]k, k = 2,4,6.
Theorem 3: Define Gk:= ak+bk+ck+dk+ek-(fk+gk+hk+ik+jk) and rk:= ak+bk+ck+dk+ek. Let {u,v,w,x} = {r2, r4, r6, r8}. If Gk = 0 for k = 2,4,6,8 then,
G10 = ± 5(a2-f2)(b2-f2)(c2-f2)(d2-f2)(e2-f2) 5 G12 = 6 G10 (u) 10 G14 = 7 G10 (u2+v) 30 G16 = 8 G10 (u3+3uv+2w) 120 G18 = 9 G10 (u4+6u2v+3v2+8uw+6x)
Corollary: "Define n as a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2, then 72 G10G14 = 35(n+1)(G12)2." Small soln: [71, 131, 180, 307, 308]k = [99, 100, 188, 301, 313]k, k = 2,4,6,8.
Theorem 4: Define Hk:= ak+bk+ck+dk+ek+fk-(gk+hk+ik+jk+lk+mk) and sk:= ak+bk+ck+dk+ek+fk. Let {u,v,w,x,y} = {s2, s4, s6, s8, s10}. If Hk = 0 for k = 2,4,6,8,10, then,
H12 = ± 6(a2-g2)(b2-g2)(c2-g2)(d2-g2)(e2-g2)(f2-g2) 6 H14 = 7 H12 (u) 12 H16 = 8 H12 (u2+v) 36 H18 = 9 H12 (u3+3uv+2w) 144 H20 = 10 H12 (u4+6u2v+3v2+8uw+6x) 720 H22 = 11 H12 (Poly1), where Poly1 is a 5th deg polynomial in terms of u,v,w,x,y.
Corollary: "Define n as a4+b4+c4+d4+e4+f4 = n(a2+b2+c2+d2+e2+f2)2, then 49 H12H16 = 24(n+1)(H14)2." Small soln: [22, 61, 86, 127, 140, 151]k = [35, 47, 94, 121, 146, 148]k, k = 2,4,6,8,10.
And so on for other k = 2,4,..2n. While an infinite number of solns are known for Theorems 1 to 4, none are known for higher systems. (Q: Anyone with the programming skills and computing power to find a few?)
Note 1: The coefficient of Hi as {6, 12, 36, 144, 720} divided by 6 is equal to {1, 2, 6, 24, 120}, or the sequence of factorial n!, so it is easy to extrapolate that the next system involves the sequence 7(n!). Note 2: As an overview, we have the sequence,
32 E6E10 = 15(n+1)(E8)2 50 F8F12 = 24(n+1)(F10)2 72 G10G14 = 35(n+1)(G12)2 98 H12H16 = 48(n+1)(H14)2
This has the general form,
2m2 X2(m-1) X2(m+1) = (m2-1)(n+1)(X2m)2
starting with m = 4,5,6… so one can see the next in the series. However, by taking square roots of the variables, these theorems are also valid for systems with k = 1,2,3,…n, or those satisfying the Prouhet-Tarry-Escott problem. See also the section on Equal Sums of Like Powers.
II. Even Powers (Part 2)
Theorem 1: Define {a,b,c,d} = {m+p, m-p, m+q, m-q}. If p2+q2 = m2, then 9(a2+b2-c2-d2)(a4+b4-c4-d4) = 7(a3+b3-c3-d3)2.
Theorem 2: Define,
{a,b,c,d} = {m+p, m-p, m+q, m-q} {e,f,g,h} = {m+r, m-r, m+s, m-s}
If,
p2+q2 = r2+s2 (eq.1) p2+q2 = 4m2 (eq.2)
then 25(a4+b4+c4+d4-e4-f4-g4-h4)(a6+b6+c6+d6-e6-f6-g6-h6) = 21(a5+b5+c5+d5-e5-f5-g5-h5)2.
Also, for any m, then ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,2,3. Hence it is a symmetric ideal solution (see Theorem 1 and 2) of the Prouhet-Tarry-Escott problem. Note that eq.1 can be completely solved as {p,q,r,s} = {ux+vy, vx-uy, ux-vy, vx+uy} and substituting this into eq.2 yields,
(u2+v2)(x2+y2) = 4m2
which can be solved by using the formula for Pythagorean triples separately on {u,v} and {x,y}.
Theorem 3: Define,
{a,b,c,d,e,f} = {m+p, m-p, m+q, m-q, m+r, m-r} {g,h,i,j,k,l} = {m+s, m-s, m+t, m-t, m+u, m-u}
If,
pk+qk+rk = sk+tk+uk, for k = 2,4 (eq.1) p2+q2+r2 = vm2 (eq.2)
then, 49(a6+b6+c6+d6+e6+f6-g6-h6-i6-j6-k6-l6)(a8+b8+c8+d8+e8+f8-g8-h8-i8-j8-k8-l8) = (4/3)(v+21)(a7+b7+c7+d7+e7+f7-g7-h7-i7-j7-k7-l7)2.
And, for any m, an+bn+cn+dn+en+fn = gn+hn+in+jn+kn+ln, for n = 1,2,3,4,5. When v = 1, only two primitive solns are known with terms < 16,000,
{p,q,r,s,t,u} = {2, 289, 610, 170, 223, 614} (by "Martin") {p,q,r,s,t,u} = {1575, 11522, 11850, 4647, 8890, 13230} (by James Waldby)
after this author posted the problem on the newsgroup sci.math.symbolic. However, when v = 2, there are small solns starting with {p,q,r,s,t,u} = {3,5,8,0,7,7} and it can be shown there are an infinite number. Let {p,q,r,s,t,u} = {z1+z2, -z1+z2, 2z2, z3+z4, -z3+z4, 2z4} and eq.1 and eq.2 become,
z12+3z22 = z32+3z42 (eq.1) 2(z12+3z22) = 2m2 (eq.2)
Eq. 1 can be completely solved as {z1, z2, z3, z4} = {ux+3vy, vx-uy, ux-3vy, vx+uy} and eq.2 becomes,
(u2+3v2)(x2+3y2) = m2
and it is easy to solve this using a variant of the formula for Pythagorean triples separately on {u,v} and {x,y}. The procedure can be extrapolated to higher systems such as,
pk+qk+rk+sk = tk+uk+vk+wk, for k = 2,4,6 (eq.1) p2+q2+r2+s2 = ym2 (eq.2)
which, for y = 1, then solves 81C8C10 = 65C92 where the Ci are analogous 16-term expressions to the 12-term ones above, though it is unknown if both eq.1 and eq.2 (for y = 1) will have non-trivial rational solns.
III. Odd Powers
The difficulty with the system,
a1k + a2k + … + amk = b1k + b2k + … + bmk, k = 1,3,5,…2n+1 (eq.1)
valid only for odd powers is that terms can be transposed or moved around to the form,
c1k + c2k + … + cpk = 0, k = 1,3,5,…2n+1
and it is more challenging to find relations beyond k = 2n+1. The trick then is to “fix the variables in place” by finding a balanced partition such that,
a) a1+a2+… +am = b1+b2+… +bm = 0, or b) eq.1 is valid for k = 2 as well
and hence terms can no longer be transposed arbitrarily. (For the system k = 1,3,5, the two conditions are equivalent.)
a) Condition: a1+a2+… +am = b1+b2+… +bm = 0.
Third powers: “If ak+bk+ck = dk+ek+fk, for k = 1,3, where a+b+c = d+e+f = 0, then 9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9).”
Fifth powers: “If ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,3,5, where a+b+c+d = e+f+g+h = 0, (or equivalently, k = 1,2,3,5), 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).”
Note: As these two systems have complete solns, given here (Form 13) and here (Form 5.3b), one can substitute those into the given equations and see that they are true. For higher than fifth powers, there are only isolated results with this condition so it harder to come up with general statements.
b) Condition: Valid for k = 2 as well.
As Bastien’s Theorem forbids a non-trivial result for the system k = 1,2,3 with only three terms on either side of the eqn, this condition yields a “general” result valid only for those with higher than third powers, starting with,
Define Fk: = x1k+x2k+x3k+x4k – (y1k+y2k+y3k+y4k). If Fk = 0 for k = 1,2,3,5, then (9F7)(x12+x22+x32+x42) = 7F9.
Define Fk: = x1k+x2k+x3k+x4k+x5k – (y1k+y2k+y3k+y4k+y5k). If Fk = 0 for k = 1,2,3,5,7, then (11F9)(x12+x22+x32+x42+x52) = 9F11.
Define Fk: = x1k+x2k+x3k+x4k+x5k+x6k – (y1k+y2k+y3k+y4k+y5k+y6k. If Fk = 0 for k = 1,2,3,5,7,9, then (13F11)(x12+x22+x32+x42+x52+x62) = 11F13.
Note: All known solns satisfy this and the form is easily extrapolated for higher systems. However, it is proven only for fifth powers and a general proof is needed. |