(Update, 1/25/10): Piezas
In the previous section, it was shown 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
[pa+qb+c, pa+qb-c, ra+sb+c, ra+sb-c, ta+wb] [ua+vb+2c, ua+vb, ua+vb-2c, xa+yb+d, xa+yb-d]
where m
Crussol
Given the system of eqns for indefinite
x
Chernick used certain expressions to solve the case k = 1,2,3 while Lander focused on k = 1,3,5, (which Choudhry showed could be extended to k = 1,3,7). But prior to this, Crussol took the case k = 1,2,4,6 and used the version,
(a+x)
which in one sense is a simpler form and gave a
x 5(a 3(a
^{2}-b^{2}) = y^{2}-z^{2}The terms of eq.0, expressed as {
x
x
y To solve the first condition, Crussol used {x,y,z,t} = {de+cf, ce-df, de-cf, ce+df} which reduced the other two to,
5(a 3(a
Note that eq.1 is of the form m(a
{a,b,c,d} = {fu-ev, eu+fv, 2u+v, -u+2v}
for free variables {e,f,u,v}. Using these expressions on (eq.2) results in a quadratic in
{e,f} = {3u
where {u,v,w} satisfy the quartic elliptic curve,
(3u
two small solns of which are {u,v} = {5,3} or {9,16}, etc.
(a+x)
where now it has x
31
which has 31+126 = 38+119, 62+107 = 51+118 and apparently is a particular instance of a parametric identity though I haven’t been able to derive it yet. (
Note 3: Using a different approach, the complete soln to k = 1,2,4,6 with x_{1}+x_{2 }= y_{1}+y_{2}, x_{3}+x_{4 }= y_{3}+y_{4} will be discussed in the subsequent sections.(Update, 1/18/10): Piezas
Crussol’s soln can be given in a more aesthetic manner as two quadratics to be made squares. Let,
[c+be+af, c-be-af, d+ae-bf, d-ae+bf] [c+be-af, c-be+af, d+ae+bf, d-ae-bf]
for k = 1,2,4,6, where,
ma na
with {m,n} = {(4e
(a+13b+c) (-a+13b+c)
where 45a
{4m+n, m+4n} = {e
hence, if the two expressions themselves are squares, then it can be used for a k = 1,2,4,6. (
(Update, 2/1/10) This update covers a family that includes 6th, 8th, and 10th power multi-grades. Whether it can be extended for 12th powers with a minimal number of terms remains to be seen.
Piezas (
Given the general form,
[xy+ax+3by-c, xy-ax-3by-c, xy-3bx+ay+c, xy+3bx-ay+c] [xy+ay+3bx-c, xy-ay-3bx-c, xy-3by+ax+c, xy+3by-ax+c]
Notice how {x,y} just swap places or alternatively, “
p p p
We can make this valid for k = 1,2,4,6. For k = 4, let
10a
This is an elliptic “curve” in disguise. Let {x,y} = {u, v/(10u
(a
which is a quartic polynomial in
u = 9b(a
from which one can then successively compute an infinite sequence of rational polynomials.
Chris Smyth (
This is a variation of the Letac-Sinha 8th power multi-grade. For k = 1,2,4,6,8,
[xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy] [xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]
where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn,
x
One can note its affinity with the 6th power multi-grade and see also in the general form how “
p p
If we set that,
p
then one must have 2a
p p
for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity. Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m
x
Let {x,y} = {u, v/(u
(u
Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7). A possible addition, given in italics, may be,
[xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, [xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy,
which is the same pair successfully added to the (k.4.4). For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.
Choudhry, Wroblewski (
The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms,
[-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), [-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d),
where again, {x,y} merely swap places. (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.) This obeys the same basic constraints (up to sign changes),
p p (-p p
If we assume,
p p
then it must be the case that 2(b-c) = e-f (eq.1). Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns,
{a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5}, where 2x {a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5}, where 8x
Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are,
{a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10}, where 125x {a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x
though this author is not certain if these are the same 8th power multi-grades found by Wroblewski. (
Piezas
To find an infinite family of
(a+bh)
the big brother of six terms with k = 2,4 and k = 1,2,6). To make F_{2} valid for k = 1,2, set g = -(ab+cd+ef) and f = -(1+b+d). Expanding for k = 4,6, we get,
(Poly11)h (Poly21)h
respectively, where the
(a+ab-c+cd-be-de) (-a+ab+c+cd-be-de) (a+ab+c+cd-2e-be-de) = 0
plus an irreducible cubic factor. (One can get more solns if a linear root of the cubic can be found such as for special cases like
(a+bh)
where g = -(ab+cd+ef) and f = -(1+b+d), then for k = 1,2,4,6, if,
A. Linear factors:First factor:
b = (1+d)(a+2e)/(a-e), c = (2+d)(a+e)/(1-d), where {a,d,e} satisfy
(Poly1)(-1+d)
Poly1:= a
For arbitrary {a,d,e,h}, this already solves k = 1,2,4. For k = 6, the quadratic in
{a,e} = {9d(41+80d+41d
though presumably smaller polynomials may exist. Using this initial rational point on the curve D = y
a = (2+b)(c+e)/(1-b), d = (1+b)(c+2e)/(c-e), where {b,c,e} satisfy
(Poly3)(-1+b)
Poly3:= bc
so to solve (Poly3)(Poly4) = y
b = -(a+2c)(a+c-2e)/(a
(Poly5)h
Poly5:= (a+c)
so to solve (Poly5)(Poly6) = y
^{2}. This again has a square constant term so is easily made a square.B. Alternative method
Still more solns can be found by solving Poly11 = Poly12 = 0 alternatively. Instead of eliminating
(2+b+d)(1+b)(1+d)(b+d) = 0
The first is a special case of Poly11 = Poly12 = 0, giving a simple soln to F
(a+bh)
still with g = -(ab+cd+ef) and f = -(1+b+d), where,
{c,d,e,h} = {-ab(1-b)/w, -2-b, a(1-b)(3+2b)/w, ay/w},
with w = (2+b)(3+b) and b
This gives terms as quartic polynomials. (Alternatively, one can set
d = 1 to get similar results.) The next three factors are variations on the same theme, namely a system that also satisfies x_{1}+x_{2} = y_{1}+y_{2} and x_{3}+x_{4} = y_{3}+y_{4} which imply an F_{2} with d = -b; f = -1. Solving Poly11 = Poly12 = 0 gives a soln but is trivial. However, by using the previous method of eliminating h between eq.1 and eq.2, the resultant gives three factors (squared this time), the first two trivial but not the third. This, in fact, is the linear root of the irreducible cubic which reduces for the special case d = -b. Using this root, eq.2 factors into two quadratics, one identical to eq.1 so the problem is reduced to just solving this. Thus, the
complete soln to,
x
for k = 1,2,4,6 where x
with {d, f} = {-b, -1} where {a,g,h} = {v
and {b,c,e} satisfying the quartic,
(2v
where {v
{a,g,h} = {(b
with u = b(b
nx x
for k = 2,4,6, using our previous results for k = 1,2,4,6 it can be proven this has a soln for
(a+bh) + (c+dh) + (e+fh) + (g+h) = (a-bh) + (c-dh) + (e-fh) + (g-h)
where f = -1-b-d, is also good for,
(-a-bh) +
if
{a,e} = {9d(41+80d+41d
for
where a = 1, f = -1-bn-d, g = -(b+cd+ef), valid for k = 2,4,6 if,
1) {b,c,d,e,h} = {-2/(1+n), -2(-2+n)(3+n)/w, 2/(1+n), (1+n)(3+n)/w, (1+n)y/w}, where w = 2n(-5+n) and y
2) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-1+3n)/w, -1, -2(3+n)/w, (1+n)y/w}, where w = 2n(1+3n) and y
3) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-2+3n)/w, -n/(1+n), -(3+n)/w, (1+n)y/w}, where w = 2n(-1+3n) and y
which also are true for both,
(a+bh) + (-c-dh) + (-e-fh) + (g+h) = (a-bh) + (-c+dh) + (-e+fh) + (g-h)
nx x
for k = 2,6 and
D a square. As a polynomial in the variable a, D is a quartic with a square leading term so it is quite easy to do this. A simpler case is when b = -2 since the constant term of D also becomes a square. Set c = 1 without loss of generality to get,
D: = (3-4e+2e
Three small non-trivial solns to D = y
a
giving three families solving the specialized form of F
(a-2h)
where g = -2+2a+e and either,
1) a = 8(2e-e
2) a = -(21-22e+8e
3) a = (13-18e+8e
second approach is to find b,c such that it factors as P = (x^{2}+mx+n)(x^{2}-mx+n) = 0 in which case it generally does not have a D that is a square. |