Updates Page 03: Mar

 
 
 

 

(Update, 3/28/10):  A summary of three quartic Diophantine equations:

 

1.  Any primitive integral soln to,

 

a4+b4+c4+d4 = e4        (eq.1)

 

must obey the congruential constraint that there is an addend such that either e2+d2 or e2-d2 is divisible by 54.  The smallest was found by Norrie in 1911 as [30, 120, 315, 272] = [353].  Note that,

 

e+d = 353 + 272 = 54 

e-d  = 353 – 272 = 34 

 

2.  For the eqn,

 

a4+b4+c4 = d4            (eq.2)

 

Morgan Ward proved the added constraint that either c+d or -c+d is divisible by 210.  The smallest is [95800, 414560, 217519] = [422481] found by Roger Frye in 1988.  Note that,

 

c+d = 217519 + 422481 = 21054

 

3.  Finally, for the Swinnerton-Dyer quartic equation,

 

a4+2b4 = c4+4d4        (eq.3)

 

the smallest is {a,b,c,d} = {1484801, 1203120, 1169407, 1157520} found by Andreas Elsenhans and Jorg Jahnel in 2004, and it can be seen that,

 

a+c = 1484801 + 1169407 = 21534

 

Why the smallest solns of these three eqns have sums that obey beautifully simple factorizations, I do not know.  Eq.[1] and [2] can be reduced to an elliptic curve, hence have an infinite number of primitive solns.  However, for [3], this is the only one found after an exhaustive computer search below a bound.  Determining whether [3] has a finite or infinite number of primitive solns would settle a famous open problem in the arithmetic of K3 surfaces (Poonen and Tschinkel).   In general,

 

Problem:  “If there is one non-trivial primitive soln to the quartic Diophantine equation x4+ay4 = z4+bt4, are there an infinite number of primitive solns?”

 

For the special case when a = b, the answer is yes.  But when they are dissimilar, it is still an open problem.  This author knows of only three instances where the answer is known, namely,

 

1)  x4+y4 = z4-t4        (Elkies)

3)  x4+y4 = z4+12t4   (Elkies)

4)  x4+ny4 = z4-4nt4   (Carmichael, Choudhry)

 

If you know of any other case, kindly send an email.  (Update, 4/6/10):  Seiji Tomita has given a fourth class by showing that x4+ay4 = z4+bt4 where {a,b} are integers such that ab = 12n4 for some integer n also has an infinite number of solns.  See April updates.

 

 

(Update, 3/28/10):  By Fermat’s Last Theorem, the quartic equation,

 

x4+y4 = z4

 

has no non-trivial rational solns.  In fact, the same can be said for x4+y4 = z2.  So how are we to explain the near-equalities,

 

245764 + 487674 = (49535.000000000006…)4

 

4199044 + 12577674 = (1261655.000000000000001…)4

 

A search for others with z < 8,000,000 will not yield better approximations.  Noam Elkies showed that an identity is behind it, namely,

 

(192n7)4 + (192n8-24n4-1)4 = (192n8+24n4-1)4 + 12(2n)4

 

See his Table of Fermat Near-Misses.  Since the second term in the RHS is small compared to the three others, then this gives an excellent near-miss of FLT.  However, unlike the even closer x3+y3 = z3±1, it is unknown if,

 

x4+y4 = z4+12t4

 

has an infinite number of formulas.  Q:  Any other identities for x4+ay4 = z4+bt4 with a ≠ b?

 

 

(Update, 3/28/10):  Seiji Tomita gave some pretty identities expressing 1 as a sum/difference of integral powers.  Let a = ±1, then,

 

(2x2+a)2 - (2x2)2 - a(2x)2 = 1

 

(3x2+a)3 - (3x3)3 - a(3x2)3 - (2x)3 - x3 = a

 

(4x4+a)4 - (4x4)4 - a(4x3)4 - a(2x)4 - (3x2)4 - (2x2)4 + (x2)4 = 1

 

For more details, see http://www3.alpha-net.ne.jp/users/fermat/dioph68e.html.

 

 

(Update, 3/23/10):  Choudhry: “Given an initial soln to x4+ny4+4nz4 = t4, then an infinite more can be found.”

 

Choudhry's paper dealt only with the case n = 1, but a small tweak can make it valid for any rational nProof:  Let,

 

x4+ny4+4nz4 = t4    (eq.1)

 

where {x,y,z,t} = {cu+dv,  au-(a-2b)v,  bu-(a-b)v,  du-cv},

 

u = 6n(a2-2ab+2b2)2

v = 4n(a2-2b2)(a2-2ab+2b2)-4cd(c2+d2)

 

and {a,b,c,d} satisfies na4+4nb4+c4 = d4.  (The expressions {u,v} have been modified slightly by this author.)  Since Carmichael’s Identity,

 

(p4-2q4)4 + (2p3q)4 + 4(2pq3)4 = (p4+2q4)4

 

gives an initial soln for the case n = 1, one can then generate an infinite sequence of formulas for eq.1, with the next two having polynomials of deg 8, 16, and so on.  Note that Sophie Germain’s Identity,

 

a4+4b4 = (a2+2ab+2b2)(a2-2ab+2b2)

 

explains why the algebraic form a2-2ab+2b2 appears in the {u,v}.  Choudhry used a similar approach to prove that an initial soln to,

 

x4+4y4 = z4+4t4    

 

non-trivially generates an infinite more.  Source:  On a quartic Diophantine equation, The Mathematics Student, Vol 70, 2001. 

 

Note:  Alain Verghote pointed out that Carmichael’s Identity can be derived from the fact that,

 

(x+y)4 - (x-y)4 = 8x3y + 8xy3

 

Let {x,y} = {p4, 2q4}, then (p4+2q4)4 - (p4-2q4)4 = (2p3q)4 + 4(pq3)4.

 

 

(Update, 3/23/10):  Choudhry:  "If ak+bk+ck = dk+ek+fk,  for k = 1,3, (eq.1 & 2) and abc = def (eq.3), then a+b+c = d+e+f = 0."

 

Proof:  One can use the identity,

 

(a+b+c)3 + 2(a3+b3+c3)-6abc = 3(a+b+c)(a2+b2+c2)

 

and an identical one in the variables {d,e,f}.  If eqs.1,2,3 hold, then

 

(a+b+c)(a2+b2+c2) = (d+e+f)(d2+e2+f2)     (eq.4)

 

For a+b+c = d+e+f ≠ 0, this implies a2+b2+c2 = d2+e2+f2But Bastien’s Theorem excludes a non-trivial soln to,

 

ak+bk+ck = dk+ek+fk,  for k = 1,2,3,

 

Thus, for eq.4 be non-trivially true, it must be the case that a+b+c = d+e+f = 0.  Eqs 1,2,3 are then satisfied if,

 

ak+bk+(-a-b)k = dk+ek+(-d-e)k,  for k = 1,3

 

where ab(a+b) = de(d+e), parametric solns of which are easily found.  (End proof.)  Source: Triads of cubes with equal sums and equal products, The Mathematics Student, Vol. 70, 2001.  Note: This author would independently find an alternative proof using resultants, as well as interesting relations between the terms involving its 6th and 9th powers.

 

(Update, 3/23/10):  While looking for simple identities to u4+v4 = x4+y4+z4, I came across another numerical curiosity about the prime number p = 239.  In addition to being involved in three known Diophantine eqns, there is a fourth and fifth one.  The system is given by,

 

1)  a2-b2 = p      (Factorization)

2)  a2 + b2 = c4     (special Pythagorean triple)

3)  p2 - dc4 = -1      (special Pell equation)

4)  c4 - (ab/8)d4 = 1     (special Pell equation)

5)  a4 + c4+ (cd)4 = b4 + (2cd)4     (ESLP)

 

where {a,b,c,d,p} = {120, 119, 13, 2, 239}.  Since 120 is divisible by 8, then [4] involves the Pell equation x2 - 1785y2 = 1.  And of course, there is the trigonometric identity,

 

 π/4 = 4arctan(1/5) – arctan(1/239)

 

Since p = 239 is a prime, then {a,b} is the unique soln in positive integers.  It is quite easy to resolve this system of four eqns in five variables (excluding [4])into the hyperelliptic curve,

 

2d-30d5+225d9 = (c2d)2

 

and thus prove there is only a finite number of possible integral p, with the only known soln as the unusual prime p = 239.

 

Note 1:  Given x2 - 1785y2 = 1, if {x,y} = {u, ±v} is a soln, then so is {x,y} = {c2u+ab/2v,  d2u+c2v}, with {a,b,c,d} as defined above, and can be recursively applied to find all its integral solns.

 

Note 2:  Incidentally, in the context of Waring's Problem, the number 239 is unique in that it needs the maximum number of nine 3rd powers and nineteen 4th powers in its representation, namely,

 

a3+3b3+4c3+d3 = 239   where {a,b,c,d} = {5,3,2,1}.

2a4+4b4+13c4   = 239   where {a,b,c} = {3,2,1}.

 

  

(Update, 3/22/10):  There are 24 papers/articles added to the Ramanujan Pages.
 

 

(Update, 3/22/10):  Alain Verghote gave,

 

(an+b+c)k + (-an-b+c)k + (a-bn+c)k + (-a+bn+c)k =

(an-b+c)k + (-an+b+c)k + (a+bn+c)k + (-a-bn+c)k,  for k = 1,2,3

 

and pointed out similar identities can be found for 4th and higher powers.  Note how the variable b has simply been negated in the RHS.  Since appropriate pairs from the same side have the common sum 2c, the basis for this is,

 

(a+bn)2 + (an-b)2 = (a-bn)2 + (an+b)2

 

and where Theorem 5 of ESLP has been applied, though Verghote used a different approach.

 

 

(Update, 3/22/10):  Alain Verghote found the nice identity,

 

(x+y)(x+2y)(x+3y)(x+4y) + y4 = (x2+5xy+5y2)2      (eq.1)

 

Thus, the product of 4 numbers in arithmetic progression, plus the 4th power of their common difference, is a square.  Examples,

 

(1)(2)(3)(4) + 14 = 52

(5)(7)(9)(11) + 24 = 592

 

and so on.  Letting y = 1, one can see this is related to Brocard’s Problem which asks to find values of n for which,

 

n!+1 = m2

 

This is an unsolved problem, the only known values for n < 109 are {n,m} = {4, 5}, {5, 11}, {7, 71}, and it is conjectured by Erdos et al that these are the only ones.  More generally, we can find identities like eq.1 by solving,

 

(ax+1)(bx+2)(cx+3)(dx+4) + 1 = (ex2+fx+5)2

 

by collecting powers of x, and choosing {a,b,c,d,e,f} such that the coefficients vanish.  This is an under-determined system of six unknowns in four equations, and which this author found has a final equation that is only linear in one variable.  Thus, this system has an infinite number of rational solns, some integral like Verghote’s or this one,

 

(ax+y)(bx+2y)(cx+3y)(dx+4y) + y4 = (15x2+19xy+5y2)2     

 

where {a,b,c,d} = {1, 3, 5, 15}, and abcd = 152.  We can increase the number of factors by considering the more general, n!+k = m2 for some even n.  Since 6!+9 = 272, then the next is,

 

(ax+1)(bx+2)(cx+3)(dx+4)(ex+5)(fx+6) + 9 = (gx3+hx2+ix+27)2

 

though it is still unknown if there are non-trivial integral {a,b,c,…} that solves this eqn.  (Update, 5/21/10):  Piezas:

 

(ax+y)(bx+2y)(cx+3y)(dx+4y)(ex+5y)(fx+6y) + 9y6 = (12x3+54x2y+72xy2+27y3)2   

 

where {a,b,c,d,e,f} = {2, 2, 3, 2, 2, 3}, and other equivalent forms where abcdef = 122.  Is there a soln to the next level n = 8,

 

(ax+y)(bx+2y)(cx+3y)…(hx+8y) + 81y8 = (px4+qx3y+rx2y2+sxy3+201y4)2   

 

where the coefficients are integers?

 

 

(Update, 3/22/10):  The smallest a4+b4+c4+d4+e4 = f4 is given by,

24+24+44+34+44 = 54

 

and it is hard to miss the Pythagorean triple (3,4,5) tucked away at the side.  Pythagorean triples explain other equations like the pair,

 

24+324+344+134+844 = 854

164+224+384+134+844 = 854

 

since 132+842 = 852.  All three have a common form, and was given by R. Norrie as,

 

(x+y)4 + (-x+y)4 + (2y)4 + (u2-v2)4 + (2uv)4 = (u2+v2)4

 

where 2uv(u2-v2) = x2+3y2.  (This was independently rediscovered by this author and discussed in one section of his paper, Ramanujan and the Quartic Equation 24+24+34+44+44 = 54.)  Thus, if the product of the legs of a right triangle can be expressed as x2+3y2, then it gives a soln to the above equation.  Interestingly, primes of form x2+3y2 can be factored over the Eisenstein integers (which form a triangular lattice in the complex plane), in contrast to the primes of form u2+v2 that factor over the Gaussian integers (which is a square lattice).

 
 

(Update, 3/19/10): The smallest nth power that is the sum of n positive nth powers in two non-trivial ways are,

 

[7, 24]2 = [15, 20]2 = 252

 

[2, 17, 40]3 = [6, 32, 33]3 = 413

 

[2260, 4870, 17386, 30335]4 = [2495, 11998, 16430, 30320]4 = 311274

 

[14, 95, 545, 586, 644]5 = [100, 210, 414, 629, 651]5 = 7445

 

where the last two are by Jarek Wroblewski and James Waldby, respectively.  There is yet no (6.1.6), but there are (6.1.7) and the smallest in two ways is,

 

[258, 600, 2838, 3384, 3441, 5172, 1720]6 = [495, 3018, 3300, 3858, 3912, 4884, 2654]6 = 53276

 

Remarkably, there are also three-way sums,

 

83236 = [547, 7398, 7032, 5748, 4050, 4002, 3792]6

83236 = [547, 7692, 6414, 6168, 2862, 1884, 1206]6

83236 = [6500, 6813, 7344, 1980, 1794, 1272, 648]6

 

and,

 

88696 = [2309, 7368, 7260, 7086, 5952, 4566, 810]6

88696 = [2309, 8598, 6534, 3792, 3468, 894, 264]6

88696 = [5710, 1755, 8028, 7368, 5382, 1278, 330]6

 

all found by Peter Ansell, and the only ones known so far.  Interestingly, they not only share a common sum, but each three-way has a common addend (marked in blue).  Such phenomenon can occur in polynomial identities where a term has only even exponents and hence is immune to sign changes, as in the identity,

 

(9t4)3 + (1-9t3)3 + (3t-9t4)3 = 1

 

For example, for t = {2, -2}, this explains the common terms of,

 

[144, -71, -138]3 = 1

[144, 73, -150]3 = 1

 

though whether an explanation for the form of Ansell’s (6.1.7) triplets can ever be found is uncertain.

 

 

(Update, 3/19/10):  It turns out the (4.1.4) and (6.1.7) equations,

 

a4+b4+c4+d4 = (a+b+c+d)4                              (eq.1)

 

a6+b6+c6+d6+e6+f6+g6= (a+b+c+d+e+f+g)6     (eq.2)

 

have non-trivial solns in the integers given by,

 

Simcha Brudno:  {a,b,c,d} = {-2634, 955, 1770, 5400}

Peter Ansell:  {a,b,c,d,e,f,g} = {-4170, -3187, -888, 1854, 3300, 3936, 4230}.

 

L. Jacobi and D. Madden used the fact that (eq.1) is equivalent to the special Pythagorean triple,

 

(a2+ab+b2)2 + (c2+cd+d2)2 = ((a+b)2+(a+b)(c+d)+(c+d)2)2

 

to reduce it to an elliptic curve and prove it has an infinite number of unscaled integer solns.  See also Sums of Four Fourth Powers.  However, no such reduction is yet known for eq.2 and while Ansell has found three solns (given here), it is unknown: a) if there is an infinite number of them, or b) if one variable can be g = 0.  It would be interesting to consider the higher generalization (8.1.n),

 

x18+x28+…+xn8 = (x1+x2+…+xn)8

 

for some minimal n since there are (8.1.8), (8.1.9), (8.1.10), etc, but nobody apparently has checked if they can be placed in this form if terms are signed.  (MathWorld has a list for 8th powers).

 
 

(Update, 3/16/10):  In response to my post about simple (4.3.3), (5.3.3), and (6.3.3) identities, Seiji Tomita gave a couple he had discovered  previously. Comparing two (4.3.3),

 

Norrie:  (x4y-2y)4 + (xy4+2x)4 + (2x3y)4 = (x4y+2y)4 + (xy4-2x)4 + (2xy3)4

 

Tomita:    (2x4+y4)4  +  (x3y)4 + (2x3y)4 =  (2x4-y4)4 + (3x3y)4 + (2xy3)4

 

Notice how they share a pair of terms.  Others by Tomita include a (4.3.2) and a (4.4.2).

 

(6xy8)4 + (x9-4xy8)4 + (x8y+3x4y5-4y9)4 = (x9+2xy8)4 + (x8y-3x4y5-4y9)4

 

(3xy4)4 + (3x4y)4 + (2x5-2xy4)4 + (2x4y-2y5)4 = (2x5+xy4)4 + (x4y+2y5)4

 

Note:  A broad class of multi-grade (k.3.2) for k = 1,2,4 can be given as ak+bk+(-a-b)k = ck+(-c)k where a2+ab+b2 = c2.  However, to find a uni-grade (4.3.2) like the one above is more difficult.

 
 

(Update, 3/16/10, 3/28/10):  Given the Pythagorean triple {a,b,c} = {x2-1, 2x, x2+1}, then,

 

ak+bk+ck = 2(x8+14x4+1)k/4,  for k = 4,8

 

(a4+b4+c4)3 - 54(abc)4 = 8(x12-33x8-33x4+1)2

 

These two, x8+14x4+1 and x12-33x8-33x4+1, are not ordinary polynomials as, equated to zero, they are octahedral equations.  Why Pythagorean triples, when expressed in a constrained manner, spell these out, I do not know.  See Pythagorean Triples for more details.  Alain Verghote found another nice feature of the two polynomials.  For the first, multiplied by the sum of two squares, it is also the sum of two squares,

 

((x2+y2)x+2xy2)2 + ((x2+y2)y+2x2y)2 = (x2+y2)(x4+14x2y2+y4)

 

where the variables {x,y} are easily changed.  As Verghote pointed out, a small tweak of sign gives a more familiar identity,

 

((x2+y2)x+2xy2)2 - ((x2+y2)y+2x2y)2 = (x2-y2)3

 

and another tweak using the complex unit i =√-1 gives,

 

((x2-y2)x-2xy2)2 + ((x2-y2)y+2x2y)2 = (x2+y2)3

 

For the second polynomial, multiplied by the difference of two squares, it is also the difference of two squares,

 

(x2-y2)(x6-33x4y2-33x2y4+y6) = (x4-17x2y2)2 - (y4-17x2y2)2

 

Q:  Can the second octahedral polynomial be expressed similarly?

 
   
(Update, 3/15/10):  As was previously posted, Duncan Moore has established that more than 90% of solns found by exhaustive search to,

 

x1k+x2k+x3k = y1k+y2k+y3k    (eq.1)

 

for k = 6, with terms below a certain bound, are good for both k = 2,6.  In response to an email, Seiji Tomita found all solns to eq.1 for k = 4 with terms < 500 (all 60,000+ of them! Thanks, Tomita!) and it was seen that only about 28% within that bound are for k = 2,4. 

 

Q:  Why is the percentage of solns to eq.1 for k = 2,6 vs k = 6 much higher than k = 2,4 vs k = 4?

 

The reason why eq.1 is multi-grade can't be merely random chance, otherwise there should be similar percentages.  Furthermore, as the bound increases, the percentage for k = 2,4 is going down,

 

< 100:   34.2 %  

< 200:   31.4 %

< 300:   29.9 %

< 400:   29.0 %

< 500:   28.3 %

 

while for k = 2,6, it is going up,

 

<   6000:   88.9 %  

< 12000:   91.1 %

< 18000:   91.8 %

 

though one can see, at least for the present search radius, that the rate of % decrease or increase is slowing.  It is tempting to speculate if these values taper off to a limit eventually.  Most of the multi-grades for k = 2,4 have the form,

 

ak + bk + (-a-b)k = ck + dk + (-c-d)k       (eq.2)

 

where a2+ab+b2 = c2+cd+d2.  The Ramanujan-Hirshhorn Theorem can apply on these, and it can also be shown, by using the general substitution,

 

(p+q)k + (r+s)k + (t+u)k = (p-q)k + (r-s)k + (t-u)k      (eq.3)

 

that all k = 1,2,4 have the form of eq.2 since solving eq.3 requires only linear equations which then yield terms as a+b+c = d+e+f = 0.

 

 

(Update, 3/15/10):  The Ramanujan-Hirshhorn Theorem is given by 

 

If ak+bk+ck = dk+ek+fk,  k = 2,4 where a+b+c = d+e+f = 0, then,

 

64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2   

 

25(a3+b3+c3-d3-e3-f3)(a7+b7+c7-d7-e7-f7) = 21(a5+b5+c5-d5-e5-f5)2          

 

There are also interesting analogous relations using odd powers k = 1,3, or k = 1,3,5,

 
Piezas
 

Theorem 1:  If ak+bk+ck = dk+ek+fk, k = 1,3 where a+b+c = d+e+f = 0,  then,

 

9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9)     

 

Theorem 2:  If ak+bk+ck+dk = ek+fk+gk+hk,  k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then,

 

7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7)

 

These are discussed more in Fourth Powers.  While parameterizations can be given to both systems, solns are also known to the higher version  ak+bk+ck+dk+ek = fk+gk+hk+ik+jk,  k = 1,3,5,7 where a+b+c+d+e = f+g+h+i+j = 0, but I have not yet been able to tease out similar relations.

    

 

(Update, 3/15/10)  Piezas

 

"If ak+bk+ck = dk+ek+fk,  for k = 2,6, then,

 

(a10+b10+c10-d10-e10-f10)(a4+b4+c4-d4-e4-f4)2 = 20 f1 f2 f3

 

where,

 

f1 = (a2-d2)(a2-e2)(a2-f2)

f2 = (b2-d2)(b2-e2)(b2-f2)

f= (c2-d2)(c2-e2)(c2-f2

 

One can test it with the smallest [23, 15,10] = [22, 19, 3], but it will work for all solns k = 2,6.  Note:  For the analogous 8th power version,

 

x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 2,8   

 

the ratio,

 

Product[(xi2-yi2)] / (x1h+x2h+x3h+x4h-y1h-y2h-y3h-y4h)

 

for h = 12 (or h = 14) does not seem to be the product of simple rational polynomials in the {xi, yi}, which makes the case k = 2,6, in light of its abundance of solns, even more unusual.

 

 

(Update, 3/15/10):  We have the beautiful identities,

 

R. Norrie:

 

(x4y-2y)4 + (xy4+2x)4 + (2x3y)4 = (x4y+2y)4 + (xy4-2x)4 + (2xy3)4

 

Sastry, Chowla:

 

(x5+25y5)5 + (x5-25y5)5 + (10x3y2)5 = (x5+75y5)5 + (x5-75y5)5 + (-50xy4)5

 

though a simple analogous version for,

 

x16+x26+x36 = y16+y26+y36   

 

is not yet known.

 

 

(Update, 3/15/10)  A useful theorem in elementary Diophantine equations is,

 

Theorem:  “Given an initial integral soln {x,y} to,

 

ax2+bxy+cy2 = M      (eq.1)

 

then, in general, an infinite more can be found.”  This has been long known and a proof by Euler for b = 0 (which eq.1 can be transformed to) is,

 

"Given an initial soln {p,q}, if mp2-nq2 = c, then mx2-ny2 = c where,

 

x = (pu2+2nquv+mnpv2)/(u2-mnv2),  y = (qu2+2mpuv+mnqv2)/(u2-mnv2)

 

and an infinite number of integral {x,y} can be found by solving u2-mnv2 = ±1."  See also Pell Equations.  There is also a simple proof for non-zero b using recursions and Pell equations.  Proof (Piezas):  Given (eq.1), use the recursion,

 

{xn+1, yn+1} = {pxn+qyn, 2avxn+ryn}

 

with constants {p,q,r} = {u-bv, -2cv, u+bv} and {u, ±v} is a soln to the Pell equation,

 

u2-(b2-4ac)v2 = 1

 

Example:  Let 3x2+2xy-7y2 = 11 with initial soln {xn, yn} = {3, 2}.  Since {a,b,c} = {3,2,-7}, the associated Pell equation is u2-88v2 = 1, where the fundamental soln is {u,v} = {197, 21}.  Therefore,

 

Using +v:  {xn+1, yn+1} = {155xn+294yn,  126xn+239yn}

Using -v:   {xn+1, yn+1} = {239xn-294yn,  -126xn+155yn}

 

Such recursions are also given by an applet, the Quadratic two integer variable equation solver, in Dario Alpern’s remarkable Alpertron.

 

 

(Update, 3/8/10):  There is a family of identities,

 

Piezas:  (2x+1)2 + (2x2+2x-1)2 - (2x2+2x)2 = 2

 

Dean Hickerson:  (6x3+1)3 + (-6x3+1)3 + (-6x2)3 = 2

 

Seiji Tomita(8x5-2x)4 + (8x4+1)4 + (8x4-1)4 - (8x5+2x)4 - (4x2)4 = 2

 

Choudhry:  (-8x6+2x)5 + (-8x6-2x)5 + (8x5+1)5 + (-8x5+1)5 + 2(8x6)5 = 2

 

which show that "2" is the sum/difference of kth powers of integers for k = 2,3,4, or 5 in an infinite number of ways.  (Hickerson's entry is from Kevin Brown’s Mathpages, Sums of Three Cubes.)  Q:  Can anyone find the analogous versions for 6th, 7th, and higher powers with a minimal number of terms?

 
 

(Update, 3/8/10):  The pair of simultaneous equations,

 

p2+12q2 = r2      (eq.1)

12p2+q2 = s2      (eq.2)

 

call this P12, is at the heart of the beautifully simple Letac-Sinha (8.5.5) Identity,

 

(p+r)k + (p-r)k + (3q+s)k + (3q-s)k + (4p)k = (q+s)k + (q-s)k + (3p+r)k + (3p-r)k + (4q)k,   for k = 1,2,4,6,8

 

where {p,q,r,s} must satisfy P12.  The ratios p/q = {1/2, 2} is trivial, with the smallest non-trivial soln as {p,q} = {218, 11869}.  Since this pair of quadratic conditions defines an elliptic curve, then there is an infinite number of rational points.  The identity can be derived from more general ones in at least three ways since P12 can solve three systems, namely,

 

1.  2(a2+b2+c2) = m2+11n2;  and 8(a4+b4+c4) = m4+54m2n2+89n4 

 

Soln:   {a,b,c; m,n} = {3q+s, 3q-s, 2q; 2r, 2p}

 

2.  ak+bk+ck = dk+ek+fk,  for k = 2,4 

 

Soln:  {a,b,c; d,e,f} = {p+r/3, p-r/3, 2p/3; q+s/3, q-s/3, 2q/3}

 

3.  ak+bk+ck+dk = ek+fk+gk+hk,  for k = 2,4, and abcd = efgh

 

Soln:  {a,b,c,d; e,f,g,h} = {p-r, p+r, 2p, 4p; q-s, q+s, 2q, 4q}

 

where, for all three, {p,q,r,s} satisfies P12, though other parametric solns also exist.  Note how they are simple and symmetric expressions in the {p,q,r,s}.  The more general identities are,

 

1.  Wroblewski  (8.6.6)

 

If 2(a2+b2+c2) = m2+11n2, and 8(a4+b4+c4) = m4+54m2n2+89n4, then,  

 

(2a)k + (2b)k + (2c)k + (m+n)k + (m-n)k + (4n)k = (a+b+c)k + (a-b+c)k + (a+b-c)k + (a-b-c)k + (m+3n)k + (m-3n)k,  for k = 2,4,6,8

 

2.  Birck-Sinha  (8.7.7)

 

If ak+bk+ck = dk+ek+fk,  k = 2,4, where a+b ≠ c; d+e ≠ f, then,

 

(a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2d)k + (2e)k + (2f)k =

(d+e+f)k  + (-d+e+f)k +  (d-e+f)k + (d+e-f)k +  (2a)k + (2b)k + (2c)k,   for k = 2,4,6,8

 

3. Piezas  (8.8.8)

 

If ak+bk+ck+dk = ek+fk+gk+hk,  for k = 2,4, where a+b ≠ ±(c+d); e+f ≠ ±(g+h), and abcd = efgh, then,

 

(a+b+c-d)k + (a+b-c+d)k + (a-b+c+d)k + (-a+b+c+d)k + (2e)k + (2f)k + (2g)k + (2h)k =

(e+f+g-h)k  + (e+f-g+h)k  + (e-f+g+h)k  + (-e+f+g+h)k  + (2a)k + (2b)k + (2c)k + (2d)k,   for k = 2,4,6,8.

 

Each identity has more terms and is more general than the next but, with the substitutions given in {p,q,r,s}, all will reduce to the simpler Letac-Sinha (8.5.5) since some terms vanish and/or cancel out.  (There is actually a 4th way given by Wroblewski which entails a different (8.6.6) valid, in general, only for k = 2,4,8, but can also reduce to the Letac-Sinha.)  The Choudhry-Wroblewski (10.6.6) Identity, in a form given by this author, is analogous but whether this is just a special case of more general identities remains to be seen.  (End update.)

 

 

(Update, 3/8/10)Alain Verghote gave the (k.8.8) identity,

 

(a+b+c)k+(a+b+d)k+(a+c+d)k+(b+c+d)k+ak+bk+ck+dk = 0k+(a+b)k+(a+c)k+(a+d)k+(b+c)k+(b+d)k+(c+d)k+(a+b+c+d)k,   k = 1,2,3

 

Note how appropriate pairs from the same side of the equal sign all have the common sum a+b+c+d.  The basis then is the (k.4.4) identity,

 

(-a+b+c+d)2 + (a-b+c+d)2 + (a+b-c+d)2 + (a+b+c-d)2 = (a+b+c+d)2 + (a+b-c-d)2 + (a-b+c-d)2 + (a-b-c+d)2

 

and where Theorem 5 of ESLP has been applied, though Verghote derived it differently using integration.  (End update.)

 

 

(Update, 3/7/10):  These are three forms of the same identity:

 

1. J. Zehfuss

 

(2a)k + (2b)k + (2c)k + (2d)k  = (-a+b+c+d)k + (a-b+c+d)k  + (a+b-c+d)k  + (a+b+c-d)kfor k = 1,2 

 
This can also be expressed, after minor changes in signs as,

 

2. M. Hirschhorn

 

(-4p-1)k + (4q+1)k + (4r+1)k + (4s+1)k = 4(p+q+r+s+1)k + 4(-p+q+r-s)k + 4(-p-q+r+s)k + 4(-p+q-r+s)k,  for k = 1,2 

 

or, Hirschhorn's Odd-Even Identity, proving that the sum of four distinct odd squares is the sum of four distinct even ones.  Proof:  This can easily be shown true by equating the terms on the LHS of both forms, solving for {p,q,r,s}, and substituting into the RHS of the second.  A third equivalent form, more symmetrical in one sense, is,
 
3. J. Wroblewski
 

(t+x+y+z)k +  (t-x-y+z)k + (t-x+y-z)k + (t+x-y-z)k = (t-x-y-z)k +  (t+x+y-z)k + (t+x-y+z)k + (t-x+y+z)k,  for k = 1,2

 

where {x,y,z} in the LHS is just negated in the RHS.

 
 
(Update, 3/2/10):  Duncan Moore has done an exhaustive search of  a6+b6+c6 = d6+e6+f6 within a radius of 17,800.  Out of about 400 solns, it turns out that 92% (!) are good for k = 2,6.  Why that is the case -- and how high the percentage gets the bigger the radius -- no one knows, though it may have to do with the algebraic form x2+xy-y2.  (See Bremner and Kuwata's work below.)  Data results are hereThe smallest is,

 

23k + 10k + 15k = 3k + 19k + 22k

 

which is for k = 2,6.  This deceptively simple-looking eqn has a lot of structure.  If expressed as,

 

{-23, 10, 15} = {3, 19, -22}

 

labelled {a,b,c,d,e,f} respectively, then,

 

a2+ad-d2 = -(b2-be-e2)

b2+be-e2 = -(c2-cf-f2)

c2+cf-f2  = -(a2-ad-d2)

3a+b+c = d+e+3f

 

Not surprisingly, this is just the smallest instance of a parametric soln.  It was seen that for 3rd and 4th powers, a lot of identities involved equivalent forms of F3:= x2+xy+y2.  For 5th and 6th powers, it is now the form F5:= x2+xy-y2 that is implicit in some identities.  It has been shown that solns to a2+ab+b2 = c2+cd+d2 also solve ak+bk+(a+b)k = ck+dk+(c+d)k  for k = 2,4, and vice versa.  For k = 2,6, one needs a system of three equations,
 

Theorem 1:  (Bremner, Kuwata)  If there are {a,b,c,d,e,f} such that, call this system S1,

 

a2+ad-d2 + (b2-be-e2) = 0

b2+be-e2 + (c2-cf-f2) = 0

c2+cf -f2 + (a2-ad-d2) = 0

 

is true, then so is system S2,

 

ak+bk+ck = dk+ek+fk,  for k = 2,6

 

Proof:  Simply add the expressions together to get the identically true statement,

 

(a2+ad-d2)k + (b2-be-e2)k + (b2+be-e2)k  + (c2-cf-f2)k + (c2+cf-f2)k + (a2-ad-d2)k  = 2(a2k+b2k+c2k-d2k-e2k-f2k),  for k = 1,3

 

If the LHS is zero, then so is the RHS, which proves the theorem.  (Masato Kuwata, Equal Sums of Sixth Powers and Quadratic Line Complexes, 2007. End proof.)  (Note:  This probably contributes to the reason why more than 90% of solns are multi-grade for k = 2,6, since being valid for one power also means the other.  Unfortunately, no similar expressions are yet known for 8th or higher powers!) 

 

Corollary 1: (Piezas) Solutions to S1 imply the ff sums are squares,

 

4(a2+ad-d2) + 5b2 = (b+2e)2,   4(b2+be-e2) + 5c2 = (c+2f)2,   4(c2+cf-f2) + 5a2 = (a+2d)2

 

4(a2-ad-d2) + 5c2 = (c-2f)2,      4(b2-be-e2) + 5a2 = (a-2d)2,   4(c2-cf-f2) + 5b2 = (b-2e)2

 

In Unsolved Problems in Number Theory, R. Guy asked if the reverse was true: does S2 imply S1?  This was answered in the negative by Delorme who gave one more condition, and Choudhry who gave a parametric example that solved S2 but not S1.

 

Theorem 2: (Delorme) The system S2 implies S1 if there is a signed combination of the terms such that,

 

ad(a2-d2) + be(b2-e2) + cf(c2-f2) = 0    

 

S. Brudno

 

This is a simple example that solves both systems,

 

ak + bk + ck = dk + (b+c)k + (b-c)k,   k = 2,6

 

Note that for k = 2, one must solve a2 = b2+c2+d2.  The complete soln for k = 2,6 is then,

 

{c,d} = {ax/(a2+9b2),  3bx/(a2+9b2)}

 

where {a,b,x} satisfies the elliptic curve,

 

(a2-b2)(a2+9b2) = x2

 

Labelled as xi, yi, the terms also satisfy x1y1(x12-y12) + x2y2(x22-y22) + x3y3(x32-y32) = 0.  One small soln, among infinitely many, is {a,b} = {5/4, 1} which yields,

 

65k + 52k + 15k = 36k + 67k + 37k 

 

Piezas

 

This is another simple example that also solves both,

 

(ad+b)k + (c-2d)k + (de+f)k = (de-f)k + (c+2d)k + (ad-b)k,   k = 2,6

 

where {a,b,c,d,e,f} = {-4x+11y,  y,  2(x-3y)(x-2y)+y2,  x-y,  2x-3y,  2x-5y},  and {x,y} satisfies x2-6y2 = 1. 

 

It is quite interesting that this particular Pell equation appears in the context of 6th powers.  However, since the eqn is homogeneous, one can just as well solve it in the rationals.  See also Sixth Powers.  (End update.)

   

 

(Update, 3/1/10):  Last Jan, Wroblewski gave 3 formulas for the multi-grade (k.10.10) with k = 2,4,6,8,10 using the form,

 

[pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb, ta+ub, ua-tb, va+wb, wa-vb]k =

[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb, ta-ub, ua+tb, va-wb, wa+vb]k

 

where b is just negated in the RHS and independent variables {a,b} must satisfy the quadratic conditions ma2+nb2 = c2 and na2+mb2 = d2.  These two define an elliptic curve and one rational point {a,b} can be used to generate an infinite more.  He found three non-trivial solns,

 

{p,q,r,s; t,u,v,w; m,n} = {4, 4, 3, 11; 5, 7, 7, -3; 45, -11}

{p,q,r,s; t,u,v,w; m,n} = {4, 4, 9, 17; 15, 13, 13, 7; 189, 85}

{p,q,r,s; t,u,v,w; m,n} = {3, 3, 11, 17; 16, 14, 14, 10; 220, 136}

 

though certain ratios a/b must be avoided and which can determined by expanding at k = 12.  Note that the first has the same conditions as the only known formula for (k.6.6) with k = 2,4,6,8,10.

 

Piezas

 

This author noticed that, other than the obvious relations p = q, u = v, the above also obey,

 

r = -q+v

s = q+v

mn = rstw

 

Combined with the conditions k = 2,4,6,8,10, these were enough to find a formula for these variables.  Thus,

 

{p,q,r,s; t,u,v,w} = {y, y, -y+z, y+z; x+y, z, z, x-y}

{m,n} = {x2+yz+y2,  x2-yz+y2}

 

where {x,y,z} satisfies the simple quadratic condition x2+3y2 = z2.  Wroblewski’s were the cases,

 

{x,y} = {1, 4}; {11, 4}; {13, 3}

 

However, there is an infinite number of them given by {x,y,z} = {e2-3f2, 2ef, e2+3f2}.  So a fourth one using {x,y} = {11, 5} gives,

 

{p,q,r,s; t,u,v,w; m,n} = {5, 5, 9, 19;  16, 14, 14, 6;  216, 76}

 

with one soln to 216a2+76b2 = c2 and 76a2+216b2 = d2 as,

 

{a,b} = {495753715, 352750681}

 

though presumably smaller solns may exist.  From this initial rational point, one can then compute an infinite more.  And so on for an infinite number of other formulas using different {m,n} and other elliptic curves, though expanding this family at k = 12 or 14 generates only trivial solns.  See also Tenth Powers.  (End update.)

 

 

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

 

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