001b: Assorted Identities, Part 2 (Waring-like Problems)

 
 
 
IV.  Waring-like Problems
 
1. Summary
 
The following identities are good for Waring-like problems since they involve rational, instead of integral, terms.  Known results are, 
 
"Any non-zero rational N, in an infinite number of non-trivial ways, is the sum/difference:
 
1) of 2 rational 2nd powers (easy)
2) of 3 rational 3rd powers (Ryley)
3) of 4 rational 4th powers (Norrie)
4) of 6 rational 5th powers (Choudhry)
5) of 8 rational 6th powers (Piezas)
6) of 8 rational 7th powers (Choudhry)
7) of 12 rational 8th powers (Choudhry)."
 
It would be nice if it would be proven in a general way that, "...any rational N is non-trivially the sum/difference of k rational kth powers", but these are the best results so far.  If you know of others, pls submit them.  (An update on sums of increasing kth powers is given at the end of this section.)
 
 
2. Form: a2+b2+c2+d2 = N
 
By Langrange's Four-Square Theorem, any positive integer N is the sum of four integral squares.  If we extend this to rational numbers, then,
 
Theorem:  "Any positive rational N is the sum of four rational squares in an infinite number of ways."
 
Proof:  x12+x22+x32+x42 = (y12+y22+y32+y42)(z12+z22+z32+z42)2

 

where {x1, x2, x3, x4} = {uy1-vz1,  uy2-vz2,  uy3-vz3, uy4-vz4} and {u,v} = {z12+z22+z32+z42,  2(y1z1+y2z2+y3z3+y4z4)}

 

with initial solution yi to y12+y22+y32+y42 = N and four arbitrary zi.  Since there is always yi for any positive rational N, then by dividing the eqn with the square factor of the RHS, one can find an infinite number of xi. This is just a particular case of a general identity involving n sums of squares discussed more here.

 

Note:  It is easily proven that the rational number p/q is the sum of four rational squares.  Multiplying this by q/q, the denominator becomes a square and the problem is reduced to expressing pq as the sum of four squares.  Also, four squares can't generally be reduced to three since one can't solve a2+b2+c2 = Ny2,  where N = 8m+7.  Thus, Lagrange's Four-Square Theorem applies whether the variables are integral or rational.   

 

 

3a. Form: x3+y3+z3 = N

 

Ryley's Theorem: “Any non-zero rational number N is the sum of three rational cubes in an infinite number of non-trivial ways.”  (S. Ryley, 1825)
 
Proof(p3+qr)3 + (-p3+pr)3 + (-qr)3 = N (6Nvp2)3,  where {p,q,r}= {N2+3v3,  N2-3v3,  36N2v3},  for arbitrary v.
 
Corollary:  "Any positive rational N is the sum of three positive rational cubes in an infinite number of ways."
 
Since v is an arbitrary variable, one can choose it such that all the terms are positive.  For small integral N, the ranges roughly are:
 
N = 1:  v = {1.32 - 1.42}
N = 2:  v = {1.20 - 1.40}
N = 3:  v = {2.8 - 3.0}
N = 4:  v = {3.4 - 3.7}
 
and so on.  Q:  Does anyone know how to calculate v and express it in terms of N such that all terms are positive?
 
(Update, 11/11/09):  Laurent Bartholdi from sci.math.research gave the answer.  Setting x = N2 and y = v3 for ease of notation, one gets the constraints,

 

x-3y < 0

x2-30xy+9y2 < 0

x3+45x2y-81xy2 +27y3 > 0

 

which define 6 lines in the {x,y} plane delimiting two regions.  These regions are,

 

a)  1 < 3v3/N2 < 1+Sec[2π/9], 

b)  1+Sec[4π/9] < 3v3/N2 < 5+2√6

 

where the exact values in terms of the secant function was pointed out by Warut Roonguthai from the NMBRTHRY mailing list. Or, v3 is approx. within,

 

a)  0.333N2 < v3 < 0.768N2,

b)  2.253N2 < v3 < 3.299N2,

 

For ex, for N = 163, then 20.7 < v < 27.3, and 39.2 < v < 44.4, and one can choose an infinite number of rational v within those two ranges.  (End update.)

 

(Update, 4/19/10):  It is easily seen that Ryley's Identity is a 6th-deg polynomial in the constant N.  A 3rd-deg polynomial, in fact, is possible.  William Ellison cited one example from the book Cubic Forms by Yuri Manin as,

 

(m3-36n9)3 + (-m3+35mn6+36n9)3 + (33m2n3+35mn6)3 = m(32m2n2 +34mn5+36n8)3

 

for arbitrary n.  Note how this has neat coefficients that are only powers of 3.  Robert Israel pointed out that a small tweak can reduce the powers as,

 

(27m3-n9)3 + (-27m3+9mn6+n9)3 + (27m2n3+9mn6)3 = m(27m2n2 +9mn5+3n8)3

 

(End update.)

 

3b. Form: (a3+b3)(c3+d3) = N

 

Theorem: “Any rational number N is the product of two sums of rational cubes."

 

Proof:  ((1+18N-27N2)3 + (-1+18N+27N2)3) ((1+3N)3 + (1-3N)3) = 63N (1+27N2)3

 

where the RHS is then divided by its cubic factor.  Q: Who discovered this identity?  Euler? Lebesgue?

 

4. Form: x4+y4-(z4+t4) = N

 

Norrie's Theorem: “Any rational number N is the sum and difference of four rational fourth powers in an infinite number of ways.”  (R. Norrie)

 

((2a+b)c3d)4 + (2ac4-bd4)4 - (2ac4+bd4)4 - ((2a-b)c3d)4 = a(2bcd)4

 

where b = c8-d8, for arbitrary {c,d}.
 
 
5. Form: x15+x25+x35+x45+x55+x65 = N
 
(Update, 11/17/09)Choudhry's Fifth Powers Theorem: “Any rational number N is the sum of six rational fifth powers in an infinite number of non-trivial ways.”  Proof:
 

(x+a)5 + m5(x+b)5 + m10(x+c)5 - (x+d)5 - m5(x+e)5 - m10(x+f)5 = -40v35(v30-1)(v10+1)x + 2v30(-v30+12v20+8v10+1)(v30-1)(v10+1)

 

where,

 

{a,b,c,d,e,f,m} = {-v20,  v5,  -v15-v5-1,  v20,  -2v15-v5,  -v15-v5+1,  v3}

 

Since the RHS is only linear in x, one can equate that side to any N, easily solve for x, and express the LHS in terms of N and the free variable v.  (End proof.) 

 
Note:  Method 1 below was when I didn't had access yet to Choudhry's paper, "Representation of every rational number as an algebraic sum of fifth powers of rational numbers", so I came up with my own.  As this update is after the one for Seventh Powers, this incorporates some insights from that section.  Method 2 is Choudhry's (yielding the identity above) after he gave me a copy of his paper.
 
Method 1.  Proof (Piezas):  Expand the system,
 
F(x): = (x+a)5+(x+b)5+(mx+mc)5-(x+d)5-(x+e)5-(mx+mf)5
 
and collecting powers of x, this resolves to,
 
F(x): = 5P1x4 + 10P2x3 + 10P3x2 + 5P4x + P5 
 
where Pk = ak+bk+m5ck-(dk+ek+m5fk),  for k = {1,2,3,4,5}. The objective is to eliminate all terms of F(x) other than the linear term x, or to find {a,b,c,d,e,f,m} such that,
 
ak+bk+m5ck = dk+ek+m5fk,   for k = 1,2,3,5    (eq.1)
 
which is just an Equal Sums of Like Powers problem.  Once found, since x is arbitrary, let x = 54P44 N.  Thus,
 
F(x): = 55P45 N 
 
Dividing by the numerical factor, any rational N then is the sum of six rational fifth powers as claimed.  The problem, of course, is if there are non-trivial values such that eq.1 can be solved.  It turns out this system can be reduced to an elliptic curve, hence N in fact is the sum of six rational fifth powers in an infinite number of ways.  To see this, let,
 
{a,b,c,d,e,f,m5} = {p+q, r+s, 1+t, -p+q, -r+s, -1+t,  n}
 
where the symmetry of terms simplifies the system considerably.  Then let {q, t, r} = {(-rs-nt)/p,  (rs-h)/r,  -(n+p)} with,
 
s = (-5n2-4np+16p2+5n4+16n3p-4n2p2-60np3-60p4) / (60hn+120hp)
 
and the multi-grade system (eq.1) is solved if {n,p} satisfies the curve,
 
p(n+p)(n2+3np+3p2-1) = 3h2
 
where n must be the 5th power of a rational.  One value I found is n = 25 = 32 with an initial rational point p = 50/9, from which others can then be calculated.  This gives an initial set of 9-digit solns to the eqn,
 
ak+bk+32ck = dk+ek+32fk,   for k = 1,2,3,5 
 
as {a,b,c,d,e,f} = {-225,478,523;  -172,632,729;  112,165,576;  -291,996,273;  -277,027,261;  100,192,381}
 
thus providing an algebraic identity proving Choudhry's Fifth Powers Theorem.  (End proof.)  The approach can be extended to higher odd powers, with a corresponding increase of the complexity of the system to be solved, but Choudhry found an elegant soln for Seventh Powers discussed in the next section.  Note:  It is not really necessary to eliminate the constant term P5 of F(x) or, equivalently, to solve eq.1 for k = 5.  However, since it can be done, then might as well do it anyway.
 
(Update, 12/21/09):  Method 2:  Choudhry sent me a copy of his paper and it turns out he used a similar approach.  After some modification, the method is essentially equivalent to using the form,
 

F(x): = (x+a)5 + m5(x+b)5 + m10(x+c)5 - (x+d)5 - m5(x+e)5 - m10(x+f)5

 

Collecting powers of x, this is,

 

F(x): = 5P1x4 + 10P2x3 + 10P3x2 + 5P4x  + P5

 

where Pk = ak+m5bk+m10ck-(dk+m5ek+m10fk),  for k = {1,2,3,4,5}.  To eliminate the three highest terms of F(x), one is to solve,

 

ak+m5bk+m10ck = dk+m5ek+m10fk,   for k = 1,2,3

 

Choudhry found a remarkable identity to do this, given by,

 

{a,b,c,d,e,f,m} = {-v20,  v5,  -v15-v5-1,  v20,  -2v15-v5,  -v15-v5+1,  v3}

 

which reduces F(x) to the simple form,

 

F(x): = -40v35(v30-1)(v10+1)x + 2v30(-v30+12v20+8v10+1)(v30-1)(v10+1) = N

 

Given a constant N, one can easily solve for x.  And since v is arbitrary, then N is the sum of six 5th powers in an infinite number of ways.  Also, if N = 0, this provides, after scaling, a concise 36-deg polynomial identity to six 5th powers with a zero sum.  (End update.)

 
Q: Can it be reduced to any N as five rational 5th powers in an infinite number of ways?  And if the theorem can be reduced for any N, this would imply that x15+x25+x35+x45+x55 = 0 has an infinite number of non-trivial rational solns, since only 3 are known so far.
 
 
6a. Form: x16+x26+x36-(y16+y26+y36) = N  
 
(Update, 11/30/09)To prove that any N is the sum and difference of six 6th powers, analogous to Norrie's Theorem for 4th powers, one way is to use a similar method as in 5th powers.  Expanding, 

 

F(x): = (x+a)6+(mx+mb)6+(nx+nc)6-(x+d)6-(mx+me)6-(nx+nf)6

 
and collecting powers of x, this resolves to,
 
F(x): = 6P1x5 + 15P2x4 + 20P3x3 + 15P4x2 + 6P5x + P6 
 
where Pk = ak+m6bk+n6ck-(dk+m6ek+n6fk),  for k = {1,2,3,4,5,6}.  The objective is to eliminate all terms of F(x) other than the linear and constant term, or to find {a,b,c,d,e,f,m,n} such that,
 
ak+m6bk+n6ck = dk+m6ek+n6fk,   for k = 1,2,3,4    (Sys.1)
 
Thus, what will be left is,
 
F(x): = 6P5x + P6 = N
 
where x is then easily solved for any N.  However, this author has not been able to find non-trivial {a,b,c,d,e,f} and {m,n}, if any, that solves Sys.1, though it can be shown if one of {m,n} = 1, then the system is trivial.  Note:  To simplify Sys.1, one can again do the substitution,
 
{a,b,c,d,e,f} = {p+q, r+s, t+u, p-q, r-s, t-u}, 
 
though it still yields no non-trivial factor of small degree.  However, another approach might prove (or disprove) the conjecture that any N is the sum and difference of six 6th powers.  (End update.)
 

 

6b. Form: x16+x26+x36+x46-(y16+y26+y36+y46) = N

 

(Update: 12/11/09): Theorem: “Any rational number N is the sum and difference of eight rational 6th powers in an infinite number of ways.”  (Piezas)

 

Proof:  Using a similar approach as Choudhry's for 7th powers, expand,

 

F(x): = (x+a)6+(mx+mb)6+(x-c)6+(mx-md)6-(x-a)6-(mx-mb)6-(x+c)6-(mx+md)6
 
and collecting powers of x, this resolves to,
 
F(x): = 12P1x5 + 40P3x3 + 12P5x
 
where Pk = ak+m6bk-ck-m6dk,  for k = {1,3,5}.  To get rid of the x5 and x3 terms, find {a,b,c,d,m} such that
 
a+m6b  =  c+m6d       (eq.1) 
a3+m6b3 = c3+m6d3   (eq.2)
 
Eq.2 can be completely solved using the modified Binet formula given by,
 
{a,b} = {1-m6n(p-3q),  m6n2-(p+3q)}
{c,d} = {1-m6n(p+3q),  m6n2-(p-3q)}
 
where n = p2+3q2 for arbitrary m,p,q.  To solve eq.1 as well, m = 1 must be avoided as the system becomes trivial but {p,q} must be chosen such that p2+3q2 = 1.  This is given by,
 
{p,q} = {(u2-3v2 )/(u2+3v2),   2uv/(u2+3v2)}
 
for arbitrary u,v, and m, and avoiding m = 1.  After the x5 and x3 terms are eliminated, since x is arbitrary, let x = (12P5)5 N.  Thus,
 
F(x): = (12P5)6 N
 
Dividing by the numerical factor, N then is the sum/difference of eight rational 6th powers in an infinite number of ways as claimed.  (End proof)
 
 
 
7. Form: x17+x27+x37+x47+ ... +x87 = N
 
(Update, 11/10/09)Choudhry's Seventh Powers Theorem: “Any non-zero rational number N is the sum of eight rational 7th powers in an infinite number of non-trivial ways.”  
 
Proof:  Expand the equation,
 
F(x): = (x+a)7+(x-a)7+(mx+b)7+(mx-b)7-(x+c)7-(x-c)7-(mx+d)7-(mx-d)7
 
and collecting powers of x, this resolves to,
 
F(x): = 42(a2+m5b2-c2-m5d2)x5 + 70(a4+m3b4-c4-m3d4)x3 + 14(a6+mb6-c6-md6)x  
 
To get rid of the x5 and x3 terms, find {a,b,c,d,m} such that,
 
a2+m5b2 = c2+m5d2
a4+m3b4 = c4+m3d4
 
and only the linear term is left.  Since x is arbitrary, let x = 146(a6+mb6-c6-md6)6 N.  Thus,
 
F(x): = 147(a6+mb6-c6-md6)7 N
 
Dividing by the numerical factor, N then is the sum of eight rational seventh powers as claimed. (End proofChoudhry chose m = 2 and found {a,b,c,d} with 33-digits!  I'm assuming there might be smaller solns, so I posted the problem in sci.math.symbolic to see if someone can find one, for any m > 1. 
 

Note 1:  It can be shown that the general system,

 

a2+mb2 = c2+md2   (eq.1)

a4+nb4 = c4+nd4     (eq.2)

 

can be reduced to solving an elliptic curve.  Euler’s complete soln to eq.1 (see Form 10 of Sums of Two Squares) is given by,

 

(pr+mqs)2 + m(ps-qr)2 = (pr-mqs)2 + m(ps+qr)2

 

If we apply these values for {a,b,c,d} on eq.2 as well, it reduces to the eqn,

 

(mp2-nq2)r2 = (np2-m3q2)s2

 

which have rational solns if (mp2-nq2)(np2-m3q2) = y2.  For the special case relevant to 7th powers when {m,n} = {h5, h3}, the curve is,

 

(h2p2-q2)(p2-h12q2) = y2   (eq.3)

 

For h = 2, Choudhry found the 16-and-17-digit soln {p,q} = {5911167604843137, 12317476831120126} which gives his 33-digit {a,b,c,d}.  It might be interesting to know if eq.3 has small solns for some other rational h > 1.
 
Note 2Update (11/16/09)  D. Rusin gave a very thorough analysis here and, as a first step, simplified it to a problem of Equal Sums of Like Powers.  By letting {b,d} = {Bm, Dm}, one is to solve,
 
ak+m7Bk = ck+m7Dk,   for k  = 2,4     (eq.4)
 
Then, by a series of clever transformations, he reduced this system to the rather simple elliptic curve,
 
V2 = U(U+1)(U+q2)
 
where q = (1-m7) / (1+m7) and using Maple's APECS and Magma, eventually gave an explicit soln {U,V}, apparently the smallest, and the same as what Choudhry found.  From this initial point, an infinite more can be calculated proving that any rational N is the sum of eight rational 7th powers in an infinite number of ways.  For other m such that eq.4 has non-trivial solns, because of its symmetry, it suffices to look within the interval m = {0 to 1}.  In addition to m = 2 (or equivalently m = 1/2), Rusin found possible candidates m = {3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/7} though solns, if any, must be large.  (End update.)
 
Note 3:  When N = 0, the theorem gives a trivial result, hence does not prove there are an infinite number of non-trivial solns to eight 7th powers equal to zero, though many have been found already.
 
Q:  Can this be reduced to seven rational seventh powers? 
 
 
(Update, 1/30/10):  Choudhry
     
8. Form: x18+x28+...+x68 (y18+y28+...+y68) = N
 
Choudhry's Eighth Powers Theorem: “Any non-zero rational number N is the sum and difference of twelve rational 8th powers in an infinite number of non-trivial ways."

 

Proof:  Choudhry’s method is equivalent to solving the equation,

 

U1 – U2 = M

 

U1:= (ax+y)8 + (p18+p38)(x+by)8 + (p28+p48)(cx-y)8 + (x-dy)8

U2:= (ax-y)8  + (p18+p38)(x-by)8 + (p28+p48)(cx+y)8 + (x+dy)8

 

for variables {a,b,c,d} and {p1, p2, p3, p4} such that, when expanded and collecting {x,y}, all terms vanish except M which is only linear in x.  He found a wonderfully simple and symmetric soln given by,

 

{p1, p2, p3, p4} = {t, t3, t5, t7}

{a,b,c,d} = {t12, t12, t4, t20}

 

such that,

 

M = 24 t8(t4-t20-t52+t84+t116-t132) xy7

 

for arbitrary {t,x,y}.  Note how the exponents match up as 1+7 = 3+5 = 8, 12+12 = 4+20 = 24, as well as 4+132 = 20+116 = 52+84 = 136.  (Q. What is the reason for the simplicity of the soln, and can this be generalized?)  Doing the substitution {x,y} = {24N,  t4-t20-t52+t84+t116-t132}, we then get the form,

 

M = (2ty)8 N

 

Dividing by the factor, then any non-zero N is the sum and difference of twelve rational 8th powers in an infinite number of ways as claimed.  (Using another approach, it may be possible to reduce this to ten or even eight rational 8th powers.)  Of course, if N is itself an 8th power, then this provides a polynomial identity for the 13-term,

 

x18+ x28+… x68 = y18+ y28+… y78

 

Note:  My thanks to Choudhy for providing a copy of his paper "On Sums of Eighth Powers", Journal of Number Theory, Vol 39, Sept 1991.  (End update.)
 
 
9. Form: x111+x211+x311+x411+ ... +x1211 = N
 
Using Choudhry's and Rusin's approach for 7th powers, if there is a non-trivial soln {xi, yi} to the system S11,
 
ax1k+bx2k+cx3k = ay1k+by2k+cy3k,   for k = 2,4,6,8
 
for some rational {a,b,c} = {p11, q11, r11}, then any rational N is the sum of 12 rational 11th powers.  Multi-grade systems valid for k = {2,4,6,8} or even k = {2,4,6,8,10} are known (see the section on Tenth Powers) so it might be possible, with certain assumptions, that S11 can be reduced to an elliptic curve as well.  As a numerical experiment, one may test this by using small {p,q,r} < 11, like Choudhry's m = 2, excluding the case p = q = r, and testing values {xi, yi} below some reasonable bound.  In general, the conjecture would be that any rational N is the sum of at most 4m rational (4m-1)th powers (with Ryley's Theorem reducing the cubic case to just three cubes), with the problem reducible to a system of equal sums of like powers.
 
 

(Update, 4/20/10):  Enrico Jabara gave a polynomial soln to x13+x24+x35+x47+x58 = z8 as,

 

(288747n4)3 + (292749n)4 + (232717n4)5 + (2874n4)7 + (n4-252728)8 = (n4+252728)8

 

More generally, as a variant of the easier Waring’s Problem, he proved that any integer N is the sum of consecutive unlike powers,

 

N = ± x12 ± x23 ± x34 ± x45

 

N = ± x13 ± x24 ± x35 ± x46 ± x57 ± x68 ± x79 ± x810

 

where the xi are also in the integers.  However, for the next in the series that starts with 4th powers,

 

N = ± x14 ± x25 ± … ± (xk-3)k

 

it is yet unknown what is the ending power k.

 

Source: Representations of numbers as sums and differences of unlike powers, (by Enrico Jabara, Università di Ca’ Foscari).

 
 
  
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