Piezas
We can also consider the more general system,
na+b+c = nd+e+f (eq.1) a
with k = 2,6 for F that gave the complete soln for the case k = 2,4 is also useful here,_{1}
(p+qu)
hence {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} and we will solve the system,
a+b-c = d+e-f na+b+c = nd+e+f a
giving {t,s,q} = {-pq-rs, -1-nq, 2/(1-n)}. Using these, F
(Poly1)u
where Poly1 is linear in {p,r} and Poly2 is cubic given by,
Poly1:= (-3+n)(5-2n+n Poly2:= (-3+n)(5-2n+n
Thus, again, one should solve,
y
a situation similar to the previous section. Trivial solns should be avoided such as y = Poly1 = 0 or r = -p(n-3)/(2n) with the latter, using Fermat’s method to compute a second rational point, just yielding the former. (Note that the constraint excludes n = 3, though this in fact is solvable using another method.) There
y
and with {p,r} = {-104, 17} this gives,
(-127)
From this initial soln, subsequent ones can then be found proving that the system,
a+b-c = d+e-f 12a+b+c = 12d+e+fa
^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}
for k = 2,6 has an infinite number of solns. Later, it will be shown that this approach can be
ma+b+c+d = me+f+g+h a
for
{x {y
discussed previously which is already true for k = 1,2, a linear combination of terms involving
nx
where n = ( Note 4: If there is an initial soln for a certain n, is there always an infinite number for this n?
-a+b+c = -d+e+f a+b+nc = d+e+nf a
giving,
{t,s,q} = {-pq-rs, -n-q, (1-n)/2} and F
(Poly11)u
though now defined diffently,
Poly11:= (-3+n)(5-2n+n Poly12:= (-3+n)(5-2n+n
As before, solve y
y
with non-trivial soln {p,r} = {-39, 17} which has the same
Notice that the value for
D
_{1} = (5-2n+n^{2})(5+2n+n^{2}) D
_{2} = (5-2n+n^{2})(1+n^{2})
respectively. Do the two methods then for the same
Q: Any other forms for equal sums of sixth powers with six terms?
It is unknown if the unbalanced multi-grade eqn,
x
_{1}^{k}+x_{2}^{k}+x_{3}^{k} = x_{4}^{k}+x_{5}^{k}+x_{6}^{k}+x_{7}^{k}, for k = 2,4,6 has non-trivial solns.
Q: Anyone can provide one, preferably parametric?(Update, 2/22/10): In response to a post in
sci.math, Giovanni Resta made a search and found that there are no solns with terms < 845. James Waldby would later extend the search radius to < 1280 with still no soln. However, there are a few that are good only for k = 2,6, namely,[73, 58, 41] = [70, 65, 32, 15]
Some exhibit interesting side relations between its terms, such as the smallest one with x
Some general identities analogous to the quadratic-quartic system Q
Define {x,y,z} = {a
a
4(a
8(a
24(a
The first one is true regardless of whether
a
then,
25(a
This is analogous to Ramanujan’s 6-10-8 Identity though since Ramanujan was able to use a linear relationship between the terms a
a
does have a parametric soln for n = ½, given in the section on
Chernick (and Escott, independently)
There is the aesthetically-pleasing identity,
(-u+7w) (u+7w)
for k = 1,2,4,6, if u
with signs changed by this author so it would be valid for k = 1. This was derived using the equally simple quartic identity,
u
for k = 2,4 with the same condition. Since the eqn is homogeneous, one can do the substitution {u,v,w} = {y+z, y-z, x} and the condition becomes,
7x
where the binary quadratic form has square-free discriminant
[u, v, u+v, -u, -v, -u-v] = [w, 2w, 3w, -w, -2w, -3w], k = 1,2,..5
If the Tarry-Escott theorem is then used, generally this doubles the number of terms. But for this particular instance, using T = w, the number of terms on either side of the equation instead of doubling, increases only up to eight but the range is now k = 1,2,3,4,5,6. A
Using a similar method, Tarry earlier found the first ideal soln (symmetric) for k =1,2,..7. Starting with [1, 9] = [4, 6] which is only for k = 1, he used
[1, 5, 10, 16, 27, 28, 38, 39] = [2, 3, 13, 14, 25, 31, 36, 40], k = 1,2,…6
Note that this is a symmetric soln with the common sum
[1, 5, 10, 24, 28, 42, 47, 51] = [2, 3, 12, 21, 31, 40, 49, 50], k = 1,2,…7
It turns out this can be generalized with solns depending on the nice eqn x
Piezas
To simplify matters we will start with the sixth deg. Rearranging terms, this is
[5, 16, 27, 38, 28, 39, 1, 10] = [3, 14, 25, 36, 2, 13, 40, 31], k = 1,2,…6
which makes it easy to see that the first six terms (with a skip after the fourth) of each side involve an arithmetic progression, differing by 11. This is then an example of a “special” property. To recall, by Frolov’s theorem we can always set a
[0, a, 2a, 3a, a+c, 2a+c, x
with unknowns {a,b,c,x
[0, a+c, x
now for k = 1,2,…7. But since we wish this to be symmetric, pairs must have a common sum,
0 + (4a+b) = (a+c) + (3a+b-c) = (x b + (4a) = (a+b-c) + (3a+c) = (y
where the first two pairs have the sum
(t+u) + (a+(t-u)) = 4a+b
so t = (3a+b)/2. We have reduced our unknowns to just five: {a,b,c,u,v}. (In fact, it can be assumed c = 1 without loss of generality.) Using Theorem 5 in reverse, if we subtract (4a+b)/2 from the first four terms on each side, namely {0, a+c, x
(4a-b)
for k = 2,4,6. To simplify, let {a,b,c} = {2p, -2q, -q+r} and rearranging terms, we get,
(p+u)
Using this simpler system, we can then solve for the unknowns. This is given by,
{p,q,r,u,v} = {2x
where x,y must satisfy,
y
with integral solns
x
with initial soln {x,y,z}= {1,1,1} and is easily solved parametrically. By recovering {u,v} there is a second quadratic condition to fulfill hence the elliptic curve. As was alluded to earlier, this symmetric sixth degree implied a fifth degree identity. Thus, in summary, if,
(3p+q)
(p-u)
^{k} + (p-v)^{k} + (2p-r)^{k} + (-4p+q)^{k} = (-p-u)^{k} + (-p-v)^{k} + (-2p-r)^{k} + (4p+q)^{k}, for k = 1,2,4,6. Note that the second system satisfies x
_{1}-y_{1} = x_{2}-y_{2} = (x_{3}-y_{3})/2 = -(x_{4}-y_{4})/4 and a minor change in signs made it valid for k = 1 as well. From this one can find an ideal seventh degree soln such as the beautifully concise,
q (-q)
for k = 1,2,…7 where {p,q,r,u,v} are as defined above. (Update, 8/4/09): Alternatively, the system can be expressed in a more symmetric form,
(a+b+c+d)
^{k} + (a+b+c-d)^{k} + (-a+b+2c)^{k} + (b-4c)^{k} = (a+b-c+d)^{k} + (a+b-c-d)^{k} + (-a+b-2c)^{k} + (b+4c)^{k}, for k = 1,2,4,6, where {a,b,c,d} satisfy a
(an+p)
for some constants
x
and an inspection reveals this is just the Chernick-Escott identity in disguise. I am not aware of another
a,b such that the form has a non-trivial soln though there may be more. The next example also involves a quartic elliptic curve. |