(Update, 2/26/10) There is a family of identities starting with Birck's,
(-a+b+c)^{k} + (a-b+c)^{k} + (a+b-c)^{k} + (a+b+c)^{k} = (2a)^{k} + (2b)^{k} + (2c)^{k}, for k = 1,2
This was generalized by J. Zehfuss as,
(-a+b+c+d)^{k} + (a-b+c+d)^{k} + (a+b-c+d)^{k} + (a+b+c-d)^{k} = (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k}, for k = 1,2
where Birck’s was d = 0. Generalizing this with a fifth variable e by adding just one term to the RHS will no longer work, but Alain Verghote modified this to the two (2.5.6) identities, Form 1:
(-a+b+c+d+e)^{k} + (a-b+c+d+e)^{k} + (a+b-c+d+e)^{k} + (a+b+c-d+e)^{k} + (a+b+c+d-e)^{k} = (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k} + (2e)^{k} + (a+b+c+d+e)^{k}, for k = 1,2
Form 2: (-a+b+c+d+e)^{2} + (a-b+c+d+e)^{2} + (a+b-c+d+e)^{2} + (a+b+c-d+e)^{2} + (a+b+c+d-e)^{2} = (-a+b+c+d)^{2} + (a-b+c+d)^{2} + (a+b-c+d)^{2} + (a+b+c-d)^{2} + (a+b+c+d+e)^{2} + (2e)^{2}
Note how for e = 0, Form 1 reduces to Zehfuss', while Form 2 reverts into a tautology. These can be further generalized as,
(-a+b+c+d+e+f)^{k} + (a-b+c+d+e+f)^{k} + (a+b-c+d+e+f)^{k} + (a+b+c-d+e+f)^{k} + (a+b+c+d-e+f)^{k} + (a+b+c+d+e-f)^{k} = (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k} + (2e)^{k} + (2f)^{k} + 2(a+b+c+d+e+f)^{k}, for k = 1,2
and so on. (Setting f = 0 reduces to Form 1.) The significance of Birck's and Zehfuss' symmetric versions is that they can give rise to the multi-grade identity,
(-a+b+c+d)^{k} + (a-b+c+d)^{k} + (a+b-c+d)^{k} + (a+b+c-d)^{k} + (2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k }= (-e+f+g+h)^{k} + (e-f+g+h)^{k} + (e+f-g+h)^{k} + (e+f+g-h)^{k} + (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k}, for k = 1,2,4,6,8
if abcd = efgh and a^{n}+b^{n}+c^{n}+d^{n} = e^{n}+f^{n}+g^{n}+h^{n}, for n = 2,4. This has an infinite number of solns and one can also let d = h = 0 for a simpler version. However, to create an analogous multi-grade using Verghote’s Form 1 identity would entail a more complicated set of conditions on ten variables. (End update.) (Update, 2/23/10) Wroblewski, Piezas
Let a^{2}+b^{2} = c^{2} and a^{2}+52b^{2} = d^{2}, a pair of conditions which is a concordant form. Then a [k,7,7] for k = 1,2,4,6,8,10 is, (8b)^{k} + (-5a-4b)^{k} + (5a-4b)^{k} + (-a-2d)^{k} + (a-2d)^{k} + (-12b+4c)^{k} + (12b+4c)^{k} =
(-16b)^{k} + (-4a+8b)^{k} + (4a+8b)^{k} + (-a+4c)^{k} + (a+4c)^{k }+ (-3a-2d)^{k} + (3a-2d)^{k}
though ratios such as a/b = {4, 12, 12/5} should be avoided as the system becomes trivial, with a non-trivial example the smallest Pythagorean triple {a,b,c} = {3, 4, 5}, with d = 29. The two quadratic polynomials define an elliptic curve, and there is an infinite number of solns. A pair of algebraic forms {a^{2}+b^{2}, a^{2}+Nb^{2}} that is to be made squares is called a concordant form. (However, found throughout this book, is the more general case of {pa^{2}+qb^{2}, ra^{2}+sb^{2}} to be made squares.) This identity is a special case of a more general one involving 16 terms. (End update.)
(Update, 2/22/10, 3/1/10): Piezas
Theorem: “The eqn a^{4}+b^{4} = c^{6}+d^{6} has an infinite number of unscaled, integral solns.”
Proof: The first known example by Giovanni Resta has the form,
(m^{2}n)^{4}(x^{4}+y^{4}) = (mn)^{6}(u^{6}+v^{6}) (eq.1)
with primitive soln {m,n} = {5, 73}; and {u,v,x,y} = {3, 4, 18, 31}. Eq.1 can be simplified as,
m^{2}(x^{4}+y^{4}) = n^{2}(u^{6}+v^{6})
Hence the situation is reduced to the easier problem of finding,
(x^{4}+y^{4})(u^{6}+v^{6}) = w^{2} (eq.2)
where gcd(x,y) = 1 and gcd(u,v) = 1. However, the second factor can also be treated as a constant, so this becomes,
25(193)(x^{4}+y^{4}) = w^{2}
which is an elliptic curve. Starting with the known {x,y} = {18, 31}, an infinite number of other rational points can then be computed. Hence, another soln to eq.1 is:
{m,n} = {5, 228420709564570748140960777}; and {u,v,x,y} = {3, 4, 56328091105014, 7382523294847}
and so on, ad infinitum. It seems quite interesting that the Pythagorean triple {3, 4, 5} appears in this problem. Note: A computer search can probably find smaller co-prime solns to eq.2, though the relation u^{3}y^{2}-v^{3}x^{2} = 0 should be avoided as it is trivial. Using two generators of an elliptic curve, Seiji Tomita found smaller solns to x^{4}+y^{4} = 193z^{2} as,
{x,y}= {1266, 181}; {19021, 10806}; {890858057, 810357546}; {288871421574, 44739797383}
with the next the same as the one found by this author. (End update.)
(Update, 2/22/10): Gerardin gave the beautiful 4th power identity with small coefficients {1,2,3},
(a+3a^{2}-2a^{3}+a^{5}+a^{7})^{4} + (1+a^{2}-2a^{4}-3a^{5}+a^{6})^{4} = (a-3a^{2}-2a^{3}+a^{5}+a^{7})^{4} + (1+a^{2}-2a^{4}+3a^{5}+a^{6})^{4}
and Choudhry gave a 5th power version,
(a-a^{3}-2a^{5}+a^{9})^{5} + (1+a^{2}-2a^{6}+2a^{7}+a^{8})^{5} + (2a^{3}+2a^{4}-2a^{7})^{5} = (a+3a^{3}-2a^{5}+a^{9})^{5} + (1+a^{2}-2a^{6}-2a^{7}+a^{8})^{5} + (-2a^{3}+2a^{4}+2a^{7})^{5}
which in fact is good for k = 1,5. (Differences between RHS and LHS have been highlighted in blue.) While there are identities for k = 6, none are known with such small coefficients. (Anyone can find one?) Other 5th deg identities can be found using the ff. two methods by Choudhry. 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)
where,
{x_{1}, x_{2}, x_{3}, x_{4}} = {a+b+c, a-b-c, -a+b-c, -a-b+c} {y_{1}, y_{2}, y_{3}, y_{4}} = {d+e+f, -d+e-f, -d-e+f, d-e-f}
Expanding one side of the eqn, we get,
∑ x_{i}^{1} = 0 ∑ x_{i}^{5} = 80abc(a^{2}+b^{2}+c^{2})
and similarly for the y_{i}. Eq.1 can be reduced to just 6 terms if x_{4} = y_{4}. Thus, if,
Method 1: a+b-c = -d+e+f ; ab = de; c(a^{2}+b^{2}+c^{2}) = f(d^{2}+e^{2}+f^{2}) Method 2: a+b-c = -d+e+f ; c = f; ab(a^{2}+b^{2}+c^{2}) = de(d^{2}+e^{2}+f^{2})
then ∑ x_{i}^{k} = ∑ y_{i}^{k}, for k = 1,5.
Choudhry showed that given an initial point, even a trivial one, using appropriate substitutions and solving a system of linear equations, then an infinite number of non-trivial solns can be found. In fact, by directly solving Method 1, one can end up with a quartic equation which has a rational root if a certain quartic polynomial is made a square, hence can be treated as an elliptic curve. Note: Choudhry also used Eq.1 and its substitutions to solve k = 1,3,7.
Source: “On Equal Sums of Fifth Powers”, Indian Jour. Pure & Applied Math, Nov. 1997.
(Update, 2/22/10): 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? 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 this 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 73-58 = 15, 73-41 = 32, though with appropriate substitutions, they are still not enough to reduce it to a parametric form. (End update.)
(Update, 2/18/10): On a hunch, this author re-checked Wroblewski’s soln for (k.5.5) for k = 1,3,9 to see if there was a partition such that it would be good for k = 2 as well. It turns out there was. Thus,
[51, 253, 412, -621, 600]^{k} = [187, 100, 429, 603, -624]^{k}, for k = 1,2,3,9
with terms obeying the constraints,
1. x_{1}+x_{2}+x_{3} = y_{1}+y_{2}+y_{3} 2. x_{4}+x_{5} = y_{4}+y_{5} 3. (x_{1}-y_{1})(x_{2}-y_{2}) = -(x_{3}-y_{3})(x_{4}-y_{4}) 4. (x_{1}-y_{1})(x_{2}-y_{2}) = (x_{3}-y_{3})(x_{5}-y_{5})
with the last two suggested by Wroblewski, though [4] is just a consequence of [2] and [3]. Why this “numerical curiosity” obeys such symmetric constraints is unknown. (Note that the difference x_{i}-y_{i} for all five pairs is divisible by 17.) (End update.)
(Update, 2/16/10): As observed by Bremner and Delorme, the smallest [k.6.6] for k = 9, found by Lander et al through computer search back in 1967, is in fact good for k = 1,2,3,9,
[18, 23, 13, -10, -15, -5]^{k} = [21, 22, 9, -13, -14, -1]^{k}
Notice also how appropriate pairs all have the same sum,
18-10 = 23-15 = 13-5 = 21-13 = 22-14 = 9-1 = 8
This “numerical curiosity” has an explanation, which lies in the simple eqn,
14^{2 }+ 19^{2} + 9^{2} = 17^{2} + 18^{2} + 5^{2}
and is just a special instance of Theorem 5 of the Prouhet-Tarry-Escott Problem.
Define, F_{n} = a^{n}+b^{n}+c^{n}+d^{n}-(e^{n}+f^{n}+g^{n}+h^{n}) and the [k.8.8],
[x+a, x+b, x+c, x+d, x-a, x-b, x-c, x-d]^{k} = [x+e, x+f, x+g, x+h, x-e, x-f, x-g, x-h]^{k}
1) If F_{2} = 0, and a variable x such that 42F_{4}x^{4}+28F_{6}x^{2}+3F_{8} = 0, then the system is for k = 1,2,3,9.
2) If F_{2} = F_{4} = 0 and 28F_{6}x^{2}+3F_{8} = 0, then it is for k = 1,2,3,4,5,9.
Notice how appropriate pairs of terms all have the same sum 2x. Also, it reduces to a [k.6.6] if a pair from opposite sides, say d = h = 0. Lander's soln was simply the case,
{a,b,c,d} = {14, 19, 9, 0} {e,f,g,h} = {17, 18, 5, 0}
and x = 4. Bremner and Delorme found a relationship between the terms such that the quartic in x factored into two quadratics. Making its discriminant a square involved an elliptic curve, so there are an infinite number of solns to [1]. This is discussed more in Section 9.3 of Ninth Powers.
For non-zero {d,h}, no example is yet known for [1], but for [2], Wroblewski directly found a nonic [k.8.8] for k = 1,2,3,4,5,9, which yields,
{a,b,c,d} = {240, 63, 197, 122} {e,f,g,h} = {167, 10, 168, 243} and x = 52. Since these variables can be reduced in size with appropriate transformations given in the previous update, it is possible there are others. (End update.)
(Update, 2/15/10): Wroblewski, Piezas
A multi-grade [k.4.4] for k = 2,4 can be used to find a nonic [k.8.8] for k = 1,2,3,4,5,9. Given six variables {p,q,s,t,y,z},
System 1: [10p+2q-s+t, -10p-2q-s+t, 2p-10q+y-z, -2p+10q+y-z]^{k} = [10p-2q-s+t, -10p+2q-s+t, 2p+10q+y-z, -2p-10q+y-z]^{k}, for k = 2,4
System 2: [5p+q+s, -5p-q+s, 5p+q+t, -5p-q+t, p-5q+y, -p+5q+y, p-5q+z, -p+5q+z]^{k} = [5p-q+s, -5p+q+s, 5p-q+t, -5p+q+t , p+5q+y, -p-5q+y, p+5q+z, -p-5q+z]^{k}, for k = 1,2,3,4,5
is identically true using the substitutions,
{p,q,s,t,y,z} = {u/2, v/2, a+b, a-b, a+c, a-c} {b,u,c,v} = {eg-2fh, eh+fg, eg+2fh, eh-fg}
for arbitrary {a,e,f,g,h}. It can be shown that [2] is derived from [1] using Theorem 5 of the Prouhet-Tarry-Escott problem.
A) One can make the first system valid for k = 6 and the second for k = 6,7 as well if,
(10g^{2}-13h^{2})e^{2}-(13g^{2}-40h^{2})f^{2} = 0
for arbitrary a. This is reducible to an elliptic curve, with one soln as {e,f,g,h} = {171, 97, 7, 47}, and an infinite more can then be computed.
B) However, if System 2 is to be true only for k = 1,2,3,4,5,9, then one must solve a quadratic in a with a sextic discriminant that must be made a square. The only known soln found by Wroblewski is {a,e,f,g,h} = {104, 7, 3/2, -18, 17}, though others may exist. Whether with the right constraints this can be reduced to a quartic polynomial to be made a square, hence an elliptic curve, is unknown. (End update)
(Update, 2/15/10): Piezas, Wroblewski
Using a numerical search with constraints, Wroblewski found hundreds of solns to a multi-grade octic chain of form,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}-(x_{5}^{k}+x_{6}^{k}+x_{7}^{k}+x_{8}^{k}) = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}-(y_{5}^{k}+y_{6}^{k}+y_{7}^{k}+y_{8}^{k}) = z_{1}^{k}+z_{2}^{k}+z_{3}^{k}+z_{4}^{k}-(z_{5}^{k}+z_{6}^{k}+z_{7}^{k}+z_{8}^{k})
valid for k = 2,4,6,8. An example is,
[221, 93, 73, 9]^{k}-[219, 117, 31, 17]^{k} = [198, 116, 96, 32]^{k}-[192, 144, 58, 44]^{k} = [189, 125, 105, 23]^{k}-[175, 161, 75, 27]^{k}
Note that it is also true x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = x_{5}^{k}+x_{6}^{k}+x_{7}^{k}+x_{8}^{k}, for k = 2,4 and x_{1}x_{2}x_{3}x_{4} = x_{5}x_{6}x_{7}x_{8}, relationships which also hold for the y_{i} and z_{i}. It turns out the general case could be explained in the context of an old result by the first author, namely,
“If a^{n}+b^{n}+c^{n}+d^{n} = e^{n}+f^{n}+g^{n}+h^{n}, for n = 2,4,6 and abcd = efgh, then,
(a+b+c-d)^{k} + (a+b-c+d)^{k} + (a-b+c+d)^{k} + (-a+b+c+d)^{k} + (2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k} = (e+f+g-h)^{k} + (e+f-g+h)^{k} + (e-f+g+h)^{k} + (-e+f+g+h)^{k} + (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k}
for k = 1,2,4,6,8,10.” (If n is only up to n = 4, then k is only up to k = 8.) Moving half of the terms of one side to the other yields,
(2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k} - ((2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k}) = (a+b+c-d)^{k} + (a+b-c+d)^{k} + (a-b+c+d)^{k} + (-a+b+c+d)^{k} - ((e+f+g-h)^{k} + (e+f-g+h)^{k} + (e-f+g+h)^{k} + (-e+f+g+h)^{k})
and since either sign of {±d, ±h} can be used and still satisfy abcd = efgh, then the RHS has two distinct octuplets for a given {a,b,c,d,e,f,g,h}. Thus, given, p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}, k = 2,4
where p_{1}p_{2}p_{3}p_{4} = q_{1}q_{2}q_{3}q_{4}, then this automatically leads to an octic chain of length 3. Wroblewski’s solns satisfies the constraints,
p_{1}p_{2} = q_{3}q_{4} p_{3}p_{4} = q_{1}q_{2} p_{1}^{2}+p_{2}^{2} = q_{1}^{2}+q_{2}^{2} p_{3}^{2}+p_{4}^{2} = q_{3}^{2}+q_{4}^{2} p_{1}^{4}+p_{2}^{4}+p_{3}^{4}+p_{4}^{4} = q_{1}^{4}+q_{2}^{4}+q_{3}^{4}+q_{4}^{4}
The first author found the complete soln to this system as an elliptic “curve”, namely,
{p_{1}, p_{2}, p_{3}, p_{4}} = {ar+bc, ac-br, ds+ef, df-es} {q_{1}, q_{2}, q_{3}, q_{4}} = {ar-bc, ac+br, ds-ef, df+es}
r = (abc^{2}+def^{2})(d^{2}-e^{2})(f/y), s = (abc^{2}+def^{2})(a^{2}-b^{2})(c/y)
and {a,b,c,d,e,f} must satisfy,
(abc^{2}+def^{2})(c^{2}de)(a^{2}-b^{2})^{2 }+ (abc^{2}+def^{2})(abf^{2})(d^{2}-e^{2})^{2} = y^{2}
which is only a quartic polynomial in {c,f} to be made a square. Given one soln, then an infinite more can be found.
Piezas
Another set of conditions is,
p_{1}p_{4} = q_{1}q_{4} p_{2}p_{3} = q_{2}q_{3} p_{1}+p_{2} = q_{1}+q_{2} p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}, k = 2,4
The soln can be given as,
{p_{1}, p_{2}, p_{3}, p_{4}} = {a(p+q), b(r-s), c(r+s), d(p-q)} {q_{1}, q_{2}, q_{3}, q_{4}} = {a(p-q), b(r+s), c(r-s), d(p+q)}
{p,q,r,s} = {a(b^{2}-c^{2}), be, (a^{2}-d^{2})b, ae}
and where {a,b,c,d,e} satisfy,
a^{2}(3b^{2}-c^{2})+e^{2} = (b^{2}+c^{2})d^{2}
This can be easily solved parametrically as quadratic forms. (End update)
(Update, 2/9/10): Lee Jacobi, Daniel Madden
The complete soln to a^{4}+b^{4} = c^{4}+d^{4} and a^{4}+b^{4}+c^{4} = d^{4} can be reduced to solving an elliptic curve. It turns out that for a special case of four 4th powers equal to a 4th power namely,
a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4} (eq.0)
with variables as ±, then one can do so as well, a small example of which is,
955^{4}+1770^{4}+(-2634)^{4}+5400^{4} = (955+1770-2634+5400)^{4} = 5491^{4}
Jacobi's and Madden’s clever soln depended on the identity,
a^{4}+b^{4}+(a+b)^{4} = 2(a^{2}+ab+b^{2})^{2}
an identity also useful for solving a^{4}+b^{4}+c^{4}+d^{4}+e^{4} = f^{4} as quadratic forms. (See Form 21 here.) Their method starts by adding (a+b)^{4} + (c+d)^{4} to both sides of eq.0,
a^{4}+b^{4}+(a+b)^{4}+c^{4}+d^{4}+(c+d)^{4} = (a+b)^{4}+(c+d)^{4}+(a+b+c+d)^{4}
and, using the identity, it is seen that eq.0 is equivalent to the special Pythagorean triple,
(a^{2}+ab+b^{2})^{2} + (c^{2}+cd+d^{2})^{2} = ((a+b)^{2} + (a+b)(c+d) + (c+d)^{2})^{2}
Sparing the reader the intermediate algebraic manipulations, using the transformation, {a,b,c,d} = {p+r, p-r, q+s, q-s}, the eqn,
(p+r)^{4}+(p-r)^{4}+(q+s)^{4}+(q-s)^{4} = (2p+2q)^{4}
can be solved if {p,q,r,s} satisfy two quadratics to be made squares,
(m^{2}-7)p^{2}+4(m^{2}-1)pq+4(m^{2}-1)q^{2} = (m^{2}+1)r^{2} (eq.1) 8mp^{2}+8mpq-(3m^{2}-8m+3)q^{2} = (m^{2}+1)s^{2} (eq.2)
for some constant m, which is easily proven by solving {r,s} in radicals and substituting it into the eqn to see that it holds. These two quadratic conditions define an elliptic curve. One can then try to find a suitable m and find rational {p,q,r,s}. Or, given a known {p,q,r,s}, a value m can be derived as,
m((p+2q)^{2}-r^{2}) = 3q^{2}+s^{2}
For example, given {a,b,c,d} = {5400, 1770, -2634, 955}, then {p,q,r,s} = {3585, -1679/2, 1815, -3589/2}, and m = 961/61. But this initial soln set can be used to find an infinite more. Since eq.1 and eq.2 are homogeneous, we can assume q = 1 without loss of generality. Starting with a soln to eq.1 as p_{1} = -2(3585)/1679, a complete parameterization is,
p = (30/1679) (4291863+112196282u+110805419u^{2})/(448737-463621u^{2})
which makes eq.1 a square. But substituting this to eq.2, and still with q = 1, produces a rational quartic polynomial in u that still has to be made a square. To find an initial rational point u, equating p-p_{1} = 0 yields the factor 55770003+56098141u = 0, hence this is one suitable value. Treating the quartic polynomial as an elliptic curve, an infinite more can be computed, thus proving that the eqn,
a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4}
has an infinite number of distinct, rational solns. (End update.)
(Update, 2/8/10): Piezas
If p^{4}+q^{4}+r^{4} = 1, then (p+q)^{4} - r^{4} + 1 = 2(p^{2}+pq+q^{2})^{2} also holds.
(Update, 2/8/10): Demjanenko, Elkies
The complete soln to
p^{4}+q^{4}+r^{4} = 1 (eq.0)
can be given in the form,
(x+y)^{4} + (x-y)^{4} + z^{4} = 1
where,
ay^{2} = (8mn-3a)x^{2}-2bx-2mn (eq.1) ±az^{2} = 4bx^{2}+8mnx-b (eq.2)
and {a,b} = {2m^{2}+n^{2}, 2m^{2}-n^{2}}, for some constants {m,n} hence is a problem of making two quadratics in x as squares and, for appropriate {m,n}, reduces to solving an elliptic curve. One can solve {y,z} in the radicals to see that it holds and it is easily proven that this is the complete soln. Proof: Let {x+y, x-y, z} = {p,q,r}. As eq.1 and eq.2 are also quadratics in {m,n}, solve for m, assume eq.0 as true to rationalize the discriminant, and take the positive sign of the square root to get,
m_{1} = n((p+q)^{2}+r^{2}-1) / (2(p^{2}+pq+q^{2}+p+q)) m_{2} = n(p^{2}+pq+q^{2}-p-q) / ((p+q)^{2}-r^{2}-1)
One can use the identity given by the author to rationalize the disciminant of m_{2}. It will then be seen that m_{1}-m_{2} = 0 if eq.0 is true, hence one can always find rational {m,n} in terms of {p,q,r}. For example, using the smallest {p,q,r} = {217519/s, 414560/s, 95800/s} and s = 422481, we get {m,n} = {6007, 30080}. (End proof) However, one can also start with {m,n} as it is possible they are relatively small. Elkies’ first soln used only {m,n} = {8,-5}, and (eq.1) becomes,
153y^{2} = -779x^{2}-206x+80
which has initial soln {x_{1},y_{1}} = {3/14, 1/42}. This yields a parametrization for x as,
x = (51p^{2}-34p-5221)/(238p^{2}+10906)
which, when substituted into (eq.2), becomes the problem of finding a value such that the quartic polynomial in p is a square. This can be treated as an elliptic curve and since x_{1 }= x = 3/14 has p = -3779/17, from this initial point one can calculate more, proving that eq.0 has an infinite number of distinct solns. Clever transformations may then reduce the size of the coefficients. Explicitly, a particular case can be given as the identity,
(85v^{2}+484v-313)^{4} + (68v^{2}-586v+10)^{4} + (2u)^{4} = (357v^{2}-204v+363)^{4}
if u^{2} = 22030+28849v-56158v^{2}+36941v^{3}-31790v^{4},
a soln of which is v = -31/467. This gives, after removing common factors, Elkies' soln. One can then use this initial point to generate an infinite more. (End update.)
(Update, 2/3/10): Choudhry
Gave a soln to a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 1,2,4, for integer {a,b,c,d,e,f} with the interesting property that one term is equal to 1.
(-p+v)^{k} + (-q+v)^{k} + (p+q-2v)^{k} = (-p-q+2uv)^{k} + (p+q-2uv-1)^{k} + 1^{k}
where
{p,q} = {2u^{2}v+u-v, 4uv+3v+1}
Source: On Representing 1 as the sum or difference of kth powers of integers, The Mathematics Student, Vol. 70, 2001.
Note: Since this also has a+b+c = d+e+f = 0, by the Ramanujan-Hirschhorn Theorem, its terms obey the beautiful relations,
64(a^{6}+b^{6}+c^{6}-d^{6}-e^{6}-f^{6})(a^{10}+b^{10}+c^{10}-d^{10}-e^{10}-f^{10}) = 45(a^{8}+b^{8}+c^{8}-d^{8}-e^{8}-f^{8})^{2}
25(a^{3}+b^{3}+c^{3}-d^{3}-e^{3}-f^{3})(a^{7}+b^{7}+c^{7}-d^{7}-e^{7}-f^{7}) = 21(a^{5}+b^{5}+c^{5}-d^{5}-e^{5}-f^{5})^{2}
(End update)
(Update, 2/3/10): Choudhry
Sophie Germain’s Identity is given by,
a^{4}+4b^{4} = (x^{2}+2xy+2y^{2})(x^{2}-2xy+2y^{2})
Choudhry solved the eqn,
a^{4}+4b^{4} = c^{4}+4d^{4} (eq.1)
using the transformation,
{a,b,c,d} = {(x^{2}+2x-2)z+xy, 2xz+y, (x^{2}+2x+2)z+xy, y}
so that, substituted into eq.1, it had a linear factor in y. This first soln can be used to compute an infinite sequence of polynomial solns, as eq.1 is an elliptic curve in disguise. The transformations, {a,b,c,d} = {p+q, r-s, p-q, r+s}, then {p,q,r,s} = {u, mv, nu, v} reduce eq.1 to the form,
(m-4n^{3})u^{2}+(m^{3}-4n)v^{2} = 0
so one is to solve,
-(m-4n^{3})(m^{3}-4n) = z^{2}
One soln is,
{p,q,r,s} = {8x+2x^{3}, -4+2x^{2}, 4+4x^{2}, -2x+x^{3}}
from which others can then be computed. Source: The Diophantine Equation A^{4}+4B^{4} = C^{4}+4D^{4}, Indian Journal of Pure and Applied Mathematics, Vol. 29, 1998. (End update.)
(Update, 2/3/10): Choudhry
(-8m^{6}+2mn^{5})^{k} + (8m^{5}n+n^{6})^{k} + 2(8m^{6})^{k} = (8m^{6}+2mn^{5})^{k} + (8m^{5}n-n^{6})^{k} + 2(n^{6})^{k}, for k = 1,5
A soln in positive terms can be acquired if n > m and within the range 2^{2/5} < n/m < 2^{3/5}, or approx. 1.32 < n/m < 1.51. With terms transposed to one side,
(-8m^{6}+2mn^{5})^{k} + (8m^{5}n+n^{6})^{k} - (8m^{6}+2mn^{5})^{k} - (8m^{5}n-n^{6})^{k} + 2(8m^{6})^{k} = 2(n^{6})^{k}
For n = 1, the identity proves that the integer 2 is the sum/difference of six integral 5th powers in infinitely many ways. Source: The Diophantine Equation x_{1}^{5}+x_{2}^{5}+2x_{3}^{5} = y_{1}^{5}+y_{2}^{5}+2y_{3}^{5}, Ganita, Vol. 48, No. 2, 1997, 115-116.
Note: In one sense, this is a 5th power version of,
(-6x^{3}+y^{3})^{3} + (6x^{3}+y^{3})^{3} - (6x^{2}y)^{3} = 2(y^{3})^{3}
where, for y = 1, proves that 2 infinitely is the sum is the sum/difference of three integral 3rd powers.
Q: Anyone can give a similar identity expressing 2 as integral 7th powers? (End update.) (Update, 2/2/10): Piezas
More generally, for the equivalent form (disregarding signs),
[a+b, -a+b, b+u, -b+u, c+v, -c+v, 2d]^{k} = [c+d, -c+d, a+u, -a+u, d+v, -d+v, 2b]^{k}, for k = 1,2,4,6,8,
the {a,b,c,d,u,v} must satisfy the simultaneous eqns,
a^{2}+2b^{2}+u^{2} = c^{2}+2d^{2}+v^{2} (a^{2}-b^{2})(b^{2}-u^{2}) = (c^{2}-d^{2})(d^{2}-v^{2})
excluding signed combinations of trivial cases (a+b)(c+d)(b+d)(a+b+c+d)(a^{2}+3b^{2}-c^{2}-3d^{2}) = 0. For the special case u = v, the first condition reduces to,
a^{2}+2b^{2} = c^{2}+2d^{2}
the complete soln of which {a,b,c,d} = {pr+2qs, ps-qr, pr-2qs, ps+qr} reduces the second condition to the easy problem of making a quadratic polynomial into a square. (End update.)
(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 (6th powers)
Given the general form,
[xy+ax+3by-c, xy-ax-3by-c, xy-3bx+ay+c, xy+3bx-ay+c]^{k} = [xy+ay+3bx-c, xy-ay-3bx-c, xy-3by+ax+c, xy+3by-ax+c]^{k}
Notice how {x,y} just swap places or alternatively, “a” in one side is replaced by “3b” in the other side. With terms expressed as {p_{i}, q_{i}}, this form has the constraints,
p_{1}+p_{2} = q_{1}+q_{2}, p_{3}+p_{4} = q_{3}+q_{4}, p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}, k = 1,2
We can make this valid for k = 1,2,4,6. For k = 4, let c = ab. For k = 6 and some constants {a,b}, then {x,y} must satisfy the simple eqn,
10a^{2}b^{2}-(a^{2}+9b^{2})(x^{2}+y^{2})+10x^{2}y^{2} = 0
This is an elliptic “curve” in disguise. Let {x,y} = {u, v/(10u^{2}-a^{2}-9b^{2})} and one is to solve,
(a^{2}+9b^{2}-10u^{2})(10a^{2}b^{2}-a^{2}u^{2}-9b^{2}u^{2}) = v^{2}
which is a quartic polynomial in u to be made a square. Trivial ratios are a/b = {0,1,3,9}. It can be proven that for any rational {a,b} that do not have a trivial ratio, one can always find an infinite number of rational {x,y}. Proof: A polynomial soln can be given starting from the trivial point u = 3b. This yields the non-trivial,
u = 9b(a^{2}+15b^{2})/(5a^{2}-117b^{2})
from which one can then successively compute an infinite sequence of rational polynomials.
Chris Smyth (8th powers)
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]^{k} = [xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]^{k}
where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn,
x^{2}y^{2}-13(x^{2}+y^{2})+11^{2 }= 0
One can note its affinity with the 6th power multi-grade and see also in the general form how “a” in one side has just been replaced with “b” in the other side. With terms expressed as {p_{i},q_{i}}, this obeys the constraints,
p_{1}-p_{2} = q_{1}-q_{2} p_{3}-p_{4} = q_{3}-q_{4}
If we set that,
p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k}+p_{5}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}+q_{5}^{k}, k = 1,2
then one must have 2a^{k}-2b^{k}+c^{k} = 0, for k = 1,2, and the consequence is that,
p_{1}^{k}+p_{2}^{k}+p_{5}^{k} = q_{1}^{k}+q_{2}^{k}, p_{3}^{k}+p_{4}^{k} = q_{3}^{k}+q_{4}^{k}+q_{5}^{k}
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^{2}} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,
x^{2}y^{2}-13(x^{2}+y^{2})+11^{2 }= 0
Let {x,y} = {u, v/(u^{2}-13)} and this is the elliptic curve,
(u^{2}-13)(13u^{2}-11^{2}) = v^{2}
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, ex+fy, fx-ey]^{k} = [xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]^{k}
which is the same pair successfully added to the (k.4.4) to find the first (k.6.6) identity for 10th powers. For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.
Choudhry, Wroblewski (10th powers)
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), ex+fy, fx-ey]^{k} = [-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]^{k}
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_{1}-p_{2} = q_{1}-q_{2}, p_{3}-p_{4} = q_{3}-q_{4}, (-p_{1})^{k}+p_{2}^{k}+p_{3}^{k}+(-p_{4})^{k} = (-q_{1})^{k} +q_{2}^{k}+q_{3}^{k}+(-q_{4})^{k}, k = 1,2 p_{5}^{2}+p_{6}^{2} = q_{5}^{2}+q_{6}^{2}
If we assume,
p_{1}+p_{2}+p_{5} = q_{1}+q_{2}+q_{5}, p_{3}+p_{4}+p_{6} = q_{3}+q_{4}+q_{6},
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^{2}y^{2}-17(x^{2}+y^{2})+392 = 0 {a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5}, where 8x^{2}y^{2}-17(x^{2}+y^{2})+98 = 0
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^{2}y^{2}-53(x^{2}+y^{2})+45 = 0 {a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x^{2}y^{2}-221(x^{2}+y^{2})+605 = 0
though this author is not certain if these are the same 8th power multi-grades found by Wroblewski. (End update.)
You can email author at tpiezas@gmail.com. |