020: Fifth Powers (4 or 6 Terms)

(√p+√q)⁶ + (√p-√q)⁶ = (√r+√s)⁶ + (√r-√s)⁶

or after expanding,

105⁵ + 1153⁵ + (-105-1153)⁵ = 455⁵ + 582⁵ + (-455-582)⁵

Note 1: Incidentally, given the monic quartic of form,

y² = x⁴+(Poly1)x³+(Poly2)x²+(Poly3)x+(Poly4)

to be made a square and where the Poly_i are polynomials in a variable p of degree 1,2,3,4 respectively (like eq.1 above), what is the maximum number possible of linear solns of form x = (ap+b) or x = (ap+b)/(cp+d)?

Note 2: The complete criterion of x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+x₃^k, k = 1,5, where x₁-x₂ = y₁-y₂ has also been given by Bremner though I haven’t seen that work yet. I will update this section to compare methods when I’ve read it.

Note 3: It can be proven that k = 1,2,5 yields only trivial rational results. The complete soln to k = 1,2 is by L. Dickson, (Form 4 in Sums of Three Squares)

(ad+e)^k + (bc+e)^k + (ac+bd+e)^k = (ac+e)^k + (bd+e)^k + (ad+bc+e)^k

and expanding this at k = 5 gives merely a quadratic polynomial in e with discriminant D,

D = -2(a²-ab+b²)(c²-cd+d²)

which must be made a square. But let {a,b,c,d} = {u+v, 2v, x+y, 2y} and it becomes,

D = -2(u²+3v²)(x²+3y²)

and it is obvious there are no real values such that D becomes a positive real square. The same conclusion was arrived at by Choudhry in his paper, The system of simultaneous equations x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, k = 1,2,5 has no non-trivial solns in integers, The Mathematics Student, Vol 70, 2001. (It is not known, however, if the case k = 2,5 has a soln or not, though it is easily proven using Moore's database that there are none within a radius of 17000.)

Note 4: Is a⁵+b⁵+c⁵ = d⁵+e⁵+f⁵ where a+b+c = d+e+f = 0 solvable? (The analogous eqn for seventh powers does have solns.) This can be completely parametricized as,

p⁵+q⁵+(-p-q)⁵ = r⁵+s⁵+(-r-s)⁵

which when expanded is,

pq(p+q)(p²+pq+q²) = rs(r+s)(r²+rs+s²)

though it seems to be unknown if this is non-trivially solvable. (Update: Wroblewski has proven this has non-trivial solns, the smallest of which is {p,q,r,s} = {105, 1153, 455, 582}, yielding,

(1-2q+2r)^k + (-3-2q)^k + (1+2q)^k = (2+q+r)^k + (-2+q-r)^k + (-1-4q+2r)^k

Again, it is already true for k = 1 and, when expanded for k = 5, yields a quadratic in r. Its discriminant is a quartic in q and, if made a square is,

y² = -(384+1112q+1297q²+738q³+171q⁴)

with a small rational point at q = -16/15. Q: Any other simple soln to system M₁ for some integer n? (End note.)

Since the u_i are in terms of {a,b}, then the unknowns are {a,b,v₁,v₂,v₃}. Expanding the above and collecting powers of p, one then has a system of five eqns in five unknowns, and the final eqn in this case has two rational solns. A similar approach for q = (ap+b)/(cp+d) will give a single rational soln.

By adding yet another constraint, one can also find a simple form for S₅. An example is,

(v+8)^k + (u-8)^k + (-u+2v+7)^k = (-u+2v+8)^k + (v-8)^k + (u+7)^k

This is already true for k = 1. Expanding for k = 5 yields only a quadratic in v and is easily solved. For rational v, one must make its discriminant, a quartic polynomial in u, as a square,

y² = -43281+40110u-4597u²+390u³-31u⁴

with one small soln as u = 3/2, giving [389, -208, 442] = [474, -123, 272]. From this initial point, one can then compute an infinite number of rational points. This, in fact, is the complete soln to the system, call this M₁,

x₁-x₂ = y₁-y₂

x₂-nx₃ = -ny₁+y₃

x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k

for k = 1,5 and n = 15. Other than the case n = 1 which is trivial, I haven’t checked yet for what other n this system is solvable.

(Update 7/12/09): Another one obeying the system M₁ but with n = 2 is,

This is more easily made a square being a polynomial with a square leading and constant term, and illustrates how using the appropriate form can simplify a problem. This quartic is also rather special since its solns are plentiful enough that they can be parametricized linearly in three ways, namely {q₁, q₂, q₃} = {-(80p+40)/(57p+80), -(p+2), -(3p+2)/3}, though the last two are trivial with respect to the original eqn. However, we can always derive non-trivial points. Three parametric solns will then be given, each of which involves a distinct elliptic curve. Thus,

(p+r)^k + (p-r)^k + (2q+1)^k = (q+r)^k + (q-r)^k + (2p+1)^k, k = 1,5

has solns,

q = – (80p+40)/(57p+80), r = z/(57p+80), 40²+10160p+22999p²+14478p³+3249p⁴ = z²

q = 3(13p²+10p+8)/(9p-16), r = z/(9p-16), 8²+656p+1999p²+2922p³+1629p⁴ = z²

q = (13p²+14p+8)/(3p+16), r = z/(3p+16), 24²+1232p+1183p²+542p³+181p⁴ = z²

where one must solve the elliptic curve in p. The first q was found was found using Fermat’s method. There are trivial p like {-1/2, -80/57, 0}, but these can lead to non-trivial ones, like p = -15/38, etc. The next two q’s were found using Fermat’s method on the two trivial linear solns. These do not completely give all values that make the discriminant a square but it can be shown that the three q's as linear polynomials are the only ones of form q = (ap+b) or q = (ap+b)/(cp+d) for some rational {a,b,c,d}. To show this, given eq.1, substitute q = (ap+b) into the eqn and assume the resulting polynomial to be a square,

u₁p⁴+u₂p³+u₃p²+u₄p+u₅ = (v₁p²+v₂p+v₃)²

Lander would later find a three-parameter family which in general would involve 9th degree polynomials. This author found that for special cases, it can be reduced to 7th or 8th deg. Whether there is a 6th deg is unknown. We have already seen that given an initial soln to a₁^k+a₂^k = b₁^k+b₂^k for either k = 2, 3, or 4, one can use this to find more solns. It turns out there is a quintic version of this, albeit with a small condition,

Theorem (Lander): "Given a rational solution to a₁^k+a₂^k+a₃^k = b₁^k+b₂^k+b₃^k for both k = 1,5, call this system S₅, then this can generate subsequent ones."

In “Geometric Aspects of Diophantine equations involving Equal Sums of Like Powers”, Lander started out with the form,

(u+x)^k + z^k + (v+y)^k = (v+x)^k + y^k + (u+z)^k, (call this L₀)

a general soln for k = 1. This can also be true for k = 5 using appropriate values. (Note: Incidentally, it should be pointed out that the eqn, x₁^k + x₂^k + …+ x_k+1^k = 0 simultaneously valid for three odd powers k = 1,3,n only has trivial solns when n = 5. Using the form above and Mathematica’s Resultant[] function to eliminate one variable at k = 3 and 5, it will be seen that there are only trivial solns. In contrast, there are non-trivial solns when k = 1,3,7 as found by Choudhry and k = 1,3,9 by Wroblewski. Whether there is a k = 1,3,11 remains to be seen.)

Anyway, L₀ when expanded out for k = 5 has the nice symmetrical form,

(x-z)u⁴+2(x²-z²)u³+2(x³-z³)u²+(x⁴-z⁴)u = (x-y)v⁴+2(x²-y²)v³+2(x³-y³)v²+(x⁴-y⁴)v

where all fifth powers vanish and as such is a quartic surface. After a clever geometric analysis, Lander gave a method to derive polynomial expressions for all the variables. Given an initial soln {a_i, b_i} to L₀, this leads to a new one, typically much larger, in two ways. First,

1. Using point P₁(x₁,y₁,z₁)

Define {u,v,x,y,z} = {-a₃+b₃, a₂-b₂, x₀+(x₁-x₀)t , y₀+(y₁-y₀)t, z₀+(z₁-z₀)t},

where,

x₁ = y₁ = z₁ = (x₀d₁+y₀d₂+z₀d₃)/(d₁+d₂+d₃)

{d₁, d₂, d₃} = {a₁⁴-b₁⁴, a₂⁴-b₂⁴, a₃⁴-b₃⁴}

{x₀, y₀, z₀} = {a₁+a₃-b₃, b₂, a₃}

However, it still remains to find t. When all these expressions are substituted into L₀, if {a_i, b_i} is a soln, then the eqn is a quartic with four roots {0,0,1,t} so is just a linear eqn in t which is then easily solved. One can also use a second approach,

2. Using point P₂(x₂,y₂,z₂)

Define {u,v,x,y,z} = {-a₃+b₃, a₂-b₂, x₀+(x₂-x₀)t , y₀+(y₂-y₀)t, z₀+(z₂-z₀)t},

where,

{x₂, y₂, z₂} = {x₁-d₀, x₁-d₀+u, x₁-d₀+v}

x₁ = (x₀d₁+y₀d₂+z₀d₃)/(d₁+d₂+d₃)

d₀ = (d₂u+d₃v)/(d₁+d₂+d₃)

and {d₁, d₂, d₃}, {x₀, y₀, z₀} are defined the same way as in the first point. Using these, eq.0 again reduces to a linear eqn in t. It is also possible to find a polynomial soln using a trivial initial one. Let,

{a₁, a₂, a₃, b₁, b₂, b₃} = {p, -p, q, q, r, -r}

{a₁, a₂, a₃, b₁, b₂, b₃} = {p, -p, q, r, q, -r}

for the first and second points respectively and these will yield 9th degree polynomials.

Theorem (Piezas): "The complete soln to S₅, or x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, for k = 1,5 involves solving only a quadratic equation with a discriminant D that is quartic polynomial in one variable. The problem of making D a square can then be treated as an elliptic curve."

Proof: It can be shown that Lander’s result can be concisely encoded in the form L₁,

(a+bp+q)^k + (b-bp+q)^k + (c+ap+q)^k = (a+cp+q)^k + (b+ap+q)^k + (c-cp+q)^k

which we’ve come across before. This is already true for k = 1. It is a subset of Lander's L₀,

(u+x)^k + z^k + (v+y)^k = (v+x)^k + y^k + (u+z)^k

excluding the case uv(u-v) = 0. To see this, simply equate the first five terms of L₁ with that of L₀, and one can always find rational {a,b,c,p,q} using,

{a,b,c,p,q} = {(1+t)(st+y-z)/(2t), -(1+t)(st-y+z)/(2t), (1+t)(y+tu-tx+ty-z)/(2t), t/(1+t), (st-y+z+2tz)/(2t)}, where {s,t} = {u+x-y, -(u-v)/u}

hence the cases with denominator (1+t)t = uv(u-v) = 0 should be excluded, otherwise there is division by zero. Using these expressions, the sixth terms of L₁ and L₀ are also seen to be equal. The fact that L₁ is less general than L₀ does not matter since for the exceptional case, L₀ reduces to the four-term form,

n₁^k + (n₂+n₃)^k = (n₁+n₂)^k + n₃^k

which, after expanding, is easily seen to have trivial solns for k < 6 other than k = 1, hence is of no interest. With that out of the way, by expanding L₁ at k = 5, this reduces to a polynomial in p that is just a quadratic of form,

(Poly1)p²+(Poly2)p+(Poly3) = 0 (eq.1)

disregarding the trivial factor p(p-1). The Poly_i are in {a,b,c,q} and as polynomials in q are of degree 1,1,3, respectively. (Unfortunately, they are tedious to explicitly write down.) The discriminant D is,

D:= (Poly2)²-4(Poly1)(Poly3)

of deg 1+3 and so is a quartic in q, thus proving the theorem. To find a polynomial soln, there are two ways:

1st method: Since Poly1 is linear in q, all we have to do is to set Poly1 = 0 and solve for q, with the eqn given by,

2(b+c)(b²+c²)q = (a⁴-c⁴)-(a+b)(b+c)(b²+c²)

and (eq.1) reduces to the linear eqn (Poly2)p+(Poly3) = 0 where p is easily solved for. Consistent with Lander’s result, this yields ninth deg polynomials in one of the variables {a,b,c}. Incidentally, the sums of the terms as x₁+x₂, y₁+y₃ have the simple forms,

(a+bp+q) + (b-bp+q) = (a⁴-c⁴)/((b+c)(b²+c²))

(a+cp+q) + (c-cp+q) = (a⁴-b⁴)/((b+c)(b²+c²))

for {p,q} as defined above. This author found out that the degree of the polynomials can be reduced if we assume certain linear relationships between the terms. It’s known that the system for k = 1,5 may have the side condition x₁-x₂ = y₁-y₂. Let,

(a+bp+q) - (b-bp+q) = (a+cp+q) - (b+ap+q)

which is true if c = a+2b. Using this constraint, the polynomials reduce to seventh deg. Since the eqn is homogeneous, it can be set b = 1 without loss of generality and we have the identity,

(at+p+q)^k + (t-p+q)^k + (ct+ap+q)^k = (at+cp+q)^k + (t+ap+q)^k + (ct-cp+q)^k, k = 1,5,

c = a+2

t = 4a(a-2)(a+3)(a²+4a+5)²

p = (a²+1)(a²+8a+11)(a³+15a²+33a+31)

q = -2a(a+1)(a-2)(a²+4a+5)(a³+15a²+33a+31)

where the common factor (a+1) of the terms has to be removed. (A. Moessner also found a seventh deg soln in 1951 but I do not know if this is the same since Lander gave the source as an Italian journal.) Alternatively, if c = a-2b is used, they reduce to eighth deg though I haven’t been able to find a linear relationship between the terms. Q: So there are 5th, 7th, 8th, and 9th deg solns. Any for 6th deg? (Note: As was already discussed, the form L₁ can also be used for appropriate higher powers k. Later, it will be used to prove that the system k = 1,2,6 can also be reduced, again, to merely solving a quadratic.)

(Update 7/13/09) 2nd method: We directly find a q such that the discriminant D is a square. To find a trivial soln, one begins by expanding L₁ for k = 5,7 and eliminate p between them. Factoring the resultant, one finds a linear relationship between the {a,b,c,q} given by,

2(b+c)q = a²-b²-c²-ab-ac-bc

Using this q, D becomes the simple square,

D: = (a²-b²)²(a²-c²)²(b-c)²/(b+c)²

This is a trivial soln but, as was shown in the first method that D is a quartic in q, from this initial point, one can then find an infinite number of non-trivial ones. For simplicity, one can set c = a+2b and the next point yields terms for the x_i, y_i as 13th-deg polynomials after removing common factors.

Piezas

The complete criterion to solve the system x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k for k = 1,5 has already been given above, with a subset also satisfying x₁-x₂ = y₁-y₂. We can also explore other methods for the sake for the exercise. This subset can be completely determined by either of two nice forms. Let,

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

already true for k = 1 and if expanded for k = 5 reduces to the simple eqn,

(a⁴+b⁴-2c⁴) + 2(a²+b²-8c²)d² - 6d⁴ = 0

One way to solve this is to treat it as a quadratic in d² and make its discriminant (a quartic polynomial) a square, though is not easily done for this particular quartic. However, we can use the alternative form,

(p+r)^k + (p-r)^k + (2q+s)^k = (q+r)^k + (q-r)^k + (2p+s)^k

which, for convenience, after scaling can be assumed s = 1 without loss of generality. By expanding this at k = 5, this is only a quadratic in r². Solving for r²,

r² = -p²-pq-q² ± y,

where y is the square root of the discriminant,

y² = 4q⁴+(8+5p)q³+2(4+4p+3p²)q²+(4+8p+8p²+5p³)q+(1+2p+2p²)² (eq.1)

t⁶(a+20t)⁵ + t³(b-20t²-20t⁴)⁵ + (c+20t⁵)⁵ = t⁶(a-20t)⁵ + t³(b+20t²+20t⁴)⁵ + (c-20t⁵)⁵

where {a,b,c} = {t⁶+8t⁴+12t²-1, t⁶+8t⁴-8t²-1, t⁶-12⁴-8t²-1}.

Since t is arbitrary, one can set t = v⁵ to have equal sums of 5th powers. Note the nice symmetry of terms and how the coefficients, after negation, are palindromic. Surely this is not an isolated result. Anyone else knows of a similarly simple identity, as well as linear relations among the x_i, y_i, if any, for this? (End update.)

(Update, 2/22/10): Gerardin gave the beautiful 4th power identity with small coefficients {1,2,3},

(a+3a²-2a³+a⁵+a⁷)⁴ + (1+a²-2a⁴-3a⁵+a⁶)⁴ =

(a-3a²-2a³+a⁵+a⁷)⁴ + (1+a²-2a⁴+3a⁵+a⁶)⁴

and Choudhry gave a 5th power version,

(a-a³-2a⁵+a⁹)⁵ + (1+a²-2a⁶+2a⁷+a⁸)⁵ + (2a³+2a⁴-2a⁷)⁵ =

(a+3a³-2a⁵+a⁹)⁵ + (1+a²-2a⁶-2a⁷+a⁸)⁵ + (-2a³+2a⁴+2a⁷)⁵

which in fact is good for k = 1,5. (Differences between LHS and RHS are highlighted in blue.) While there are identities for k = 6, none are known with such small coefficients. (Anyone can find one?) Other 5th deg identities can be found using the ff. two methods by Choudhry. Given,

x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k (eq.1)

where,

{x₁, x₂, x₃, x₄} = {a+b+c, a-b-c, -a+b-c, -a-b+c}

{y₁, y₂, y₃, y₄} = {d+e+f, -d+e-f, -d-e+f, d-e-f}

Expanding one side of the eqn, we get,

∑ x_i¹ = 0

∑ x_i⁵ = 80abc(a²+b²+c²)

and similarly for the y_i. Eq.1 can be reduced to just 6 terms if x₄ = y₄. Thus, if,

Method 1: a+b-c = -d+e+f ; ab = de; c(a²+b²+c²) = f(d²+e²+f²)

Method 2: a+b-c = -d+e+f ; c = f; ab(a²+b²+c²) = de(d²+e²+f²)

then ∑ x_i^k = ∑ y_i^k, for k = 1,5.

Choudhry showed that given an initial point, even a trivial one, using appropriate substitutions and solving a system of linear equations, then an infinite number of non-trivial solns can be found. In fact, by directly solving Method 1, one can end up with a quartic equation which has a rational root if a certain quartic polynomial is made a square, hence can be treated as an elliptic curve. Note: Choudhry also used Eq.1 and its substitutions to solve k = 1,3,7.

Source: “On Equal Sums of Fifth Powers”, Indian Jour. Pure & Applied Math, Nov. 1997.

Update, 11/24/09): Choudhry (1999) has considered the general eqn,

ax⁵+by⁵+cz⁵ = au⁵+bv⁵+cw⁵ (eq.1)

for the case a+b+c = 0. Excluding the situation when abc(a-b)(a-c)(b-c) = 0, he gave the soln to,

ax⁵+by⁵+cz⁵ = t² (au⁵+bv⁵+cw⁵)

as,

{x,y,z} = { (p₁t²-p₁t+q₁)r₁, (p₂t²-p₂t+q₂)r₂, (p₃t²-p₃t+q₃)r₃ }

{u,v,w}= { (q₁t²-p₁t+p₁)r₁, (q₂t²-p₂t+p₂)r₂, (q₃t²-p₃t+p₃)r₃ }

where.

{p₁, p₂, p₃} = {17a²+14ab+14b², 14a²+14ab+17b², 17a²+20ab+17b²}

{q₁, q₂, q₃} = {2a²-ab-b², -a²-ab+2b², 2a²+5ab+2b²}

{r₁, r₂, r₃} = {a+2b, -2a-b, a-b}

given constant {a,b}, for arbitrary t. One can then set t = v⁵ to satisfy eq.1. As an example, for {a,b,c} = {1,2,-3}, after transposing terms, we get a soln to,

x⁵+2y⁵+3z⁵ = u⁵+2v⁵+3w⁵

{x,y,z} = {101t¹⁰-101t⁵-4, 4t¹²-88t⁷+88t², 25t¹⁰-25t⁵+4}

{u,v,w}= {-4t¹²-101t⁷+101, 88t¹⁰-88t⁵+4, 4t¹²-25t⁷+25t²}

(End update.)

(Update, 2/3/10): Choudhry

(-8m⁶+2mn⁵)^k + (8m⁵n+n⁶)^k + 2(8m⁶)^k = (8m⁶+2mn⁵)^k + (8m⁵n-n⁶)^k + 2(n⁶)^k, for k = 1,5

A soln in positive terms can be acquired if n > m and within the range 2^2/5 < n/m < 2^3/5, or approx. 1.32 < n/m < 1.51. With terms transposed to one side,

(-8m⁶+2mn⁵)^k + (8m⁵n+n⁶)^k - (8m⁶+2mn⁵)^k - (8m⁵n-n⁶)^k + 2(8m⁶)^k = 2(n⁶)^k

For n = 1, the identity proves that the integer 2 is the sum/difference of six integral 5th powers in infinitely many ways. Source: The Diophantine Equation x₁⁵+x₂⁵+2x₃⁵ = y₁⁵+y₂⁵+2y₃⁵, Ganita, Vol. 48, No. 2, 1997, 115-116.

Note: In one sense, this is a 5th power version of,

(-6x³+y³)³ + (6x³+y³)³ - (6x²y)³ = 2(y³)³

where, for y = 1, proves that 2 infinitely is the sum is the sum/difference of three integral 3rd powers.

Q: Anyone can give a similar identity expressing 2 as integral 7th powers? (End update.)

If rational {a,b} can be found such that,

20(a²-b²) = mp⁵ (eq.1)

10(a⁴-b⁴) = nq⁵ (eq.2)

then it yields an identity for (5,4,m+n). For m = n = 1, Sastry and Chowla found {a,b} = {75, 25} and {p,q} ={10, 50}. Q: Are there others not simply fifth power multiples of these two, or other small m,n?

(Update, 11/18/09): Dave Rusin showed that the system can reduced to a certain hyperelliptic curve, thus has only finitely many rational solns, and probably the known soln is the only non-trivial one. He gave the transformation,

{a,b,p,q} = {m²y/(5n), m³n, -2mx, 20m²x}, where m = (4x⁵-5)/(5n²)

and eq.1, eq.2 are true if,

-4²x¹⁰+ 5² = y²

the rational soln of which apparently is only {x,y} = {1,3}. It is hard not to notice this is essentially the smallest Pythagorean triple, -4²+5² = 3², and it is interesting that it appears in the context of 5th powers. (End update.)

(Update, 12/15/09): Tony Rizzo pointed out that if {a,b} = {75r³, 25r³}, then {m,n} ={r, r²}. For r = 2, this gives the (5,4,6) identity,

(u⁵+600v⁵)⁵ + (u⁵-600v⁵)⁵ - (u⁵+200v⁵)⁵ - (u⁵-200v⁵)⁵ = 2(2u³v²)⁵ + 4(4uv⁴)⁵

If {u,v} are chosen such that Abs[u/v] = x is within the range 200^1/5 < x < 600^1/5, then it yields a (5.1.9) with all terms positive. (End update)

Update, 11/23/09):

Moessner (1951)

(u⁵+av⁵)⁵ + (u⁵-av⁵)⁵ - (u⁵+bv⁵)⁵ - (u⁵-bv⁵)⁵ = 20(a²-b²)(u³v²)⁵ + 10(a⁴-b⁴)(uv⁴)⁵

3. x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, k = 1,5

4. x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, x₁-x₂ = y₁-y₂, k = 1,5

If you know of another, pls send it so it can be added to the collection. The first two classes have beautiful identities.

Sastry, Chowla (1934)

(u⁵+25v⁵)⁵ + (u⁵-25v⁵)⁵ + (10u³v²)⁵ = (u⁵+75v⁵)⁵ + (u⁵-75v⁵)⁵ + (-50uv⁴)⁵

If {u,v} are chosen such that Abs[u/v] = x is within the range 25^1/5 < x < 75^1/5, or roughly 1.904 < x < 2.371, then it yields a (5.1.5) with all terms positive. Outside this range, it becomes a (5.2.4), but there is no {u,v} such that it gives a (5.3.3) with all positive terms. This has the basic form,

2. x₁⁵+x₂⁵+x₃⁵ = y₁⁵+y₂⁵+y₃⁵, (Moessner's, with unknown linear relations between the x_i, y_i)