(Update, 6/29/10): The eqns,
3^{3} + 4^{3} + 5^{3} = 6^{3} 11^{3} + 12^{3} + 13^{3} + 14^{3} = 20^{3}
are relatively well known. Not so well known,
31^{3} + 33^{3} + 35^{3} + 37^{3} + 39^{3} + 41^{3} = 66^{3}
where the cubes are in arithmetic progression, differing by 2. The problem can be generalized as: “Given the sum of an n number of cubes in arithmetic progression with common difference d and initial cube a^{3}, then for what d is this sum also a cube?” As an equation, this is,
F(d,a,n) := a^{3} + (a+d)^{3} + (a+2d)^{3} + …. + (a+dnd)^{3} = y^{3} (eq.1)
Or equivalently,
F(d,a,n) := (n/4)(2ad+dn)(2a^{2}2ad+2adnd^{2}n+d^{2}n^{2}) = y^{3} (eq.2)
For the case d = 1, there is a parametric soln, given by Dave Rusin as,
{d,a,n} = {1, (v^{4}3v^{3}2v^{2}+4)/6, v^{3}}
which is integral for v ≠ 3m. Note that n is a cube, a fact which will be relevant later. However, if {d,a,n} are positive integers > 1 and coprime {d,a}, then for d < 100, a < 1000, n < 100, I was able to find several {d,a,n}. (Update, 6/30/10): Seiji Tomita extended the range to 1 < {d,a,n} < 1000. Out of the nearly 1 billion (10^{9}) possible combinations, there were only 20 primitive solns,
{2, 31, 6} {13, 28, 8} {107, 149, 3} {3, 290, 216} {13, 230, 5} {157, 553, 7} {5, 994, 344} {37, 15, 10} {158, 737, 14} {7, 194, 153} {39, 824, 27} {258, 97, 8} {11, 56, 49} {71, 435, 6} {277, 960, 16} {11, 206, 25} {71, 816, 49} {808, 317, 3} {11, 906, 36} {82, 541, 16}
Note that a disproportionate number of n are either squares or cubes. This phenomenon was also noted by Kevin Brown in the case d = 1. Why this is so does not seem to be known. Q. Can anyone find any other positive integers {d,a,n} > 1, and coprime {d,a} that solves eq. 2? (Update, 2/11/11): For 4th powers in arithmetic progression d whose sum is equal to a kth power, the only known small primitive solution is d = 2 and 64 terms, given by
29^{4} + 31^{4} + 33^{4} + … + 155^{4} = 96104^{2}
However, since the problem can be treated as an elliptic curve, there are, in fact, an infinite number of primitive solutions with 64 terms, though the initial term and d grow rapidly.
(Update, 6/26/10): Some squares can be identically expressed as the sum of both n = 3 and 4 squares,
1. Form (a^{2}+b^{2}+c^{2}+d^{2})^{2}
n = 3: (a^{2}+b^{2}c^{2}d^{2})^{2} + (2ad2bc)^{2} + (2ac+2bd)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
n = 4: (a^{2}+b^{2}+c^{2}+d^{2})^{2} + (2ab)^{2} + (2ac)^{2} + (2ad)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
by Lebesgue and Euler. Since the form is symmetric, by judicious exchange of pairs, {b,c}, {b,d}, and {c,d}, then it can be expressed as the sum of 3 squares in 3 ways. Similarly, by choosing the negative variable from {a,b,c,d}, and adjustment of the other terms, it is the sum of 4 squares in 4 ways.
2. Form (a^{2}+b^{2}+2c^{2})^{2}
n = 3: (2(a+b)c)^{2} + (2(ab)c)^{2} + (a^{2}+b^{2}2c^{2})^{2 }= (a^{2}+b^{2}+2c^{2})^{2}
n = 4: (a^{2}c^{2})^{2} + ((a+c)(b+c))^{2} + ((a+c)(bc))^{2} + (2acb^{2}c^{2})^{2} = (a^{2}+b^{2}+2c^{2})^{2}
both by Catalan.
3. Form (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2}
n = 3: ((a+b)(b+c))^{2} + ((a+c)(b+c))^{2} + (a^{2}+ab+acbc)^{2} = (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2}
n = 4: a^{2}(a+b+c)^{2} + b^{2}(a+b+c)^{2} + c^{2}(a+b+c)^{2} + (ab+ac+bc)^{2} = (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2}
by Catalan and J. Neuberg. Since this form is also symmetric, by cycling {a,b,c}, it is expressible as the sum of 3 squares in 3 ways.
4. Form ((a^{2}+b^{2})(c^{2}+d^{2}))^{2}
n = 3: ((a^{2}+b^{2})(c^{2}d^{2}))^{2} + (2cd(a^{2}b^{2}))^{2} + (4abcd)^{2} = ((a^{2}+b^{2})(c^{2}+d^{2}))^{2}
n = 4: ((a^{2}b^{2})(c^{2}d^{2}))^{2} + (2ab(c^{2}d^{2}))^{2} + (2cd(a^{2}b^{2}))^{2} + (4abcd)^{2} = ((a^{2}+b^{2})(c^{2}+d^{2}))^{2}
by Seiji Tomita. (Replacing d with b gives an old identity by Johann Euler, the son of Leonard Euler).
Q: Any other simple algebraic form that can be identically expressed both as 3 and 4 squares?
(Update, 6/23/10): 1. Saul doublets
One may observe that,
184 + 345 = 23^{2} 184^{2} + 345^{2} = 391^{2} 184^{3} + 345^{3} = 6877^{2}
A Saul doublet {a,b} is a solution to the system of eqns,
a + b = x^{2} a^{2} + b^{2} = y^{2} a^{3} + b^{3} = z^{2}
and named after J. Saul who first cited an example in the Gentleman’s Diary of 1795. There are an infinite number, and Bremner (1985) gave a method to completely solve for them,
F_{k} := (p(p+q))^{k} + ((p+q)q)^{k}
where,
F_{1} = (p+q)^{2} F_{2} = (p+q)^{2}(p^{2}+q^{2}) F_{3} = (p+q)^{4}(p^{2}pq+q^{2})
hence it suffices to find,
p^{2}+q^{2} = u^{2} p^{2}pq+q^{2} = v^{2}
Two quadratic polynomials to be made squares define an elliptic curve, and Bremner gave an explicit recurrence relation to find all its rational points. Explicitly, one can use the formula for Pythagorean triples {p,q} = {m^{2}n^{2} , 2mn} on the first condition, and the second becomes a quartic polynomial to be made a square,
m^{4}2m^{3}n+2m^{2}n^{2}+2mn^{3}+n^{4} = v^{2}
This is an elliptic curve, and the first rational point corresponds to {m,n} = {4,1} from which we can calculate the next, {m,n} = {17, 52}, and so on for an infinite number of points, though some of these may yield negative p or q. Source: Andrew Bremner, A Diophantine System.
2. Martin triples
A Martin triple {a,b,c} is a soln to the system,
a^{ }+ b^{ }+ c = x^{2} (eq.1) a^{2 }+ b^{2 }+ c^{2} = y^{2 }(eq.2) a^{3 }+ b^{3 }+ c^{3} = z^{3 }(eq.3)
The smallest in positive integers is M_{1} = {1498, 2461, 7490}, so,
1498 + 2461 + 7490 = 107^{2} 1498^{2} + 2461^{2} + 7490^{2} = 8025^{2} 1498^{3} + 2461^{3} + 7490^{3} = 7597^{3}
There are also an infinite number of distinct Martin triples unscaled by a square factor, and this can be proved via a polynomial identity or an elliptic curve. Originally, A. Martin solved only eq.2 and eq.3, and gave a polynomial identity of high degree (see Form 12 ), but any soln can be made valid for eq.1 as well. Each term is simply multiplied with a scaling factor m which is the squarefree part of the sum a+b+c = mn^{2}. For ex, the above was derived from,
14^{2} + 23^{2} + 70^{2} = 75^{2} 14^{3} + 23^{3} + 70^{3} = 71^{3}
and since a+b+c = 14+23+70 = 107, this yields M_{1} after multiplying all terms by m = 107. Other distinct solns for eq.2 and eq.3 are {3, 34, 114}, {18, 349, 426}, and so on. Elliptic curves can also solve both eq.2 and eq.3, such as k = 2 of the polynomial,
P_{k} := (8v^{2}+36v+14)^{k} + (9v^{2}25v+23)^{k} + (6v^{2}8v+70)^{k}
which uses the first soln (in blue). P_{3} is already a perfect cube for any v, but P_{2} is a quartic polynomial in v which must be made a square, just like the case for Saul doublets, hence can be treated as an elliptic curve. An initial point is v = 68/75 which yields, after removing unwanted factors, the new soln,
{a,b,c} = {6213m, 32194m, 46458m}
where m = 6213 + 32194 + 46458 = 84865. From this initial point v, an infinite more can then be computed. Any Martin triple, after removing common factors, can be used to create a distinct elliptic curve similar to the one given.
Q. Are there Martin triples with GCD(a,b,c) = 1? (If we go into quadruples, then the answer is yes.)
3. Martin quadruples:
Martin quadruples {a,b,c,d} solve in nonzero integers,
a^{ }+ b^{ }+ c + d = x^{2} a^{2 }+ b^{2 }+ c^{2} + d^{2} = y^{2} a^{3 }+ b^{3 }+ c^{3} + d^{3} = z^{3}
The smallest in positive integers has GCD(a,b,c,d) = 1:
10 + 13 + 14 + 44 = 9^{2} 10^{2} + 13^{2} + 14^{2} + 44^{2} = 49^{2} 10^{3} + 13^{3} + 14^{3} + 44^{3} = 45^{3}
(Can anyone find another small solution with no common square factor?) (Update, 6/24/10): James Dow Allen found the three smallest {a,b,c,d} < 1000:
{10, 13, 14, 44} {54, 109, 202, 260} {102, 130, 234, 318}
(Update, 6/26/10): Seiji Tomita found the next two with terms < 5000:
{198, 630, 1594, 1674} {570, 742, 1094, 1690}
Note that these five smallest Martin quadruples (with no common square factor) have very interesting sums,
10 + 13 + 14 + 44 = 3^{4 } 54 + 109 + 202 + 260 = 5^{4 } 102 + 130 + 234 + 318 = 2^{4}7^{2} 198 + 630 + 1594 + 1674 = 2^{12} 570 + 742 + 1094 + 1690 = 2^{12 }
The 6th one is still unknown though. There are also an infinite number of distinct Martin quadruples unscaled by a square factor, as each can be used in an elliptic curve to generate more. For ex, using the first, we have the polynomial,
P_{k} := (10v^{2}v+88)^{k} + (13v^{2}+68v+28)^{k} + (14v^{2}68v+26)^{k} + (44v^{2}+v+20)^{k}
Again, P_{3} is already a perfect cube for any v, but P_{2} is a quartic polynomial in v which must be made a square. An initial point is v = 165/94 which yields, after removing unwanted factors, the new soln,
{a,b,c,d} = {443794m, 1656013m, 1390130m, 1034308m}
where m is to be chosen as in Martin triples. As usual, from this initial point v, an infinite more can be computed. 4. HaldemanRamanujan quintuples:
2^{2} + 2^{2} + 4^{2} + 3^{2} + 4^{2} = 7^{2} 2^{4} + 2^{4} + 4^{4} + 3^{4} + 4^{4} = 5^{4}
A HaldemanRamanujan quintuple {a,b,c,d,e} solves the system,
a^{2 }+ b^{2 }+ c^{2 }+ d^{2 }+ e^{2} = x^{2} a^{4 }+ b^{4 }+ c^{4 }+ d^{4 }+ e^{4} = y^{4}
in the integers. There are many small solns, and it can be given an infinite number of simple identities, one of which is,
P_{k} := (2x^{2}+12xy6y^{2})^{k} + (2x^{2}12xy6y^{2})^{k} + (4x^{2}12y^{2})^{k} + (3x^{2}+9y^{2})^{k} + (4x^{2}+12y^{2})^{k}
where,
P_{2} = 7^{2}(x^{2}+3y^{2})^{2} P_{4} = 5^{4}(x^{2}+3y^{2})^{4}
HaldemanRamanujan quintuples with the constraint a+b = c (in blue) can then be used to generate similar polynomial identities (see Form 21). Since the integers include zero, there are in fact solns when a = 0, though the only one known is,
0^{2 }+ 15935^{2 }+ 27022^{2 }+ 57910^{2 }+ 59260^{2} = 88597^{2} 0^{2 }+ 15935^{4 }+ 27022^{4 }+ 57910^{4 }+ 59260^{4} = 70121^{4}
but it is unknown if this special case has an infinite number of solns.
(Update, 6/22/10): The system,
x_{1}x_{2}x_{3} = y_{1}y_{2}y_{3} x_{1}^{3}+x_{2}^{3}+x_{3}^{3} = y_{1}^{3}+y_{2}^{3}+y_{3}^{3}
has several solns:
1. Gerardin
(a^{2}p)^{3} + (b^{2}q)^{3} + (abr)^{3} = (a^{2}q)^{3} + (b^{2}r)^{3} + (abp)^{3}
where {p,q,r} = {a^{3}2b^{3}, a^{3}+b^{3}, 2a^{3}b^{3}}.
2. Choudhry gave a 3parameter version,
(ap)^{3} + (bq)^{3} + (cr)^{3} = (aq)^{3} + (br)^{3} + (cp)^{3}
where {p,q,r} = {a^{3}2b^{3}c^{3}, a^{3}+b^{3}+2c^{3}, 2a^{3}b^{3}+c^{3}}.
3. Piezas
(ap)^{3} + (bq)^{3} + (cr)^{3} = (ar)^{3} + (bp)^{3} + (cq)^{3}
where {p,q,r} = {abc^{2}, a^{2}+bc, b^{2}+ac}.
The last identity is also valid for x_{1}+x_{2}+x_{3} = y_{1}+y_{2}+y_{3} = 0, as well as,
9x_{1}x_{2}x_{3} (x_{1}^{6}+x_{2}^{6}+x_{3}^{6}y_{1}^{6}y_{2}^{6}y_{3}^{6}) = 2(x_{1}^{9}+x_{2}^{9}+x_{3}^{9}y_{1}^{9}y_{2}^{9}y_{3}^{9})
(Update, 6/21/10): For the Mar 22 update, A. Verghote gave the nice identity,
(x+y)(x+2y)(x+3y)(x+4y) + y^{4} = (x^{2}+5xy+5y^{2})^{2} (eq.1)
This has connections to Brocard’s Problem which asks to find values of n for which n!+1 = m^{2}, the smallest of which is n = 4 giving (1)(2)(3)(4) + 1 = 5^{2}. Eq.1 can be generalized, with another soln as,
(ax+y)(bx+2y)(cx+3y)(dx+4y) + y^{4} = (15x^{2}+19xy+5y^{2})^{2}
where {a,b,c,d} = {1, 3, 5, 15}, so abcd = 15^{2}, and an infinite more. This can be generalized for higher n by using the broader n!+v^{2} = m^{2} to increase the number of factors. For n = 6, I found,
(ax+y)(bx+2y)(cx+3y)(dx+4y)(ex+5y)(fx+6y) + 9y^{6} = (12x^{3}+54x^{2}y+72xy^{2}+27y^{3})^{2}
where {a,b,c,d,e,f} = {2, 2, 3, 2, 2, 3}, and other equivalent forms where abcdef = 12^{2}. Is there a soln to the next level n = 8, where n!+9^{2} = 201^{2}, so,
(ax+y)(bx+2y)(cx+3y)…(hx+8y) + 81y^{8} = (px^{4}+qx^{3}y+rx^{2}y^{2}+sxy^{3}+201y^{4})^{2}
and the coefficients are integers? (A simple Mathematica code can show there are none if the coefficients are positive integers ≤ 5, but my code takes too long if the range is extended.)
(Update, 6/20/10): Wroblewski has a database of primitive solns (when all terms do not have a common factor) to the equation,
x_{1}^{4}+x_{2}^{4}+x_{3}^{4}+x_{4}^{4}+x_{5}^{4} = y_{1}^{4}
separated into whether the sum y_{1} is integrally divisible by 5, or not. The cases x_{1} = 0, or x_{1} = x_{2} = 0, are solvable. If we consider the case where x_{1} almost vanishes, then for y_{1} not div by 5, out of more than 13000 solns with sum y_{1} < 10000, only four examples were found with x_{1} = 1,
1^{4} + 150^{4} + 3340^{4} + 6130^{4} + 6350^{4} = 7499^{4} 1^{4} + 2520^{4} + 3250^{4} + 5050^{4} + 6970^{4} = 7499^{4}
1^{4} + 920^{4} + 3120^{4} + 5410^{4} + 8870^{4} = 9193^{4} 1^{4} + 1410^{4} + 3490^{4} + 6020^{4} + 8680^{4} = 9193^{4}
Why it comes in pairs in unknown. It is also unknown if x_{1} = 0 with x_{2} = 1 is possible, though the closest is a single example for x_{2} = 2,
2^{4} + 15045^{4} + 26870^{4} + 34090^{4} = 37239^{4}
also found in Wroblewski’s database for (4.1.4) with sum y_{1} < 222000. A higher search radius might eventually lead to a soln. Other interesting results are,
1050^{4} + 1400^{4} + 1430^{4} + 1665^{4} + 1562^{4} = 3^{28} 735^{4} + 3220^{4} + 3780^{4} + 4160^{4} + 5936^{4} = 3^{32}
Note that,
3^{7}1562 = 5^{4} 3^{8}5936 = 5^{4}
Also,
2833^{4} + 3710^{4} + 7270^{4} + 11720^{4} = 23^{12} 608^{4} + 2450^{4} + 11530^{4} + 34865^{4} = 187^{8} 1630^{4} + 21019^{4} + 22340^{4} + 33940^{4} = 191^{8}
which are the only three (4.1.4) with such sums out of the about 1000 known. For 6th powers, no known (6.1.7) have similar sums. However, the eqn,
x_{1}^{6}+x_{2}^{6}+x_{3}^{6}+x_{4}^{6}+x_{5}^{6} = x_{6}^{6}+x_{7}^{6} was the subject of a recently finished distributed computing project which found 181 primitive solns with x_{7} < 7^{6} using a total of 740 MHz years! It is conjectured that either x_{5} = 0, or x_{7} = 0 is solvable, though none have yet been found. For the latter, the smallest known for the moment is x_{7} = 125 = 5^{3}, 214^{6 }+ 2968^{6 }+ 6951^{6} + 20046^{6} + 23457^{6} = 24781^{6} + 5^{18}
found by Larry Hays, hence is a relatively close approximation to five 6th powers equal to a 6th power (with a very neat excess).
(Update, 6/15/10 & 6/16/10) I. Class number h(d) = 1.
II. Class number h(d) = 2, where d = 4m.
III. Class number h(d) = 2, where d = 3m.
I. Class number h(d) = 1. It is quite wellknown that,
e^{π√7} ≈ 15^{3} + 697
e^{π√11} ≈ 32^{3} + 738
e^{π√19} ≈ 96^{3} + 743
e^{π√43} ≈ 960^{3} + 743.999… e^{π√67} ≈ 5280^{3} + 743.99999… e^{π√163} ≈ 640320^{3} + 743.99999999999…
The discriminants d under the square root are the six highest of the nine Heegner numbers. As I pointed out in a 2008 sci.math.research post, it is not so commonly known there is another internal structure,
e^{π√7} ≈ 3^{3}(3^{2}2^{2})^{3} + 697
e^{π√11} ≈ 4^{3}(3^{2}1)^{3} + 738
e^{π√19} ≈ 12^{3}(3^{2}1)^{3} + 743
e^{π√43} ≈ 12^{3}(9^{2}1)^{3} + 744 e^{π√67} ≈ 12^{3}(21^{2}1)^{3} + 744 e^{π√163} ≈ 12^{3}(231^{2}1)^{3} + 744
The reason for the squares is due to certain Eisenstein series. But there are other nice Diophantine relationships between these numbers,
7 (12^{3} + 15^{3}) = 3^{2} *63^{2} 11 (12^{3} + 32^{3}) = 4^{2} *154^{2} 19 (12^{3} + 96^{3}) = 12^{2} *342^{2}43 (12^{3} + 960^{3}) = 12^{2} *16254^{2} 67 (12^{3} + 5280^{3}) = 12^{2} *261702^{2} 163 (12^{3} + 640320^{3}) = 12^{2} *545140134^{2}
So the sums are perfect squares. It can also be seen that the equation,
x(12^{3} + Round[(e^{π√x}  744)^{1/3}]^{3}) = y^{2} where Mathematica's "Round[n] function" rounds off n to the nearest integer has integer solns only for x = {7, 11, 19, 43, 67, 163} for x < 10,000. And where else do the blue and red numbers appear? Of all places, in formulas for π! Let c = (1)^{n}(6n)! / ((n!)^{3}(3n)!), then,
1/π = 3 Σ c (63n+8)/(15^{3})^{n+1/2} 1/π = 4 Σ c (154n+15)/(32^{3})^{n+1/2} 1/π = 12 Σ c (342n+25)/(96^{3})^{n+1/2} 1/π = 12 Σ c (16254n+789)/(960^{3})^{n+1/2} 1/π = 12 Σ c (261702n+10177)/(5280^{3})^{n+1/2} 1/π = 12 Σ c (545140134n+13591409)/(640320^{3})^{n+1/2}
where the sum Σ goes from n = 0 to ∞. Beautiful, aren’t they? The modular function responsible for this is the jfunction which has the qseries expansion,
j(q) = 1/q + 744 + 196884q + 21493760q^{2} + 864299970q^{3} + ...
which explains why for these discriminants, the number e^{π√d} has an excess close to the integer "744". This is sequence A000521 of the OEIS and, other than the constant term, is the McKayThompson series class 1A of the Monster simple group. These pi formulas were discovered by the Chudnovsky brothers inspired by Ramunujan’s formulas given in the next section (which uses d with class number 2). For more details, see “Pi formulas, Ramanujan, and the Baby Monster Group”, by this author (File 07).
II. Class number h(d) = 2, where d = 4m. There are exactly 18 negative fundamental discriminants d with class number h(d) = 2, 7 of which are even, namely d = 4m, for m = {5, 13, 37} and m = {6, 10, 22, 58}. As m increases, the expression e^{π√m} gets intriguingly close to an integer with a fixed excess "104",
e^{π√5} ≈ 32^{2} + 100 e^{π√13} ≈ 288^{2} + 103.9… e^{π√37} ≈ 14112^{2} + 103.9999…
e^{π√6} ≈ 48^{2}  106 e^{π√10} ≈ 144^{2}  104.2… e^{π√22} ≈ 1584^{2}  104.001… e^{π√58} ≈ 156816^{2}  104.0000001…
Just like for d with class number h(d) = 1, there are also nice Diophantine relationships between these numbers,
5 (4^{4} + 32^{2}) = 4^{2} *20^{2} 13 (4^{4} + 288^{2}) = 4^{2} *260^{2} 37 (4^{4} + 14112^{2}) = 4^{2} *21460^{2}
6 (4^{4} + 48^{2}) = 8^{2} *3*8^{2} 10 (4^{4} + 144^{2}) = 8^{2} *2*40^{2} 22 (4^{4} + 1584^{2}) = 8^{2} *11*280^{2} 58 (4^{4} + 156816^{2}) = 8^{2} *2*105560^{2}
So the sums for the first set are perfect squares, while the second set are almostsquares. The blue and red numbers again appear in pi formulas courtesy of, who else, but Ramanujan. Let r = (4n)! / (n!^{4}), then,
1/π = 4 Σ r (1)^{n }(20n+3)/(32^{2})^{n+1/2} 1/π = 4 Σ r (1)^{n }(260n+23)/(288^{2})^{n+1/2} 1/π = 4 Σ r (1)^{n }(21460n+1123)/(14112^{2})^{n+1/2}
1/π = 8√3 Σ r (8n+1)/(48^{2})^{n+1/2} 1/π = 8√2 Σ r (40n+4)/(144^{2})^{n+1/2} 1/π = 8√11 Σ r (280n+19)/(1584^{2})^{n+1/2} 1/π = 8√2 Σ r (105560n+4412)/(156816^{2})^{n+1/2}
where n = 0 to ∞. The modular function, call it r(q), that is responsible for this has the qseries expansion,
r(q) = 1/q + 104 + 4372q + 96256q^{2} + 1240002q^{3} + …
and explains why the number e^{π√m} has an excess close to the integer "104". This is sequence A007267 and, excepting the first term, is the McKayThompson series of Class 2A for the Monster group.
III. Class number h(d) = 2, where d = 3m. There are exactly 5 discriminants of this form, namely d = 3m, for m = {5, 8, 17, 41, 89}. This time, the expression e^{π√(m/3)} gets close to an integer with a fixed excess "42", though I will focus only on the three highest m,
e^{π√(17/3)} ≈ 12^{3} + 41.6… e^{π√(41/3)} ≈ 48^{3} + 41.99… e^{π√(89/3)} ≈ 300^{3} + 41.9999…
and three nonfundamental odd m = {9, 25, 49},
e^{π√(9/3)} ≈ 3*4^{3} + 38.8… e^{π√(25/3)} ≈ 5*12^{3} + 41.91… e^{π√(49/3)} ≈ 7*36^{3} + 41.998…
Since these nonfundamental m are the squares of the first 3 odd primes, the approximations look nice, being the product of that prime and a cube. As usual, there are nice Diophantine relationships,
3*17 (2^{2}3^{3} + 12^{3}) = 6^{2} *51^{2} 3*41 (2^{2}3^{3} + 48^{3}) = 6^{2} *615^{2} 3*89 (2^{2}3^{3} + 300^{3}) = 6^{2} *14151^{2}
3 (2^{2}3^{3} + 3*4^{3}) = 6^{2} *5^{2} 3 (2^{2}3^{3} + 5*12^{3}) = 6^{2} *27^{2} 3 (2^{2}3^{3} + 7*36^{3}) = 6^{2} *165^{2}
so the sums are perfect squares. Define r_{2} = (2^{2}3^{3})^{n} (1/2)_{n}(1/3)_{n}(2/3)_{n} / (n!^{3}) where (a)_{n} is the rising factorial, or Pochhammer symbol, such that (a)_{n} = (a)(a+1)(a+2)…(a+n1). Equivalently, let r_{2} = (2n)! (3n)!/ (n!^{5}), then,
1/π = 1/(12√3) Σ r_{2} (1)^{n }(51n+7)/(12^{3})^{n} 1/π = 1/(96√3) Σ r_{2} (1)^{n }(615n+53)/(48^{3})^{n} 1/π = 1/(1500√3) Σ r_{2} (1)^{n }(14151n+827)/(300^{3})^{n}
1/π = (1/4)√3 Σ r_{2} (1)^{n }(5n+1)/(3*4^{3})^{n} 1/π = (1/36)√15 Σ r_{2} (1)^{n }(27n+3)/(5*12^{3})^{n} 1/π = (1/108)√7 Σ r_{2} (1)^{n }(165n+13)/(7*36^{3})^{n}
where n = 0 to ∞. These have been derived by this author (On Ramanujan's Other Pi Formulas, file 08), though Chan and Liaw gave a different form for the first one in their paper, "Cubic Modular Equations and New RamanujanType Series for 1/π" (2000). The modular function responsible, call it h(q), has qseries expansion,
h(q) = 1/q + 42 + 783q + 8672q^{2} + 65367q^{3} + …
This is sequence A030197 and, perhaps not surprisingly, is the McKayThompson series of class 3A for the Monster group. To summarize, there is this deep relationship between transcendental numbers of the form e^{π√d}, the modular functions j(q), r(q), h(q), pi formulas, and the Monster group. (End update.)
(Update, 6/15/10): It is also well known that,
3^{3}+4^{3}+5^{3} = 6^{3}
It seems sums of n 5th powers equal to a 5th power also like the number 6, or specifically, 12. For n = 4,5,6, the smallest solns have sums that are ALL divisible by 12, namely,
4 terms: 27^{5} + 84^{5} + 110^{5} + 133^{5} = 12^{10} 5 terms: 19^{5} + 43^{5} + 46^{5} + 47^{5} + 67^{5} = (6*12)^{5} 6 terms: 4^{5} + 5^{5} + 6^{5} + 7^{5} + 9^{5} + 11^{5} = 12^{5}
This numerical curiosity is perhaps another instance of Richard Guy’s Strong Law of Small Numbers.
(Update, 6/14/10): Duncan Moore observed that the three known solns to,
x_{1}^{5}+x_{2}^{5}+x_{3}^{5}+x_{4}^{5}+x_{5}^{5} = 0 (eq.1)
obey curious congruence properties between its terms (just like this author’s observations about certain 4th power Diophantine equations. Refer to 3/28/10 update.) Given.
1. [27, 84, 110, 133, 144]^{5} = 0 2. [220, 5027, 6237, 14068, 14132]^{5} = 0 3. [55, 3183, 28969, 85282, 85359]^{5} = 0
Scher and Seidl, discoverer of [2], proved that eq.1 obeys the constraint that the sum of either one or two pairs of terms is divisible by 5. Moore observed that the sum of one or two pairs is also divisible by 2^{5}, namely,
1. 27 + 133 = 2^{5}*5 2. 5027 + 6237 = 11*2^{10}, and 14068 + 14132 = 2^{6} 3. 55 + 28969 = 907*2^{5}, and 3183 + 85359 = 327*2^{8}
Whether this is as universally valid as the ScherSeidl congruence is not known.
(Update, 6/13/10): Alain Verghote pointed out that A. Martin’s foursquare identity,
(4pr+s)^{2} + (4prs)^{2} + (4qr+s)^{2} + (4qrs)^{2} = 4(s+2r^{2})^{2}, if s = 2p^{2}+2q^{2}r^{2}
has a second similar soln as,
(4pr+s)^{2} + (4prs)^{2} + (4qr+s)^{2} + (4qrs)^{2} = 4(s+4r^{2})^{2}, if s = p^{2}+q^{2}2r^{2}
Verghote also found the nice 4th power identity,
8(acbd)^{4} + 8(adbc)^{4} = ((ab)(c+d))^{4} + 6((a^{2}b^{2})(c^{2}d^{2}))^{2} + ((a+b)(cd))^{4}
(Update, 6/12/10): Seiji Tomita gave a new method to solve x^{3}+y^{3}+z^{3} = mt^{3} for an infinite number of constant m that is not m ≡ 4,5 (mod 9). While Ryley’s Identity is for any m, Tomita’s method involves polynomials of smaller degree (cubic). Examples are,
(2n^{3}+6n8)^{3} + (2n^{3}6n8)^{3} + (6n^{2}+10)^{3} = 3(2n^{2}2)^{3}
(n^{3}+9n^{2}+39n+91)^{3} + (n^{3}9n^{2}39n35)^{3} + (6n^{2}+36n+70)^{3} = 6(4n^{2}+24n+56)^{3}
and so on. See http://www3.alphanet.ne.jp/users/fermat/dioph74e.html for more details. He also found,
(3n^{3}+1)^{3} + (3n^{3}+2)^{3} + (3n)^{3} = (9n^{3}+3)^{2}
similar to Bouniakowsky’s,
(n^{3}+1)^{3} + (n^{3}+2)^{3} + (3n)^{3} = (3n^{3}+3)^{2}
(Update, 6/11/10): Wroblewski found three solns to the septic 3way chain,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}x_{4}^{k}x_{5}^{k}x_{6}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}y_{4}^{k}y_{5}^{k}y_{6}^{k} = z_{1}^{k}+z_{2}^{k}+z_{3}^{k}z_{4}^{k}z_{5}^{k}z_{6}^{k}
for k = 1,2,3,5,7. They also obey,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k} = z_{1}^{k}+z_{2}^{k}+z_{3}^{k} x_{4}^{k}+x_{5}^{k}+x_{6}^{k} = y_{4}^{k}+y_{5}^{k}+y_{6}^{k} = z_{4}^{k}+z_{5}^{k}+z_{6}^{k}
for k = 1,3. Two solns are:
1: {x,y,z} = {a,b,c}
{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}} = {723, 547, 813, 855, 279, 821} {b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}} = {697, 263, 323, 421, 395, 571} {c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}} = {703, 317, 383, 509, 341, 605}
2: {x,y,z} = {d,e,f}
{d_{1}, d_{2}, d_{3}, d_{4}, d_{5}, d_{6}} = {827, 757, 947, 1045, 209, 1009} {e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}} = {47, 157, 527, 205, 59, 509} {f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}} = {227, 417, 447, 349, 45, 549}
Piezas observed that these two solns obey the additional cubic 6way chain,
a_{1}^{k}+a_{2}^{k}+a_{3}^{k}a_{4}^{k}a_{5}^{k}a_{6}^{k} = b_{1}^{k}+b_{2}^{k}+b_{3}^{k}b_{4}^{k}b_{5}^{k}b_{6}^{k} = c_{1}^{k}+c_{2}^{k}+c_{3}^{k}c_{4}^{k}c_{5}^{k}c_{6}^{k} = d_{1}^{k}+d_{2}^{k}+d_{3}^{k}d_{4}^{k}d_{5}^{k}d_{6}^{k} = e_{1}^{k}+e_{2}^{k}+e_{3}^{k}e_{4}^{k}e_{5}^{k}e_{6}^{k} = f_{1}^{k}+f_{2}^{k}+f_{3}^{k}f_{4}^{k}f_{5}^{k}f_{6}^{k}
for k = 1,3. It is unknown if an identity is behind this septic 3way, in contrast to the octic 3way chain where an identity has been found.
(Update, June 2010): Okay, so this is not about math, but I thought I'd share this clip from one of my favorite films from childhood. It's Rankin Bass' Flight of Dragons (1982), an echanting tale about the battle between science and magic. The beautiful opening song is by Don McLean, the singer of that classic song, "American Pie". Sigh, they don't make animated movies like this anymore. Maybe they can make a liveaction remake; if it can be done for Lord of the Rings, it certainly can be done for this one.
You can email the author at tpiezas@gmail.com.
