Updates Page 03: Mar

(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)^k, for 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

(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+a^k+b^k+c^k+d^k = 0^k+(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.)

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,

x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k (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,

a^k + b^k + (-a-b)^k = c^k + d^k + (-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 a^k+b^k+c^k = d^k+e^k+f^k, 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 a^k+b^k+c^k = d^k+e^k+f^k, 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 a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k, 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 a^k+b^k+c^k+d^k+e^k = f^k+g^k+h^k+i^k+j^k, 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 a^k+b^k+c^k = d^k+e^k+f^k, 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)

f₃ = (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,

x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k, k = 2,8

the ratio,

Product[(x_i²-y_i²)] / (x₁^h+x₂^h+x₃^h+x₄^h-y₁^h-y₂^h-y₃^h-y₄^h)

for h = 12 (or h = 14) does not seem to be the product of simple rational polynomials in the {x_i, y_i}, 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,

{x_n+1, y_n+1} = {px_n+qy_n, 2avx_n+ry_n}

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 {x_n, y_n} = {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: {x_n+1, y_n+1} = {155x_n+294y_n, 126x_n+239y_n}

Using -v: {x_n+1, y_n+1} = {239x_n-294y_n, -126x_n+155y_n}

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. a^k+b^k+c^k = d^k+e^k+f^k, 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. a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k, 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 a^k+b^k+c^k = d^k+e^k+f^k, 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 a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k, 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.)

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

(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 a^k+b^k+(-a-b)^k = c^k+(-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,

a^k+b^k+c^k = 2(x⁸+14x⁴+1)^k/4, for k = 4,8

(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+…+x_n⁸ = (x₁+x₂+…+x_n)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, 3/22/10): 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.

(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?

(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(a^2k+b^2k+c^2k-d^2k-e^2k-f^2k), 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.