Several results for eight terms have already been discussed in the previous section. Before discussing the present section, it may be instructive to give an overview of some of the known “side conditions” for,
x
x
since there seems to be some general pattern.
1. x 2. x 3. x 4. x
1. x
1. x 2. x 3. x
1. x
Of course these are not all of the possible side conditions but small solns tend to follow them hence are the first ones noticed. They are useful since they give more “handles” to deal with the system and, in some cases, reduce it to merely solving a quadratic or making a low-degree polynomial a square. (It is tempting to speculate ninth and tenth powers may obey similar rules though, since only
Piezas
[1] 5(a
[2] 3(a
[3] 9(a
[4] (a
[5] 7(a
Combining [2] and [3] yields the same relation as combining [4] and [5], namely,
[6] 7(a
One can test it with a random example, say, {a,b,c,d} = {21, -13, 9, -17} and {e,f,g,h} = {23, -3, -21, 1}, though it will work in the general case. Note that a+b+c+d = e+f+g+h for this system implies it is also valid for k = 2. [6] has already been given in
L(k): = (a+b+c)
since the three variables {a,b,c} can suffice to express the y). Expanding L(k) for k = 1,2,3,5, we get,_{i}
L(1) = 0, L(2):= a L(3): = abc-def, L(5): = abc(a
If we wish to set all L(k) = 0, if L(3) = 0, one can see that the eqn L(5) = 0 reduces to just L(2) = 0. Eliminating
Lander
(a+b+c)
(Update, 12/13/09): Tony Rizzo showed that there are non-trivial solns to,
80abc(a
thus yielding a (5.3.3) and gave the example {a,b,c,d,e} = {32, 25, 50, 80, 100}. More generally, this gives a (5.3.3) with the side condition x
Lander (k = 1,5)
(ax (ax
This has the side condition x
{p,q,r,s} = {a
Note that {p,q,r,s} also solve,
(p+q)
To see this, if the 8-term eqn above was expanded and powers of {x,y} collected, we get a system of four auxiliary eqns in four unknowns, one of which is,
w
explaining why it has this property. In fact, by solving this system one can directly derive the
Kawada, Wooley
(h+x)
(Update, 11/2/09): Tony Rizzo pointed out we can let x
2(h+x)
where the whole eqn is now div by 2.
(x+z)
where x ^{5}+2(2)^{5}, this results in a 10-term identity,
(x+z)
with the same conditon C
(h+x)
for m = {486, 16806}, where x
(h+x)
(h+x)
Piezas
So far, most of the identities discussed have the condition x
a) x b) x c) x
I.
We can focus on a special case of this constraint by defining the polynomial, P(k): = (a+c)
E(k): = P(k) + (r)
to be simultaneously true for k = 1,3,5, then this has a
(a
so there are two identities involving P(k), namely,
(a+c)
(a+c)
both for k = 1,3,5. Also, the polynomial P
P
while for the second, let, {a,b,c} = {2(u
P
Thus, in general, for these k
x
the quintic analogue of Ramanujan’s family of solns to x
x
by setting one term as equal to ±1 and solving the resulting Pell eqn. For ex, we have,
(a+c)
where {a,b,c} = {2(p+q)(p+5q), 8(p+4q)q, p
II.
We can focus first on a special case. Define the eqn,
(a+c)
This has the common
c
where the ratio c/d = {1, 1/3} should be avoided. A small soln is {c,d} = {37, 15} and it is easy to convert the two quadratic conditions into an elliptic curve. A special case solvable as quadratic forms is,
(a+b+c)
where -a
The common difference of the first three pairs is also 2c but the sum of its first four terms is now x
(a+m)
Expanding for k = 1,3,5, we can eliminate
(a+b)
(a
One must find {a,b,c} such that these two quartic polynomials in
c = -(2+3b)/(3+4b)
It reduces the second eqn, after removing square factors, to solving,
25+24b-48b
with initial b = 3/16 from which other rational points can then be computed. (Update, 8/13/09): One can use a variant,
(p+aq)
already satisfying the constraint (after appropriate transposition of terms). Solving this system involves only making a quartic and
III.
This is a generalization of a single numerical result by Tarry. It can be the case that a k = 1,3,5 can give rise to a k = 2,4,6. Given,
(3p+q)
(4p+q)
which has a complete soln in terms of an elliptic curve as,
{p,q,r,u,v} = {2x
and where x,y must satisfy,
y
though some of the small rational solns are trivial. (Update, 8/4/09): Alternatively, the system can be expressed as, (a-2b+c)
(a+c-d)
^{k} + (a+c+d)^{k} + (-a+2b+2c)^{k} + (b-4c)^{k} = (a-c-d)^{k} + (a-c+d)^{k} + (-a+2b-2c)^{k} + (b+4c)^{k}, for k = 1,2,4,6, where {a,b,c,d} satisfy a
This is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-1/3, -1, 2, 5} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {15,1}. (
The system of eqns, x
Piezas
Use the form,
(a+c)
where {d,e,f,g} = {(cw+u)/w, (cw-u)/w, (cw+v)/w, (cw-v)/w}. Note that this is a
(a (a
where w = a
(a+c)
where a
Piezas
This is the fifth power counterpart to the third and fourth power identities found by this author. For k = 1,2,3,4,5,
(a (a
and the sum-product identity,
(a
where {a
(a+n)
where appropriate pairs of terms have the common sum
(a+n)
when
(5y)
if x |