B. Lenhart’s method
V. A General Method Theorem (Euler): Assume,
a2+b2+c2 = (bx3+dx2)2/x12 a2+c2+d2 = (bx2+dx3)2/x12 a2+b2+d2 = (ax3-cx1)2/x22 b2+c2+d2 = (ax1-cx3)2/x22
Given two Pythagorean triples {x1, x2, x3} and {y1, y2, y3}. If the product of the legs is a square x1x2y1y2 = m2, and set n = x2y1, then a quadruple is,
{a,b,c,d} = {1, m/n, d(mx2)/(nx1), (mx3y1+nx1y3)/(nx1y2-nx2y1)}
Proof: One can substitute n = x2y1, {x1, x2, x3} = {e2-f2, 2ef, e2+f2}, and {y1, y2, y3} = {g2-h2, 2gh, g2+h2}, and the four eqns are true if,
x1x2y1y2 = m2 or, equivalently,
4efgh(e2-f2)(g2-h2) = m2 (eq.1)
(End proof)
Example: Eq.1 is also discussed in “Mengoli’s Six Square Problem and Face Cuboids”. However, there are many small solutions, one of which is {e,f,g,h} = {5, 2, 6, 1} yielding,
{x1, x2, x3} = {21, 20, 29}
{y1, y2, y3} = {35, 12, 37}
{m,n} = {420, 735}
After scaling, this gives {a,b,c,d} = {105, 60, 168, 280} which is the third smallest Euler quadruple. For a parametrization, let x2+y2 = z2: i) If {x1,x2,x3} = {y,x,z}; {y1,y2,y3} = {x,y,z}; hence x1x2y1y2 = (xy)2, then,
{a,b,c,d} = {2xyz, x(x2-y2), y(x2-y2), 2xyz}.
ii) If {x1,x2,x3} = {y,x,z}; {y1,y2,y3} = {z4-4x2y2, 4xyz2, z4+4x2y2}; hence x1x2y1y2 = (2xyz(x2-y2))2, then,
{a,b,c,d} = {2pxyz, px(x2-y2), qy(x2-y2), 2qxyz}, where {p,q} = {x4-6x2y2-3y4, 3x4+6x2y2-y4}.
Both were given in the previous section.
Any other simple formulas for generalized Euler bricks and Euler quadruples?
(Note: Instead of using superscripts as I normally do, in this short
article I’ve used the hat symbol ^ so the equations can easily be copied and pasted onto Mathematica or Maple for those who wish to see the 16th-deg resolvent themselves.)
Here's a “natural” solvable 17th-deg eqn with small coefficients:
x^{17}-6 x^{16}-24 x^{15}-42 x^{14}-31 x^{13}-23 x^{12}-7 x^
{11}-x^{10}-4 x^9-11 x^8-7 x^7-13 x^6-x^5+x^3+x^2+x-1 = 0 (eq.1)
Its unique real root is exactly given by (in Mathematica) as x = zeta_48 DedekindEta[tau]/(Sqrt[2]DedekindEta[2tau]) = 9.1630942..., with the root of unity zeta_48 = exp(2Pi I/48), tau = (1+Sqrt[-d])/2, and d = 383. This d has class number h(-d) = 17.
To solve this, depress eq.1 (get rid of its xn-1 term), by letting x = (y +6)/17 to get,
y^{17}-11832 y^{15}-1124346 y^{14}-55393735 y^
{13}-1784741617 y^{12}-41171464807 y^{11}-711423456455 y^
{10}-9455898295636 y^9-99724287747103 y^8-887992943070295 y^7-7665207188897171 y^6-70479807472769473
y^5-592167373130143650 y^4-3496187093606980919 y^3-8695712981307573757 y^2+68265051092799270505 y-427806967360317821039 = 0 (eq.2)
Its 16th-deg resolvent, a polynomial with INTEGER coefficients, call this R_16, has roots,
z_k = ((y1 + w^{k}y2 + w^{2k}y8 + w^{3k}y7 + w^{4k}y16 + w^
{5k}y4 + w^{6k}y12 + w^{7k}y15 + w^{8k}y11 + w^{9k}y10 + w^{10k}y14 + w^{11k}y13 + w^{12k}y5 + w^{13k}y17 + w^
{14k}y6 + w^{15k}y9 + w^{16k}y3)/17)^17
for k = {1 to 16}, where w is any complex 17th root of unity. This
particular R_16 has all z_k as real numbers.
Note the specific arrangement of the y_n. There are 16! ≈ 2 x 10^13 possible permutations of the y_n, and out of that huge number, there are only 16 such that R_16 has integer coefficients, and we have given one of them. Of course, I used a short cut to find it, because even if your computer can check a million permutations a second, it would still take about 8 months to go through them all. The short cut took less than two hours to find R_16. The y_n follows the root object Root [poly, n] ordering in Mathematica. Approximately, these are,
y1 = 149.7726
{y2, y3} = -27.62 -/+ 18.49 i {y4, y5} = -21.61 -/+ 7.52 i {y6, y7} = -16.58 -/+ 6.34 i {y8, y9} = -10.57 -/+ 15.32 i {y10, y11} = -5.02 -/+ 13.71 i {y12, y13} = -2.34 -/+ 13.15 i {y14, y15} = 2.57 -/+ 2.60 i {y16, y17} = 6.31 -/+ 7.04 i
R_16 has extremely large integer coefficients, with the largest being the 248-digit constant term 429534618434587^17 which, naturally
enough, is a 17th power. (Note: R_16 can easily be formed using 500-digit precision or more on the y_n, and multiplying the 16 factors together to form the polynomial. However, since it involves extremely large integers of 200+ digits, it is understandable if I do not explicitly and tediously type them down here. But if you have Mathematica or
Maple, you can easily re-construct R_16 using the details described in this article.)
The polynomial R_16 can then be factored into two octics over the extension Sqrt[17]. This, in turn, can be factored into 2 quartics over Sqrt[2(17+Sqrt[17])]. This can be factored further into 2 quadratics using an expression involved in the 17th root of unity. Apparently, in general, to solve the 16-deg resolvent of an irreducible but solvable 17-deg equation, only square roots of square roots of square roots, etc, are needed.
The real root of eq.2 in radicals is then,
y1 = z1^(1/17) + z2^(1/17) + z3^(1/17) + … + z16^(1/17) = 149.7726…
The next step, of course, is degree p = 19. Unfortunately, to find the 18th-deg resolvent R_18 of an irreducible but solvable 19th-deg
equation with rational coefficients, there is a small matter of 18! ≈ 6.4 x 10^15 possible permutations of the y_n, only 18 of which will be such that R_18 is a polynomial with rational coefficients…
Keywords: moonshine functions, Monster group, prime-generating polynomials, McKay-Thompson series, pi formulas, 163.
Abstract: An expository work on 1) Conway, Norton, and Atkin’s intriguing discovery that the moonshine functions span a linear space of dimension 163; and 2) the functions’ connection to Ramanujan-type pi formulas.
I. Introduction
II. The 172 McKay-Thompson Series and Linear Dependencies
III. More Pi Formulas?
IV. A Continued Fraction Using eπ√163
I. Introduction
Consider the following prime-generating polynomials of form P(n) = an2-an+c,
P(n) = n2-n+41 P(n) = 2n2-2n+19 P(n) = 3n2-3n+23 P(n) = 4n2-4n+5
distinctly prime for n = {1 to c-1}, with the first (studied by Euler in 1772 in the form n2+n+41) for a remarkable 40 consecutive n. And the following related functions,
j(τ) = 1/q + 744 + 196884q + 21493760q2 + 864299970q3 + … (A007240)
r2A(τ) = 1/q + 104 + 4372q + 96256q2 + 1240002q3 + … (A101558)
r3A(τ) = 1/q + 42 + 783q + 8672q2 + 65367q3 + … (A030197)
r4A(τ) = 1/q + 24 + 276q + 2048q2 + 11202q3 + … (A097340)
where (throughout this article),
q = e2πiτ = exp(2πiτ)
The first is the j-function j(τ), a modular form of weight zero, but without the constant term, all these are simply the McKay-Thompson series T1A , T2A, T3A, and T4A of the Monster group. Solving the polynomials at the point P(n) = 0, then,
n = τ = (1+√-163)/2
n = τ = (2+√-148)/4 = (1+√-37)/2 n = τ = (3+√-267)/6
n = τ = (4+√-64)/8 = (1+2√-1)/2
where, consistent with the requirements for modular forms, we have chosen the root τ that is on the upper half of the complex plane. The first has fundamental discriminant -d = 163, the largest (in absolute value) with class number h(d) = 1, while the next two -d = {148, 267}
= {4·37, 3·89} have class number h(d) = 2. Plugging τ into q = exp (2πiτ), we get the very small real numbers,
q = -exp(-π√163)
q = -exp(-π/2√148) = -exp(-π√37) q = -exp(-π/3√267)
q = -exp(-π/4√64) = -exp(-2π)
which, when substituted into the appropriate series, yields the integers, j((1+√-163)/2) = -6403203
r2A((1+√-37)/2) = -141122 r3A((3+√-267)/6) = -3003 r4A((1+√-2)/2) = -29
The first integer appears in the Chudnovsky brothers’ pi formula,
where one can see the 163, as well as in the famous, eπ√163 ≈ 6403203 + 743.9999999999992…
The second integer, -141122, appears in a pi formula found by
Ramanujan in 1912 (before the Monster group was even discovered),
where 4(37) also appears, as well as in the almost-integer, eπ/2√148 = eπ√37 ≈ 141122 + 103.99997…
This, and similar formulas, inspired the one found by Chudnovsky brothers. The third, -3003, appears in another kind of pi formula,
where, one can see |d| = 267, and in the almost-integer,
eπ/3√267 = eπ√(89/3) ≈ 3003 + 41.99997…
Finally, -29 appears in a fourth kind,
Using his own approach, Ramanujan found examples for all four types. (See Pi Formulas and the Monster Group for the complete list of 36 known formulas where the denominator C is an integer.) We instead used four functions based on T1A, T2A, T3A, T4A, but there are in fact many more of them. And here’s an amazing fact: these
moonshine functions span a linear space of dimension exactly 163!
Interesting “coincidence” that it had to be that number, isn’t it?
II. The 172 McKay-Thompson Series and Linear Dependencies