Updates Page 08: Aug

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(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 j-function And Its Cousins) explaining why eπ√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, eπ√163.  Should be interesting.



(Update, 8/17/10):  Roland van den Brink gave the identity,


(A+2B)A3 + 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)A3 + 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} = {a-2b, b}, we get a soln to the equation ax3+by3 = cy3 as the symmetrical form,


a(a-2b)3 + b(2a-b)3 = (a-b)(a+b)3


The expressions {a-2b, 2a-b, a+b, a-b} are intimately connected to the algebraic form x2+3y2 since,


(a-2b)2+3a2 = (2a-b)2+3b2 = (a+b)2+3(a-b)2 = 4(a2-ab+b2)


Furthermore, by a result of E.Grebe, let {x,y,z,t} = {a-2b, 2a-b, a+b, a-b}, then the following are all squares,


-x2+2y2+2z2 = (3a)2

2x2-y2+2z2  = (3b)2

2x2+2y2-z2  = (3t)2



(Update, 8/16/10):  Euler also considered the system,


u2+v2-w2  = a2

u2-v2+w2  = b2

-u2+v2+w2 = c2


Solving for {u,v,w}, the problem is equivalent to finding generalized Euler bricks {a,b,c} such that,


a2+b2 = nu2     (eq.1)

a2+c2 = nv2     (eq.2)

b2+c2 = nw2    (eq.3)


call this system Sn, for n = 2.  Jarek Wroblewski pointed out that Sn has non-trivial solns in the integers only for n = 1,2.  Excluding the trivial a = b = c, the smallest for S2 is {a,b,c} = {1,1,7}.  These can be parameterized as,


{a,b,c} = {p2-2q2, p2-2q2, p2+4pq+2q2}


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 x2-2y2 = 2z2 as,


{a,b,c} = {(x2+2xy-y2)z,  (x2-2xy-y2)z,  (x2-3y2)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,


a2+b2+c2 = nt2     (eq.4)


for n = 2 is solvable along with eq.1,2,3 is also unknown.



(Update, 8/15/10):  The Diophantine eqn,


p4+2np2q2+q4 = r4+2nr2s2+s4      (eq.1)


is equivalent to both,


(p2-q2)2 + m1(2pq)2 = (r2-s2)2 + m1(2rs)2


(p2+q2)2 + m2(2pq)2 = (r2+s2)2 + m2(2rs)2


where {m1, m2} = {(n+1)/2, (n-1)/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 non-trivial odd n such that {m1, m2} 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 non-trivial and unscaled solns to eq.1 for any n.  If specialized to v = 2 (since v = 1 is trivial), we get,


C1 := {p,q,r,s} = {4n3-86n2-160n-133,  48n3-22n2-35n+134,  12n3+82n2-160n-59,  16n3+106n2+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 C1, after removing common factors, give the smallest non-trivial integer soln to eq.1 for n > 1?  Can anyone find a counter-example?   (Smallest being positive integers {p,q,r,s} are all below a smallest possible bound B.)  (Update, 8/22/10):   Seiji Tomita showed that C1 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)(x2+y2) = 2(n-3)z2


though this has non-trivial solns only for some constant n.  The case n = 7 is particularly nice as the condition is reduced to the Pythagorean triple,


x2+y2 = z2



(Update, 8/14/10):  The Diophantine eqn,


x2+y2+z2 = N2k


for k > 1 can be solved for non-zero and co-prime {N,x,y,z,} in terms of co-prime Pythagorean triples {a,b,c}.  If a2+b2 = c2, then,


x2+y2+z2 = c4,  {x,y,z} = {ab, b2, ac}

x2+y2+z2 = c6,  {x,y,z} = {2a2b, 2ab2, (a2-b2)c}

x2+y2+z2 = c8,  {x,y,z} = {4a2b2, 2ab(a2-b2),  (a2-b2)c2}


and so on.  For example, let, {a,b,c} = {3, 4, 5}, then,


122 + 162 + 152 = 54

722 + 962 + 352 = 56

5762 + 1682 + 1752 = 58


Equivalently, given an integer N that is the sum of two non-zero squares u2+v2 = N, then from {u,v}, one can always compute non-zero {x,y,z} such that  N2k is the sum of three squares x2+y2+z2 = N2k for k > 1.



(Updates on Mengoli's Six-Square Problem have been condensed into an article here.) 


(Updates on Euler Quadruples are here.) 



You can email the author at tpiezas@gmail.com.