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(Update, 4/27/10): A magic square is an n x n arrangement of numbers such that all rows, columns, and both diagonals sum to the same constant c. The special case of a 3x3 magic square with all 9 numbers as distinct squares is unknown. The best so far is one with 7 squares,
with sum c = 541875 found by Andrew Bremner.
Theorem (Piezas): “In general, a rational soln to the elliptic curve,
u^{2}(-1+2u^{2}) + 2(1-u^{2}+u^{4})v^{2 }+ (2-u^{2})v^{4} = t^{2} (eq.0)
leads, after scaling, to an integral soln to the 3x3 magic square with 6 distinct squares,
though some {u,v} may not yield distinct p_{i}. An infinite family of polynomial solutions exists that solves eq.0.”
Proof: The entries can be expressed as,
{x, y, z} = {p_{2}^{2}+p_{3}^{2}-p_{5}^{2}, p_{1}^{2}-p_{3}^{2}+p_{4}^{2}, p_{2}^{2}+p_{3}^{2}-p_{4}^{2}} {p_{1}, p_{2}, p_{3}} = {ac+bd, ad-bc, (1-a^{2}+b^{2})c/2+(1+a^{2}-b^{2})d/2} {p_{4}, p_{5}, p_{6}} = {ad+bc, ac-bd, (-1-a^{2}+b^{2})c/2+(-1+a^{2}-b^{2})d/2}
with the common sum 3(ad+bc)^{2}, if {a,b,c,d} satisfies the condition,
p_{3}^{2}-2p_{4}^{2}+p_{5}^{2} = 0 (eq.1)
Substituting the expressions for the p_{i} into eq.1, it is easily seen this is only a quadratic in {c,d}. Solving for d, the discriminant D of eq.1 is given by,
D_{1}: = (1+a^{2}-2ab+b^{2}) ((a+b)^{2}(a^{2}+6ab+b^{2}) + (a^{2}-6ab+b^{2}))
Note that this is symmetric in {a,b}. For rational solns {c,d}, D must be made a square. Letting {a,b} = {(u+v)/2, (u-v)/2}, one is to solve,
D_{2}: = u^{2}(-1+2u^{2}) + 2(1-u^{2}+u^{4})v^{2 }+ (2-u^{2})v^{4} = t^{2}
as stated in the theorem. (End proof)
Some remarks:
1. A simple polynomial soln can be given to D_{2} = t^{2}. This is the non-trivial v = (u+1)/(u-1). From this point, an infinite family of polynomial solns can then be computed. 2. From the expressions for the p_{i}, if either a = b, or c = d, then p_{3} = -p_{6}, hence two or more squares become non-distinct. These cases should be avoided. 3. This author tried to find positive integers such that D is a square. Since D is symmetric in {a,b}, it suffices to let a < b. Only 9 solns with {a,b} < 1000 were found,
{2, 4}, {4, 7}, {4, 17}, {7, 444}, {21, 38}, {34, 316}, {90, 772}, {117, 148}, {166, 183}
For example, {a,b,c,d} = {4, 7, 91, 207} yields the 6-square,
Since each soln leads to an elliptic curve D_{2} = t^{2}, then there are in fact infinitely many. But only one so far is known such that one of the {x,y,z} is also a square, namely {a,b,c,d} = {4, 17, 21, 17}, or {u,v} = {21, 13}, which yields y = 23^{2} and gives the 7-square soln given by Bremner.
Note: Christian Boyer is offering 1000 euros and a bottle of champagne (how delightfully French!) for a positive integer soln to a 3x3 magic square with at least seven distinct squares that is not merely a trivial version of Bremner’s. See Magic Squares Prize Problems for more details.
(Update, 4/20/10): Alain Verghote pointed out that,
(a-b)^{3} = a(a+3b)^{2} - b(3a+b)^{2}
(a-b)^{5} = a(a^{2}+10ab+5b^{2})^{2} - b(5a^{2}+10ab+b^{2})^{2}
(a-b)^{7} = a(a^{3}+21a^{2}b+35ab^{2}+7b^{3})^{2} - b(7a^{3}+35a^{2}b+21ab^{2}+b^{3})^{2}
and so on. The coefficients involve the numbers in Pascal's Triangle.
(Update, 4/20/10): Enrico Jabara gave a polynomial soln to x_{1}^{3}+x_{2}^{4}+x_{3}^{5}+x_{4}^{7}+x_{5}^{8} = z^{8} as,
(2^{88}7^{47}n^{4})^{3} + (2^{92}7^{49}n)^{4} + (2^{32}7^{17}n^{4})^{5} + (2^{8}7^{4}n^{4})^{7} + (n^{4}-2^{52}7^{28})^{8} = (n^{4}+2^{52}7^{28})^{8}
More generally, as a variant of the so-called easier Waring’s Problem, he proved that any integer N is the sum of consecutive unlike powers,
N = ± x_{1}^{2 }± x_{2}^{3 }± x_{3}^{4 }± x_{4}^{5}
N = ± x_{1}^{3 }± x_{2}^{4 }± x_{3}^{5} ± x_{4}^{6 }± x_{5}^{7 }± x_{6}^{8 }± x_{7}^{9} ± x_{8}^{10}
where the x_{i} are also in the integers. However, for the next in the series that starts with 4th powers,
N = ± x_{1}^{4 }± x_{2}^{5 }± … ± (x_{k-3})^{k}
it is yet unknown what is the ending power k.
Source: Representations of numbers as sums and differences of unlike powers, (by Enrico Jabara, Università di Ca’ Foscari).
(Update, 4/20/10): Choudhry gave general solns to two kinds of Diophantine eqns,
1. ax^{5}+by^{5}+cz^{5} = au^{5}+bv^{5}+cw^{5 } (eq.1) 2. ax^{6}+by^{6}+cz^{6} = au^{6}+bv^{6}+cw^{6} (eq.2)
where {a,b,c} are not all equal to 1. As a particular example for the first,
x^{5}+2y^{5}+3z^{5} = u^{5}+2v^{5}+3w^{5}
{x,y,z} = {101t^{10}-101t^{5}-4, 4t^{12}-88t^{7}+88t^{2}, 25t^{10}-25t^{5}+4} {u,v,w}= {-4t^{12}-101t^{7}+101, 88t^{10}-88t^{5}+4, 4t^{12}-25t^{7}+25t^{2}}
He showed that if a+b = c, then one can find a polynomial identity for eq.1 (see general case here), and an infinite number of distinct solns to eq.2, since,
Theorem (Choudhry): “The Diophantine equation ax_{1}^{k}+bx_{2}^{k}+cx_{3}^{k} = ay_{1}^{k}+by_{2}^{k}+cy_{3}^{k} for k = 2,6 where a+b = c can be reduced to solving an elliptic curve of form px^{3}+qx^{2}+rx+s = y^{2}.”
He also gave the simple identity,
(mu^{6}+nv^{6})^{6} + (nu^{6}-mv^{6})^{6} + 12mn(m^{4}-n^{4})(uv^{5})^{6} = (mu^{6}-nv^{6})^{6} + (nu^{6}+mv^{6})^{6} + 12mn(m^{4}-n^{4})(u^{5}v)^{6}
though this is unrelated to the given theorem. Source: Some Diophantine equations involving sixth powers, The Mathematics Student, Vol 74, 2005.
(Update, 4/19/10): It's long known that any non-zero rational number N is the sum of three rational cubes in an infinite number of non-trivial ways.
Proof: Ryley's Identity
(p^{3}+qr)^{3} + (-p^{3}+pr)^{3} + (-qr)^{3} = N (6Nvp^{2})^{3}, where {p,q,r}= {N^{2}+3v^{3}, N^{2}-3v^{3}, 36N^{2}v^{3}}, for arbitrary v.
This involves a 6th-deg polynomial in the constant N. However, a 3rd-deg polynomial, in fact, is possible. William Ellison cited one example from the book Cubic Forms by Yuri Manin as,
(m^{3}-3^{6}n^{9})^{3} + (-m^{3}+3^{5}mn^{6}+3^{6}n^{9})^{3} + (3^{3}m^{2}n^{3}+3^{5}mn^{6})^{3} = m(3^{2}m^{2}n^{2 }+3^{4}mn^{5}+3^{6}n^{8})^{3}
for arbitrary n. Note how this has neat coefficients that are only powers of 3. Robert Israel pointed out that a small tweak can reduce the powers of 3 as just,
(27m^{3}-n^{9})^{3} + (-27m^{3}+9mn^{6}+n^{9})^{3} + (27m^{2}n^{3}+9mn^{6})^{3} = m(27m^{2}n^{2 }+9mn^{5}+3n^{8})^{3}
See also Ryley's Theorem. (Update, 4/12/10): The Leech lattice, a lattice in 24-dimensional Euclidean space, can be constructed in a variety of ways, one of which is in terms of a Weyl vector of form,
(0, 1, 2, 3,…, 22, 23, 24; 70)
Such an integral vector of norm zero exists because of the fact that the first n = 24 non-zero squares have the sum,
1^{2} + 2^{2} + 3^{2} + … + 24^{2} = 70^{2}
This has connections to E. Lucas’ old cannonball problem and, for n > 1, is true only for n = 24. However, it can be shown that 20 consecutive cubes also neatly sum up to,
15^{3} + 16^{3} + 17^{3} + … + 34^{3} = 70^{3}
also mentioned in Kevin Brown's Sum of consecutive nth powers equal to an nth power. Additionally, the kissing number problem (which asks to find the number of identical hyperspheres in n dimensions which can simultaneously touch a central hypersphere without intersections) is, for n = 24, given by,
K = 196560 = (2^{2}+3^{2})(3^{3}+4^{3}+5^{3})(70)
Question: Is there any other soln to the simultaneous Diophantine eqns,
a^{2 }+ (a+1)^{2} + (a+2)^{2} + … + (a+m-1)^{2} = x^{2} (eq.1)
b^{3 }+ (b+1)^{3} + (b+2)^{3} + … + (b+n-1)^{3} = x^{3} (eq.2)
in positive integers and {m,n} > 1, other than {a,b,m,n; x} = {1, 15, 24, 20; 70}?
Both eq.1 and 2 have an infinite number of distinct solns given in the previous update, but it is of added interest when their value for x coincide. One approach to the problem is to use the formulas,
F(a,m): = (m/6)(6a^{2}-6a+6am+1-3m+2m^{2}) = x^{2} F(b,n): = (n/4)(2b+n-1)(2b^{2}-2b+2bn-n+n^{2}) = x^{3}
and solve,
(F(a,m))^{3} = (F(b,n))^{2} = x^{6}
A simple Mathematica code can quickly show this has no “small” soln with {a,b,m,n; x} as positive integers and {m,n} > 1 other than x = 70. Can anyone find if there is another one in a higher range? Alternatively, the formulas can be simplified by using the substitutions,
{a,m} = {(1-p+q)/2, p} {b,n} = {(1+r-s)/2, s}
to get the much simpler,
F(p,q): = p(p^{2}+3q^{2}-1)/12 F(r,s): = rs(r^{2}+s^{2}-1)/8
Thus one is to solve,
(p/12)^{3}(p^{2}+3q^{2}-1)^{3} = (rs/8)^{2}(r^{2}+s^{2}-1)^{2} = x^{6}
the only known positive soln with {p,s} > 1 is {p,q,r,s; x} = {24, 25, 53, 20; 70} which, when the substitutions are reversed, yield positive integers {a,b,m,n}.
P.S. I checked the higher power version,
u^{k}^{ }+ (u+1)^{k} + (u+2)^{k} + … + (u+v-1)^{k} = 70^{k}
for small k > 3 and found no soln in the integers, though it may have one in the rationals.
(Update, 4/10/10): I. The sum of consecutive cubes not starting with 1 can be a square,
23^{3} + 24^{3} + 25^{3} = 204^{2}
118^{3} + 119^{3} + 120^{3} + 121^{3} + 122^{3} = 2940^{2}
and others with any odd number of cubes, as instances of an identity by Henri Cohen. To recall, the sum of the first n positive integers is,
E(1,n): = 1+2+3+ … + n = (1/2)(n^{2}+n) (eq.0)
which is a triangular number. It is quite well-known that the sum of the first n cubes is a square,
F(1,n): = 1^{3} + 2^{3} + 3^{3} + … + n^{3} = (1/4)(n^{2}+n)^{2} (eq.1)
so that,
1^{3}+2^{3}+3^{3}+ … + n^{3} = (1+2+3+ … + n)^{2}
If the starting point is not 1, but some integer b, then the sum of consecutive n cubes is given by,
F(b,n): = b^{3} + (b+1)^{3} + (b+2)^{3} + …. + (b+n-1)^{3} = (n/4)(2b+n-1)(2b^{2}-2b+2bn-n+n^{2}) (eq.2)
If b = 1, then the formula for eq.2 reduces to the square in eq.1. However, as Dave Rusin pointed out, a more aesthetic form can be derived by the substitution {b,n} = {(1+r-s)/2, s} which yields the simpler formula,
F(r,s): = rs(r^{2}+s^{2}-1)/8
The objective is to find other parametric and integral {r,s} such that rs(r^{2}+s^{2}-1)/8 = y^{2}. Only three families are known:
The soln [2] is by H. Cohen, while [3] is by Piezas. Q: Can anyone give a fourth family? Source: R.J. Stroker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, Vol 97, 1995.
II. If F(r,s) = rs(r^{2}+s^{2}-1)/8 = y^{3}, then there is also a family, one with the simple form,
given by Dave Rusin which is integral for v ≠ 3m. Interestingly, this identity involves sums of n cubes for n itself a cube. The first non-trivial case yields the well-known (at least the blue part),
(-2)^{3} + (-1)^{3} + 0^{3} + 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} = 6^{3}
which is a sum of n = 2^{3} = 8 cubes equal to a cube, though the next n = 4^{3}, 5^{3}, etc no longer involves negative numbers. The blue part is also known as one of Plato's numbers. This family does not directly yield other cute equalities like,
11^{3} + 12^{3} + 13^{3} + 14^{3} = 20^{3}
This can be generalized to an arithmetic progression with a small common difference d, where consecutive numbers is simply the case d = 1. For d = 2, it seems they are rarer, as the only known primitive soln with less than 1000 terms, or positive {m,n} < 1000 is,
31^{3} + 33^{3} + 35^{3} + 37^{3} + 39^{3} + 41^{3} = 66^{3}
found by Rusin with original variables {m,n} = {31/2, 6}. A good discussion can be found in Rusin's Mathematica Atlas.
(Update, 4/7/10): The Diophantine eqn,
2x^{2}(x^{2}-1) = 3(y^{2}-1) (eq.1)
has the complete soln in positive integers as x = {1, 2, 3, 6, 91}. A few people, including Hendrik Lenstra, noticed that remarkably, this coincides with the values of the well-known Ramanujan-Nagell equation,
(2x-1)^{2} + 7 = 2^{n} (eq.2)
with the complete n = {3, 4, 5, 7, 15} yielding, what else, x = {1, 2, 3, 6, 91}! In Stroeker’s and De Weger’s paper, “On a Quartic Diophantine Equation”, they remark that this is probably a coincidence (?). Euler discussed a more general form of eq.2,
u^{2}+7v^{2} = 2^{n} (eq.3)
As this involves the quadratic field Q(√-7) which has unique factorization, he was able to find an analytic soln for odd {u,v}. Q: Is the relationship between eq.1 and 2 really mere coincidence?
(Update, 4/7/10): One broad class of the quartic Diophantine eqn, x^{4}+ay^{4} = z^{4}+bt^{4}, which can easily be given a single soln is Norrie’s Identity,
(p^{4}-q^{4})^{4} + a(2p^{3}q)^{4} = (p^{4}+q^{4})^{4} + b(2pq)^{4}
where {a,b} = {1, (p^{8}-q^{8})/2}. An example is {p,q} = {3,1} which yields b = 205(2^{4}).
Q: Does x^{4}+y^{4} = z^{4}+205t^{4}, and other members of Norrie’s identity, have an infinite number of primitive solns?
(Update, 4/6/10): Alain Verghote noticed that,
2(a^{4}+b^{4}+(a+b)^{4}) = (a^{2}+b^{2}+(a+b)^{2})^{2} (eq.1)
This is a version of Proth’s Identity,
a^{k}+b^{k}+(a+b)^{k} = 2(a^{2}+ab+b^{2})^{k/2}, k = 2,4
Proof: Since,
a^{2}+b^{2}+(a+b)^{2} = 2(a^{2}+ab+b^{2}) a^{4}+b^{4}+(a+b)^{4} = 2(a^{2}+ab+b^{2})^{2}
Combining the two by eliminating the RHS yields eq.1. The algebraic form, a^{2}+ab+b^{2}, or equivalently, x^{2}+3y^{2}, is ubiquitous when dealing with 3rd and 4th powers, especially as the complete soln to,
a^{3}+b^{3 }= c^{3}+d^{3}
(by Binet, see [2]) uses that form, while generating further solns to,
a^{4}+b^{4 }= c^{4}+d^{4}
from an initial point also employs that same algebraic form (see also Fourth Powers, [4]). E. Escott also showed that,
(a^{2}+ab+b^{2})^{k} - (ab)^{k} - (ab+b^{2})^{k} = Square
for k = 2,4. Ramanujan used a variant of Proth’s identity to find elegant solns to a^{4}+b^{4}+c^{4} = 2d^{2k} for all positive integer k starting with,
a^{4}+b^{4}+c^{4} = 2(ab+ac+b)^{2}
where a+b+c = 0. More recently, Proth’s identity was used by Jacobi and Madden to reduce,
a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4}
to a special Pythagorean triple which was then solved by an elliptic curve.
(Update, 4/6/10): In response to my question in a March 28 post regarding the quartic Diophantine eqn,
x^{4}+ay^{4} = z^{4}+bt^{4} (eq.1)
where a ≠ b, Seiji Tomita found several more identities. It is an unsolved problem in the arithmetic of K3 surfaces to determine, given an initial non-trivial primitive soln to eq.1, whether there are an infinite more. For example, the Swinnerton-Dyer quartic eqn,
x^{4}+2y^{4} = z^{4}+4t^{4}
has only one known primitive soln, a huge one with 7-digit terms. (See Mar 28 post.) In contrast, for the eqn found by Noam Elkies while considering Fermat near-misses,
x^{4}+y^{4} = z^{4}+12t^{4}
there is just one small primitive soln with terms < 100, namely {x,y,z,t} = {6,7,5,4}. But it can proven via an identity by Elkies that there is an infinity,
(192v^{8}-24v^{4}-1)^{4} + (192v^{7})^{4} = (192v^{8}+24v^{4}-1)^{4} + 12(2v)^{4}
Seiji Tomita was inspired to find similar ones,
(48v^{8}-12v^{4}-1)^{4} + 2(48v^{7})^{4} = (48v^{8}+12v^{4}-1)^{4} + 6(2v)^{4}
(12v^{8}-6v^{4}-1)^{4} + 4(12v^{7})^{4} = (12v^{8}+6v^{4}-1)^{4} + 3(2v)^{4}
Curiously, all three have ab = 12. Tomita also found two others with {a,b} = {8, 24) and {27, 36}, and these have ab = 12(2^{4}) and 12(3^{4}). One can find more identities by using the general form,
(pv^{8}-qv^{4}-r)^{4} + a(mv^{7})^{4} = (pv^{8}+qv^{4}-r)^{4} + b(nv)^{4} (eq.2)
collecting powers of v, and making the coefficients vanish. Eq.2 can be completely solved as {a, b} = {8p^{3}q/m^{4}, 8qr^{3}/n^{4}} for free variables {m,n} and where {p,q,r} satisfies 3pr-q^{2} = 0. We can assume m = n = 1 for simplicity and find the product ab as,
ab = 12 (2pr)^{4}
explaining the structure of the five identities found by Elkies and Tomita.
Question: Given the Diophantine eqns,
x^{4}+ay^{4} = z^{4}+bt^{4} (eq.1) p^{4}+cq^{4} = r^{4}+ds^{4} (eq.2)
where {a,b,c,d} are integers and ab = cd. If eq.1 has a non-trivial soln, then does eq.2 have one as well? For ab = 12, the answer is yes. For {a,b} = {2,-2} = {1,-4}, it is also yes. Since the Swinnerton-Dyer {a,b} = {2,4} has a soln (a large one), I'm conjecturing,
x^{4}+y^{4} = z^{4}+8t^{4}
has one as well, but is probably going to be large too. (A simple Mathematica code can tell in seconds this has no non-trivial soln with all terms < 2000.) (Update, 4/6/10): A comparison between three Diophantine equations:
1. Jacobi, Madden: The eqn,
a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4} (eq.1)
after some algebraic manipulation, 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}
2. Piezas: The eqn,
a^{4}+b^{4}+c^{4} = d^{4} (eq.2)
is equivalent, not to a Pythagorean triple, but to the special equal sums of two squares,
(ab-ad+bd+d^{2})^{2} + (ab+ad-bd+d^{2})^{2} = (a^{2}+b^{2}+d^{2})^{2} + (c^{2})^{2}
Proof: Expand, subtract one side from the other, and it reduces to eq.2.
3. The analogous,
a^{6}+b^{6}+c^{6}+d^{6}+e^{6} = f^{6} (eq.3)
can also be manipulated to express it as equal sums of three cubes,
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} = y_{1}^{3}+y_{2}^{3}+y_{3}^{3}
It can be done by using the imaginary unit ω = √-1 on the Bremner-Kuwata identity to get an equivalent form of eq.3 as,
(a^{2}+adω+d^{2})^{3} + (b^{2}+beω+e^{2})^{3} + (c^{2}+cf-f^{2})^{3} = (-a^{2}+adω-d^{2})^{3} + (-b^{2}+beω-e^{2})^{3} + (-c^{2}+cf+f^{2})^{3}
Q. Can anyone manipulate eq.3 such that the {x_{i},y_{i}} are quadratic polynomials in the {a,b,c…} with rational coefficients?
(Update, 4/6/10): The eqn,
x_{1}^{k}+x_{2}^{k}+…+ x_{n}^{k} = (x_{1}+x_{2}+…+ x_{n})^{k} (eq.1)
for even k > 2 now has known solns for k = 4, 6, 8, 10:
1. S. Brudno: x_{1}^{4}+x_{2}^{4}+…+x_{4}^{4} = (x_{1}+x_{2}+…+x_{4})^{4}
{-2634, 955, 1770, 5400}
2. P. Ansell: x_{1}^{6}+x_{2}^{6}+…+x_{7}^{6} = (x_{1}+x_{2}+… +x_{7})^{6}
{-4170, -3187, -888, 1854, 3300, 3936, 4230}
3. N. Kuosa, T. Piezas: x_{1}^{8}+x_{2}^{8}+…+x_{10}^{8} = (x_{1}+x_{2}+… +x_{10})^{8}
{-226, -184, 6, 16, 34, 58, 66, 142, 152, 171}
4. S. Chase, D. Lichtblau: x_{1}^{10}+x_{2}^{10}+…+x_{13}^{10} = (x_{1}+x_{2}+… +x_{13})^{10}
{-187, -179, -128, -59, -13, -6, 49, 57, 73, 85, 122, 204, 210}
Note 1: The second name in [3] and [4] found the correct combination of signs, and I'd like to thank Daniel Lichtblau of Wolfram Research for indulging my questions. In the (k.1.n) convention adopted by Lander, Parkin, Selfridge, these are then (4.1.4), (6.1.7), (8.1.10), and (10,1,13).
Note 2: For k = 8, 10, there are smaller n but they cannot be placed in the form of eq.1, though appropriate ones may eventually be found.
(Update, 4/6/10): In 1945, after Morgan Ward proved that the (4.1.3),
a^{4}+b^{4}+c^{4} = d^{4} (eq.1)
has no soln with d < 10,000, he stated, “…this result makes it probable that there are no solns [of eq.1] whatsoever, especially since several closely allied soluble Diophantine equations such as a^{4}+b^{4} = c^{4} +d^{4} are known to have comparatively small solns…”
In 1967, Lander, Parkin, and Selfridge used a computer to search for a soln to (4.1.3) with bound d < 220000 without finding anything. If they had only doubled the bound, then they would have found it, as the smallest turned out to be,
[95800, 217519, 414560] = [422481]
found by Roger Frye twenty years later in 1988. Moral of the story? There is yet no known soln to the (6.1.6),
x_{1}^{6}+x_{2}^{6}+x_{3}^{6}+x_{4}^{6}+x_{5}^{6}+x_{6}^{6} = z^{6}
with z < 730,000 (Meyrignac, Resta, 2002), but there is a tantalizing possibility it could be just around the corner.
(Update, 4/6/10): I. It has been known that,
Theorem: “Given a primitive soln to,
x_{1}^{k}+x_{2}^{k}+…+ x_{k}^{k} = z^{k} (eq.1)
or where the x_{i} have no common factor, if k+1 is prime, then there is one and only one x_{i} such that z^{k}-x_{i}^{k} is divisible by (k+1)^{k}.” Thus, there is a congruence relationship between primes and solns to k kth powers equal to a kth power. For the smallest solns,
[3, 4]^{2} = 5^{2} [30, 120, 315, 272]^{4} = [353]^{4}
then 5^{2}-4^{2} is divisible by 3^{2}, and 353^{4}-272^{4} is divisible by 5^{4} (in fact, 353 + 272 = 5^{4}), and so on. Note that for primitive solns to eq.1, z is never divisible by k+1.
II. However, if we add one more addend,
x_{1}^{k}+x_{2}^{k}+…+ x_{k+1}^{k} = z^{k} (eq.2)
then there are two classes of primitive solns: a) z is divisible by k+1, or b) z is not divisible by k+1. Smallest for both kinds are,
[1, 2, 2]^{2} = 3^{2} [2, 3, 6]^{2} = 7^{2}
[2, 2, 3, 4, 4]^{4} = 5^{4} [10, 10, 10, 17, 30]^{4} = 31^{4}
[74, 234, 402, 474, 702, 894, 1077]^{6} = 1141^{6} [8190, 10920, 18480, 19229, 21252, 24360, 33354]^{6} = 34781^{6}
As one can see, the smallest for 6th powers are much higher than its k = 2 or 4 counterpart. The smallest (4.1.5) and (6.1.7) of the second class with terms analogous to 353 + 272 = 5^{4}, is
[30, 70, 75, 90, 312]^{4} = 313^{4} [1344, 23268, 25263, 39088, 48090, 54138, 54018]^{6} = 63631^{6}
since 312 + 313 = 5^{4} and 54018 + 63631 = 7^{6}. Meyrignac and Eulernet focused on (6.1.7) for the case where z is not divisible by 7 since, with a zero term, this covers (6.1.6) as well. All known solns were found using the five congruence classes:
Note that for 1-4, we have 1(42) = 2(21) = 3(14) = 6(7) = 42. See Peter Ansell’s site: http://www.computer-man.demon.co.uk/. Solns are known for all classes 1-5, but there is none yet where a term x_{i }= 0 for z < 730,000. (If so, then z^{6}-(nx_{6})^{6} is divisible by 7^{6} as required by the theorem.) Meyrignac has stopped the search years ago with the assumption, after observing trends, that it may be still out of reach of present computer technology.
Q: Can someone explain why the 6th power congruence involves 6(7) = 42, and not just 21? To compare with 4th powers, then we have either 4(5) = 20, or just 10. The eqns,
10^{4}(x_{1}^{4}+x_{2}^{4}+x_{3}^{4}) + x_{4}^{4} = z^{4} 20^{4}(x_{1}^{4}+x_{2}^{4}+x_{3}^{4}) + x_{4}^{4} = z^{4}
both have primitive solns, the smallest namely,
[240, 340, 430, 599]^{4} = 651^{4} [380, 1660, 1880, 4907]^{4} = 4949^{4}
but one can find the smallest soln of the first much easier than the second kind. Hence, if the divisibility by 42 of some terms of (6.1.7) is unnecessarily strict, then it may have made the problem harder than it should be.
(Update, 4/5/10): Alain Verghote gave,
(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+d^{2})(d^{2}+a^{2}) = (a^{2}+b^{2}+c^{2}+d^{2})^{2}(bd)^{2}
where ac ± bd = 0. (Update, 4/4/10): Alain Verghote gave the identity,
(a^{2}-t^{2})(b^{2}-t^{2})(c^{2}-t^{2}) = ((a+b+c)t^{2}+abc)^{2} - ((ab+t^{2})t+(a+b)ct)^{2}
One can see the elementary symmetric polynomials in the RHS. With a small tweak using the imaginary unit to have a sum of two squares, this can also be expressed as,
(a^{2}+t^{2})(b^{2}+t^{2})(c^{2}+t^{2}) = (s_{1}t^{2}-s_{3})^{2} + t^{2}(t^{2}-s_{2})^{2} (eq.1)
where {s_{1}, s_{2}, s_{3}} = {-a-b-c, ab+ac+bc, -abc}. Equivalently, eq.1 is true for any constant t where {a,b,c} are the roots of the cubic equation,
x^{3}+s_{1}x^{2}+s_{2}x+s_{3} = 0
For example, given x^{3}+3x^{2}+2x+1 = 0 with roots (only approximate) as {a,b,c} = {-2.324, -0.337+0.562i, -0.337-0.562i}, then,
(a^{2}+t^{2})(b^{2}+t^{2})(c^{2}+t^{2}) = (3t^{2}-1)^{2} + t^{2}(t^{2}-2)^{2}
for any t. Going higher,
(a^{2}+t^{2})(b^{2}+t^{2})(c^{2}+t^{2})(d^{2}+t^{2}) = (t^{4}-s_{2}t^{2}+s_{4})^{2} + t^{2}(s_{1}t^{2}-s_{3})^{2}
and {a,b,c,d} are the roots of the quartic x^{4}+s_{1}x^{3}+s_{2}x^{2}+s_{3}x+s_{4} = 0. And
(a^{2}+t^{2})(b^{2}+t^{2})(c^{2}+t^{2})(d^{2}+t^{2})(e^{2}+t^{2}) = (s_{1}t^{4}-s_{3}t^{2}+s_{5})^{2} + t^{2}(t^{4}-s_{2}t^{2}+s_{4})^{2}
with {a,b,c,d,e} as the roots of the quintic x^{5}+s_{1}x^{4}+s_{2}x^{3}+s_{3}x^{2}+s_{4}x+s_{5} = 0. And so on. (End update)
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