For k = 7, the first soln to,
x
was found in 1996 by Ekl as [10, 14, 123, 149] = [15, 90, 129, 146]. In 1999, Nuutti Kuosa found [184, 443, 556, 698] = [230, 353, 625, 673] for k = 1,3,7 though it seems he didn't realize it was multigrade. (As it turns out, neither does it belong to the family given by Choudhry below.)
Update (6/26/09): Special forms:
a) x
_{1}^{7} + x_{2}^{7} + x_{3}^{7} + x_{4}^{7} + x_{5}^{7} + x_{6}^{7} + x_{7}^{7} = x_{8}^{7}Only
three solns are known to the case of 7 positive 7th powers equal to a 7th power, the first one found by Mark Dodrill in 1999:127
^{7} + 258^{7} + 266^{7} + 413^{7} + 430^{7} + 439^{7} + 525^{7} = 568^{7} b) x _{1}^{7} + x_{2}^{7} + x_{3}^{7} + x_{4}^{7} + x_{5}^{7} + x_{6}^{7} + x_{7}^{7} = 1Just like three 3rd powers and five 5th powers, when
signed, can sum to 1, so can seven 7th powers, two of which found by Kuosa,[1, 1146, 1348, 2816] = [130, 1031, 1951, 2787]
[1, 1140, 1823, 3189] = [485, 621, 1859, 3188] (
End note)c) x _{1}^{7} + x_{2}^{7} + x_{3}^{7} + x_{4}^{7} + x_{5}^{7} + x_{6}^{7} + x_{7}^{7} = 0Still unknown. (Though since five 5th powers can sum to both zero and 1, it may be a general property of odd n nth powers for n > 3.)A.J. Choudhry
Choudhry found a multi-variable cubic polynomial such that when it has a rational root, it yields a k = 1,3,7. While no polynomial soln is yet known, it can be shown that if x
_{i} and y_{i} are complex numbers with integer components, then there are families of polynomial solns (to be given later).Given, x
_{1}^{k} + x_{2}^{k} + x_{3}^{k} + x_{4}^{k} = y_{1}^{k} + y_{2}^{k} + y_{3}^{k} + y_{4}^{k}, (eq.1)
{x {y a form which gives x
_{1}+x_{2}+x_{3}+x_{4} = y_{1}+y_{2}+y_{3}+y_{4} = 0. Note that the x_{i} (or y_{i}) also satisfies, 5(x
_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2})(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}) = 6(x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5})Theorem 1: Eq. 1 is true for k = 1,3,7 if it satisfies the two conditions, abc = def
2(a
This is the
same form used by Chernick for ideal solns of k = 1,2,3 and Lander for fifth powers k = 1,3,5, though as one can see the condition is a little more complicated. For the special case d+e = f, the system simplifies to satisfying the two eqns,
abc = de(d+e)
2(a
^{4}+b^{4}+c^{4}) - 5(a^{2}+b^{2}+c^{2})^{2} = 16(d^{2}+de+e^{2})^{2}
though whether this has non-trivial solns in the rationals is not known. If there is,
then one of the terms x so would be the first counter-example to Euler’s conjecture for seventh powers. However, if this constraint is not given, after a clever substitution which reduced the problem to finding a rational root of a cubic, Choudhry found several solns to Theorem 1, one is {a,b,c,d,e,f} = {324, 5439, 893, 5217, 1539, 196} which gives,_{i} or y_{i} vanishes[3328, 2111, -3004, -2435] = [3476, 1741, -3280, -1937]
which is for k = 1,3,7 as well as satisfying x
_{1}+x_{2}+x_{3}+x_{4} = y_{1}+y_{2}+y_{3}+y_{4} = 0, and 5(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2})(z_{1}^{3}+z_{2}^{3}+z_{3}^{3}+z_{4}^{3}) = 6(z_{1}^{5}+z_{2}^{5}+z_{3}^{5}+z_{4}^{5}) where z_{i} = x_{i}, (or z_{i} = y_{i}). Note 1: Kuosa's 1999 soln for k = 1,3,7 does not belong to this family as there is no transposition of four appropriate x such that their sum is equal to zero._{i}Note 2: Theorem 1 can also be satisfied by a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,4 where abc = def. But, using resultants, it is easily proven that this restricted case has only trivial solns.Piezas
Express eq.1 in the form F
(a+bh)
To satisfy the side condition x
d = -(e
Substituting this into k = 7 we get a quartic in
h with only even exponents, call this H_{4}, with coefficients in {a,c,e,g} as,
where {v From Choudhry's result, we know H _{4} can have non-trivial rational roots, one of which is h = 98 when {a,c,e,g} = {3166, 2273, 3378, 1839}. To find a polynomial identity, as was pointed out in the previous section, there are two ways to factor quartics with only even exponents, one of which is to make its discriminant a square. One soln is a+c+e+g = 0. Let {a,c,e,g} = {p+q+r, p-q-r, -p+q-r, -p-q+r} for symmetry, and H_{4} has the non-trivial quadratic factor,
3h
^{2}(q^{2}+r^{2}) + (5p^{2}+3q^{2}+3r^{2})(q+r)^{2} = 0
Unfortunately, this can only be solved by complex values. If we let h = h
i where i is the imaginary unit, then we are to find,3(5p
^{2}+3q^{2}+3r^{2})(q^{2}+r^{2}) = y^{2}As it is only a quadratic in
p with a square constant term, this is easily done. In summary, for the moment, only particular values (extrapolated from Choudhry's cubic) are known such that H_{4} has a non-trivial rational h. If an appropriate general relation can be found between {a,c,e,g} such that H_{4} has a quadratic factor with real roots, then it may be possible to find an infinite family of polynomials for k = 1,3,7, just as there is one for k = 1,5.
If x
x
These lead to ideal solns of deg 8. Interestingly, terms can be transposed such that x
[0, 34, 58, 82, 98] = [13, 16, 69, 75, 99] [0, 63, 119, 161, 169] = [8, 50, 132, 148, 174]
(
[0, -13, -16, -69, 98] = [75, 99, -34, -58, -82]
The second one has more "structure" as it can be re-arranged in four distinct ways,
[0, 63, 119, -8, -174] = [-161, -169, 148, 50, 132] [0, 63, 119, -50, -132] = [-161, -169, 148, 8, 174]
since 8+174 = 50+132, and,
[63, -50, -174, 161, 0] = [-119, 132, 148, -169, 8] [-119, 132, 148, -161, 0] = [63, -50, -174, 169, -8]
since 169-8 = 161-0. As the left hand side involves only four terms, this (and only this side) also obeys,
5(x
a general identity first mentioned in k = 1,3,5.
[-99, -13, 0, 34, 98] = [-82, -58, 16, 69, 75]
Shuwen found this is good for one
Piezas
Conjecture: "Define F_{k}:= x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}-(y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}) where x_{1}+x_{2}+x_{3}+x_{4} = y_{1}+y_{2}+y_{3}+y_{4}+y_{5} = 0. If F_{k} = 0 for k = 1,3,5,7, then F_{2}^{2} = -2F_{4}."
This works for the two known solns. I don’t know whether this is true in general for systems of this form or is just a peculiarity of Letac’s method.
The next step is S,_{8}
x
for k = 1,3,5,7, or k = 2,4,6,8. It can be proven by computer search that the
smallest solution to the first system, at least in distinct integers, is given by,[3, 19, 37, 51, 53] = [9, 11, 43, 45, 55]
This turns to have already been known (pre-computer era) by Sinha as belonging to an infinite family. However, there are just a
few methods known to parametrically solve S (at least previously) and just _{7}one for S, which is deplorable. With the advent of computer algebra systems, the Internet, and hopefully more people working on the field of equal sums of like powers (ESLP), perhaps this can be rectified eventually. Ironically, compared to S_{8}_{6}, it seems solns to S are rather plentiful since while only _{7}one binary quadratic form identity (Chernick’s) is known for the former, several are known for the latter. (Then again, it could be merely a lack of knowledge about S_{6}.) For example, one soln found by this author is the beautifully simple, ,Sagan's Identity
1 + 5
for k = 1,3,5,7 if x
^{2}-10y^{2} = 9. Note that this also satisfies x
_{1}+x_{2} = x_{3}+x_{4} = -(x_{5}+x_{6}) as well as some other obvious ones. While the complete soln uses rational x,y, if the condition is treated as a Pell eqn, then there are integral ones giving an infinite number of solns where two terms {1, 5} are constants. Update, 11/7/09: Last year, I dedicated an algebraic identity I found to the astronomer and science writer Carl Sagan (1934-1996). I read his "
Broca's Brain" and "The Dragons of Eden" in my late teens, and read and eventually saw "Contact". Since I already dedicated one article I wrote on Degen's Eight-Square Identity to the novelist Katherine Neville, author of the amazing book "The Eight", I thought it was only fitting I name one of the identities I found after Sagan. After all, it has billions and billions of solutions. (You need to be a Sagan fan to understand the previous remark.) Today is the first Carl Sagan Day, and I decided to revive the article I wrote about Sagan's Identity that perished when Geocities was closed by Yahoo last Oct. 26. (End update.)Another is,
(3x+2y) (-3x+2y)
for k = 1,2,3,5,7 if x
^{2}+6y^{2} = 1 which has the distinction of being valid for k = 2 as well. Finally, a few others have the condition x^{2}-70y^{2} = 1. All known parametrizations for k = 1,3,5,7 in terms of binary quadratic forms involve the unsigned discriminants D = 6, 10, 70. In short, the set D only employs the small prime factors p = 2,3,5,7. Whether this is coincidence I do not know. (Update, 1/25/10): Piezas
Previously, it was seen how a k = 1,3,5 can lead to a k = 2,4,6. This can be brought higher. Given a (k.4.4) for k = 2,4,6, if its terms {x
(3a+2b+c)
which leads to the higher system,
[3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d, -2a-b-c+d] [3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d, -2a-b+c+d]
where, for both, 10a It can be shown that given the system k = 1,3,5,7,
x
if it has either of the constraints,
1) x 2) x
(and equivalent forms after transposition such as the one above), then an
infinite sequence of polynomial solns can be found by finding non-trivial rational points on a quartic polynomial that is to be made a square.I. Constraint: x (where x_{1}-x_{2} = x_{3}-x_{4} = y_{1}-y_{2}_{1}+x_{2}+x_{3}+x_{4}+x_{5} = y_{1}+y_{2}+y_{3}+y_{4}+y_{5})Here is a beautiful theorem by T. Sinha. (See update in next page, bottom section.) If,
(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} + (a+2c)^{k} + b^{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. Note that this also satisfies,
x
_{1}-x_{2} = x_{3}-x_{4} = y_{1}-y_{2} = -2c, 2(x
_{1}+x_{3}) = y_{1}+y_{2} x
_{5} = y_{3}+y_{4}-(1/3)y_{5}
a set of conditions which is enough to give the complete soln. An example is,
{a,b,c,d,e} = {43, 9, 1, 19, 37}
yielding,
[43, 45, 9, 11, 55] = [51, 53, 19, 37, 3]
which is the smallest solution mentioned previously. Sinha’s problem then is to solve,
p
for k = 2,4 where p
1. (5x+2y+z)
where x 2. (-3x+2y+z)
^{k} + (-3x+2y-z)^{k} + (2x-4y)^{k} = (8x-y)^{k} + (2x+3y)^{k} + (14x-2y)^{k}
where 121x
3. (6x+3y)
where x
4. (2x+3y)
where 49x
where D is the discriminant of the binary quadratic form. Furthermore, treated as rational points on a certain elliptic curve, it can be showed that an infinite sequence can be derived from these thus providing polynomial solns of increasing degree. (See discussion on
(2p+q+r) (-2p-q+r)
for k = 1,3,5,7, if p
[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471], for k = 1,2,4,6,8
but whether this has a parametric soln is unknown. II. Constraint: x (where x_{1}+x_{2} = x_{3}+x_{4} = y_{1}+y_{2}_{1}+x_{2}+x_{3}+x_{4}+x_{5} = y_{1}+y_{2}+y_{3}+y_{4}+y_{5})An example is one by Gloden,
[9, 45, 11, 43, 55] = [3, 51, 19, 37, 53]
based on an identity given in the next section and simple transposition of terms,
[-51, 9, 11, 55, -53] = [3, -45, -43, 19, 37]
gives the equivalent rule,
x
useful for the case k = 1,2,3,5,7, like this numerical example by Shuwen,
{-71, 143, -17, -163, 121} = {103, -31, 157, -47, -169}
One can use this to find a polynomial identity. This eqn has a lot of structure. Expressed in terms of x
x
_{1}-y_{1} = -(x_{2}-y_{2}) = x_{3}-y_{3} x
_{3}^{2}+x_{4}^{2} = y_{3}^{2}+y_{4}^{2}
the first of which, after negating, is the same as Gloden’s. Using these constraints, we find the identity given earlier,
(3x+2y) (-3x+2y)
for k = 1,2,3,5,7, if x |