017: Sums of four or more Fourth Powers

x⁴-6x²y²+y⁴ = (x²+2xy-y²)(x²-2xy-y²) = (x²-y²)² - (2xy)²

R. Norrie

(x+y)⁴ + (-x+y)⁴ + (2y)⁴ + (u²-v²)⁴ + (2uv)⁴ = (u²+v²)⁴, if 2uv(u²-v²) = x²+3y²

Note that {u²-v², 2uv} are the legs of a Pythagorean triple. Thus, if the product of the legs can be expressed as x²+3y², it then gives a soln to the above equation, the smallest of which is the particular equation given earlier, {2,2,4,3,4; 5}. For the sum of five or six biquadrates,

Sophie Germain

Sophie Germain’s Identity is given by

x⁴+4y⁴ = ((x+y)²+y²) ((x-y)²+y²) = (x²+2xy+2y²)(x²-2xy+2y²)

The quartic form can be generalized as,

x⁴+(-a²+2b)x²y²+b²y⁴ = (x²+axy+by²)(x²-axy+by²)

where Germain's was the case {a,b} = {2,2}. Another interesting case is {a,b} = {2,-1},

(50x²+300xy-150y²)⁴ + (50x²-300xy-150y²)⁴ + (100x²-300y²)⁴ + (4x²+12y²)⁴ + (15x²+45y²)⁴ = 103⁴(x²+3y²)⁴

where p = (a+b)(a+c)(b+c), q = (ab+ac+bc), and {a,b,c} satisfies the two quadratic conditions,

p(p+4abc) = u²

p(p+4abc) - 4q³ = v²

An example of a particular soln is {a,b,c} = {-21, -4, 12} which yields the eqn,

(-63)^k + (-12)^k + 36^k + (-35)^k + 10^k = (-62)^k + 37^k + (-39)^k

Q: Any general polynomial identity for this?

20. Form: x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k, k = 1,2,3,4

The complete soln to this system, call this E₄, is unknown. Use the form,

(a+bj)^k+(c+dj)^k+(e+fj)^k+(g+hj)^k+(i+j)^k = (a-bj)^k+(c-dj)^k+(e-fj)^k+(g-hj)^k+(i-j)^k

and let i = -(ab+cd+ef+gh), h = -(1+b+d+f) to make it identically true for k = 1,2.

I. Case x₂+x₃ = y₂+y₃:

Let f = -d to satisfy the constraint. It can be shown that the end result involves making a quartic polynomial a square. Expand further the form above at k = 3,4, to get the auxiliary resultants,

(Poly10)j²+(Poly20) = 0 (eq.1)

(Poly11)j⁴+(Poly21)j²+(Poly31) = 0 (eq.2)

where the Poly_i are in {a,b,c,d,e,g}. Eliminate j between eq.1 and 2 (set j = √v for convenience) and one gets a final resultant R which is only a quadratic in a,c,e,g. Solving for g, this has a discriminant D which is only a quadratic in e so to solve D = y², one uses a quadratic form e = Q(x) for some variable x. After a rational g is found, to find j, solve,

-(Poly10)(Poly20) = z²

which is a quadratic in e. Substituting the quadratic form e = Q(x) into this, it becomes,

Poly(x) = z² (eq.3)

which is a quartic in x, hence any soln to E₄ with the constraint x₂+x₃ = y₂+y₃ must satisfy eq.3, a quartic polynomial to be made a square.

II. Case x₁+x₂ = y₁+y₂; x₂+x₃ = y₂+y₃:

This is just a special case of the one above which is much simpler. Let b = f = -d to satisfy the two constraints. After eliminating j between eq.1 and 2, the final resultant is only linear in g. Substituting this into eq.1 and eq.2, one must make a quadratic polynomial in e as a square. Since its leading coefficient is already a square, then this is easily done, giving a multi-variable polynomial parametrization to E₄ .

Note: For Case 1, since the final resultant involves six variables {a,b,c,d,e,g}, the explicit soln is a mess. There might be a way to clean this up though using a more suitable form than the one given without losing its generality. But using the simpler form,

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

for example, is messier.

21. Form: x₁⁴+x₂⁴+…x_n⁴, n > 4

E. Fauquembergue

(4x⁴-y⁴)⁴ + 2(4x³y)⁴ + 2(2xy³)⁴ = (4x⁴+y⁴)⁴

C.Haldeman

(2x²+12xy-6y²)⁴ + (2x²-12xy-6y²)⁴ + (4x²-12y²)⁴ + (3x²+9y²)⁴ + (4x²+12y²)⁴ = 5⁴(x²+3y²)⁴

This gives as a first instance 2⁴+2⁴+4⁴+3⁴+4⁴ = 5⁴. Similar identities were found by A. Martin and Ramanujan. A generalization has been found by this author.

Piezas

(ax²+2v₁xy-3ay²)^k + (bx²-2v₂xy-3by²)^k + (cx²-2v₃xy-3cy²)^k = (a^k+b^k+c^k)(x²+3y²)^k

where {v₁, v₂, v₃, c} = {a+2b, 2a+b, a-b, a+b}, for k = 2,4

Thus, it suffices to decompose a sum which starts as a⁴+b⁴+(a+b)⁴+… into a sum and difference of any number of biquadrates and apply it to the identity. For example, using the eqn 50⁴+50⁴+100⁴+4⁴+15⁴ = 103⁴, this yields,

for the special case when h = a+b+c, then,

{d,e,f,g} = {(p+u)/(2q), (p-u)/(2q), (p+v)/(2q), (p-v)/(2q)}

2(d²v⁴+x⁴+y⁴+z⁴) = (dv²+x²+y²+z²)² (eq.2)

and Euler's soln can be given as {v,x,y,z} = {2pqrs(a²-t²), t(a²-t²), t(abrs-cpqt), a(abrs-cpqt)}, where {a,b,c} = {pq(r²-ds²), p²+dq², r²+ds²}, and {p,q,r,s,t} satisfies the conditional eqn,

pqrs(p²-dq²)(r²-ds²) = t²

For d = 1, one soln is {p,q,r,s} = {2m²-n², m²+n², m²+n², m²-2n²}. Note: Strictly speaking, Euler’s analysis was focused on the case d = ±1 and this author inserted the d to generalize it. Any soln for other d?

(Update, 9/13/09): I noticed we can have a better form if we assume t = rs(r²-ds²) and the conditional eqn becomes the symmetric,

pq(p²-dq²) = rs(r²-ds²)

This has appeared already in Section 013 and has many solns and is connected to the two distinct eqns,

a⁴+b⁴ = c⁴+d⁴

a^k+b^k+c^k = d^k+e^k+f^k, k = 2,4

Thus, we can also solve 2(n²v⁴+x⁴+y⁴+z⁴) = (nv²+x²+y²+z²)² as:

I. When n = -1:

{v,x,y,z} = {2(ab-cd)(ab+cd), (a²+b²+c²+d²)(a²-b²+c²-d²), 2(ac-bd)(a²+c²), 2(ac-bd)(b²+d²)}, where a⁴+b⁴ = c⁴+d⁴.

II. When n = 1:

Let a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,4, where a²+b² = f², d²+e² = c², and ab = de, then,

{v,x,y,z} = {ae-bd, (a+d)(a-d), a(c+f), d(c+f)}

A soln given by Gloden is {a,b,c,d,e,f} = {p²-q², p²+r², 2pq, p²+q², p²-r², 2pr}, where p² = q²+qr+r². This can be solved by {p,q,r} = {u²+uv+v², u²-v², 2uv+v²}. (End update.)

19. Form: x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k, k = 1,2,3,4

The system of eqns,

x₁^k+x₁^k+…+ x_n^k = y₁^k+y₁^k+…+ y_n^k, k = 1,2,3,4

has non-trivial solns only for n > 4. However, it is possible two terms on one side are equal to zero.

Piezas

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

(2vz)² + (2xy)² = (v²-x²-y²+z²)²

so is a special case of three simultaneous Pythagorean triples. More generally, for any d, we are to solve,

18. Form: 2(v⁴+x⁴+y⁴+z⁴) = (v²+x²+y²+z²)²

The equation,

2(a⁴+b⁴+c⁴) = (a²+b²+c²)²

has the simple soln a+b = c, which is central to Ramanujan's 6-10-8 Identity. Turns out when there are four terms,

2(v⁴+x⁴+y⁴+z⁴) = (v²+x²+y²+z²)² (eq.1)

it has a parametric soln as well. Note that this is a special case of Descartes' Circle Theorem,

2(x₁²+x₂²+x₃²+x₄²) = (x₁+x₂+x₃+x₄)²

where the x_i are squares. Euler’s clever soln uses the fact that eq.1 is equivalent to either,

(2vx)² + (2yz)² = (v²+x²-y²-z²)²

(2vy)² + (2xz)² = (v²-x²+y²-z²)²

x₁^k + x₂^k + x₃^k + x₄^k = y₁^k + y₂^k + y₃^k + y₄^k, k = 2,4

assume it to have the form, call it F₁,

a^k(p+q)^k + b^k(r-s)^k + c^k(r+s)^k + d^k(p-q)^k = a^k(p-q)^k + b^k(r+s)^k + c^k(r-s)^k + d^k(p+q)^k

which has x₁x₄ = y₁y₄ and x₂x₃ = y₂y₃. This sufficient but not necessary constraint is completely parameterized by F₁. To see this, use the Mathematica command,

Solve[{a(p+q), d(p-q), a(p-q)} = {x₁, x₄, y₁}, {a,p,d}]

to express {a,d,p} in terms of {x₁, x₄, y₁}. Substituting these into d(p+q) = y₄, we get,

x₁x₄/y₁ = y₄

which, if true, yields the appropriate {a,d,p}, with q a free variable, and similarly for {b,c,r}. For example, given,

16^k + 48^k + 58^k + 99^k = 22^k + 29^k + 96^k + 72^k, k = 2,4

which is the smallest in distinct integers with NO sign changes such that x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0. This has 16(99) = 22(72); 48(58) = 29(96), then,

{a,d,p,q} = {1, 9/2, 19, -3}

{b,c,r,s} = {1/2, 1, 77, -19}

with {q,s} arbitrary. F₁ takes care of the necessary requirement that x₁x₂x₃x₄ = y₁y₂y₃y₄. To simplify matters, assume further that x₁+x₂ = y₁+y₂ (not obeyed by the example above). After some algebra, we get the soln,

{p,q,r,s} = {ab²-ac², be, a²b-bd², ae}, where a²(3b²-c²) + e² = (b²+c²)d².

This conditional eqn is easily solved parametrically. First, let n = 3b²-c² and assume LHS as a²n+e² = y². One can easily find {a,e}. Then assume RHS as b²+c² = z² and find {b,c}, to get the form y² = z²d² which then gives d. For example, let {b,c} = {3,4} entails solving the conditional equation,

11a²+e² = (5d)²

which is easily done.

(Note: If the particular form F₁ is extended to k = 6, it yields only trivial solns so another form has to be used.)