Back to Index of Updates Back to July updates Go to Sept updates
(Update, 9/24/10): Page added on Article 3 (Pi Formulas and the Monster Group). Easy reading on how the largest of the finite simple groups can be a unifying framework for 4 kinds of Ramanujan's pi formulas.
(Update, 9/13/10): Page added on Article 1 (The jfunction And Its Cousins) explaining why e^{π√58} and other similar transcendental constants involving d with class number h(d) = 2 are also close to an integer.
(Update, 8/26/10): Page added on Article 0 (Pi formulas 1) involving that fascinating constant, e^{π√163}. Should be interesting.
(Update, 8/17/10): Roland van den Brink gave the identity,
(A+2B)A^{3} + B(2A+3B)^{3} = (A+B)(A+3B)^{3}
He pointed out this appears in the context of the abc conjecture as,
“If A,B,C is an ABC triple, then (A+2B)A^{3} + B(2A+3B)^{3} = (A+B)(A+3B)^{3} is a new ABC triple if mod(A,3) > 0.”
Note: Using the transformation {A,B} = {a2b, b}, we get a soln to the equation ax^{3}+by^{3} = cy^{3} as the symmetrical form,
a(a2b)^{3} + b(2ab)^{3} = (ab)(a+b)^{3}
The expressions {a2b, 2ab, a+b, ab} are intimately connected to the algebraic form x^{2}+3y^{2} since,
(a2b)^{2}+3a^{2} = (2ab)^{2}+3b^{2} = (a+b)^{2}+3(ab)^{2} = 4(a^{2}ab+b^{2})
Furthermore, by a result of E.Grebe, let {x,y,z,t} = {a2b, 2ab, a+b, ab}, then the following are all squares,
x^{2}+2y^{2}+2z^{2} = (3a)^{2} 2x^{2}y^{2}+2z^{2} = (3b)^{2} 2x^{2}+2y^{2}z^{2} = (3t)^{2}
(Update, 8/16/10): Euler also considered the system,
u^{2}+v^{2}w^{2} = a^{2} u^{2}v^{2}+w^{2} = b^{2} u^{2}+v^{2}+w^{2} = c^{2}
Solving for {u,v,w}, the problem is equivalent to finding generalized Euler bricks {a,b,c} such that,
a^{2}+b^{2} = nu^{2} (eq.1) a^{2}+c^{2} = nv^{2} (eq.2) b^{2}+c^{2} = nw^{2} (eq.3)
call this system S_{n}, for n = 2. Jarek Wroblewski pointed out that S_{n} has nontrivial solns in the integers only for n = 1,2. Excluding the trivial a = b = c, the smallest for S_{2} is {a,b,c} = {1,1,7}. These can be parameterized as,
{a,b,c} = {p^{2}2q^{2}, p^{2}2q^{2}, p^{2}+4pq+2q^{2}}
where the smallest is {p,q} = {1,1}. For distinct and primitive {a,b,c}, Wroblewski found for bound B < 6000, only 7, namely,
{329, 191, 89} {527, 289, 23} {833, 553, 97} {1081, 833, 119} {1127, 697, 17} {4991, 2263, 287} {5609, 4991, 1871}
The smallest was known to Euler as an instance of a parametric family that depended, perhaps not surprisingly, on the simple eqn x^{2}2y^{2} = 2z^{2} as,
{a,b,c} = {(x^{2}+2xyy^{2})z, (x^{2}2xyy^{2})z, (x^{2}3y^{2})y}
Let {x,y,z} = {10, 7, 1} and this gives the smallest with distinct {a,b,c}. Note also how some triples share a common term, a phenomenon also present with face cuboids and was explained by three identities that, two at a time, share a common term. However, I haven't yet found corresponding identities for these pairs of "Euler bricks" over √2. And whether,
a^{2}+b^{2}+c^{2} = nt^{2} (eq.4)
for n = 2 is solvable along with eq.1,2,3 is also unknown.
(Update, 8/15/10): The Diophantine eqn,
p^{4}+2np^{2}q^{2}+q^{4} = r^{4}+2nr^{2}s^{2}+s^{4} (eq.1)
is equivalent to both,
(p^{2}q^{2})^{2} + m_{1}(2pq)^{2} = (r^{2}s^{2})^{2} + m_{1}(2rs)^{2}
(p^{2}+q^{2})^{2} + m_{2}(2pq)^{2} = (r^{2}+s^{2})^{2} + m_{2}(2rs)^{2}
where {m_{1}, m_{2}} = {(n+1)/2, (n1)/2}. Hence, it can be interpreted as the problem of finding two triangles with equal sums of:
a) the square of one leg plus a rational multiple of the square of the other b) the square of the hypotenuse plus a rational multiple of the square of a leg.
Gerardin considered n = 0, as well as the smallest nontrivial odd n such that {m_{1}, m_{2}} are integers, namely n = 3, and gave a parametric 7th deg soln to both cases. Choudhry would later give a more general one, also of 7th deg in a free variable v, for any n, thus proving there were an infinite number of nontrivial and unscaled solns to eq.1 for any n. If specialized to v = 2 (since v = 1 is trivial), we get,
C_{1} := {p,q,r,s} = {4n^{3}86n^{2}160n133, 48n^{3}22n^{2}35n+134, 12n^{3}+82n^{2}160n59, 16n^{3}+106n^{2}+95n+158}
The first few n give {p,q,r,s} as,
n = 0; {133, 134, 59, 158} (smallest) n = 1; 125{3, 1, 1, 3} (trivial) n = 2; 45{17, 8, 1, 20} (smallest) n = 3; {1279, 1127, 523, 1829} (smallest?)
Question: Does Choudhry’s soln C_{1}, after removing common factors, give the smallest nontrivial integer soln to eq.1 for n > 1? Can anyone find a counterexample? (Smallest being positive integers {p,q,r,s} are all below a smallest possible bound B.) (Update, 8/22/10): Seiji Tomita showed that C_{1} does not necessarily give the smallest solns. For n = 3, it is {p,q,r,s} = {35, 192, 123, 116} answering in the negative the question above, while for n = 4, it is simply {1, 4, 2, 3}. The form of the latter suggests an alternative approach. Let {p,q,r,s} = {x+z, x+z, y+z, y+z} and eq.1 transforms into the 2nd deg eqn,
(n+1)(x^{2}+y^{2}) = 2(n3)z^{2}
though this has nontrivial solns only for some constant n. The case n = 7 is particularly nice as the condition is reduced to the Pythagorean triple,
x^{2}+y^{2} = z^{2}
(Update, 8/14/10): The Diophantine eqn,
x^{2}+y^{2}+z^{2} = N^{2k}
for k > 1 can be solved for nonzero and coprime {N,x,y,z,} in terms of coprime Pythagorean triples {a,b,c}. If a^{2}+b^{2} = c^{2}, then,
x^{2}+y^{2}+z^{2} = c^{4}, {x,y,z} = {ab, b^{2}, ac} x^{2}+y^{2}+z^{2} = c^{6}, {x,y,z} = {2a^{2}b, 2ab^{2}, (a^{2}b^{2})c} x^{2}+y^{2}+z^{2} = c^{8}, {x,y,z} = {4a^{2}b^{2}, 2ab(a^{2}b^{2}), (a^{2}b^{2})c^{2}}
and so on. For example, let, {a,b,c} = {3, 4, 5}, then,
12^{2} + 16^{2} + 15^{2} = 5^{4} 72^{2} + 96^{2} + 35^{2} = 5^{6} 576^{2} + 168^{2} + 175^{2} = 5^{8}
Equivalently, given an integer N that is the sum of two nonzero squares u^{2}+v^{2} = N, then from {u,v}, one can always compute nonzero {x,y,z} such that N^{2k} is the sum of three squares x^{2}+y^{2}+z^{2} = N^{2k} for k > 1.
(Updates on Mengoli's SixSquare Problem have been condensed into an article here.)
(Updates on Euler Quadruples are here.)
You can email the author at tpiezas@gmail.com.
