16. Form: v^{4}+x^{4}+y^{4}+z^{4} = nt^{k}
There is yet no polynomial identity for a^{4}+b^{4}+c^{4}+d^{4} = t^{4} so this form is the closest so far.
E. Fauquembergue
(2a^{2}bc^{3})^{4} + (2ab^{2}c^{3})^{4} + (2ab(a^{4}+b^{4}))^{4} + ((a^{2}b^{2})c^{4})^{4} = (4a^{2}b^{2}(a^{4}+b^{4})^{2}c^{12})^{2}, if a^{2}+b^{2} = c^{2}
Any other polynomial soln to a^{4}+b^{4}+c^{4}+d^{4} = nt^{k}?
(Note: Yes, there is for n = 18, 63, etc. Add the ones by Haldemann.) (Update, 6/23/13): JeanJoël Delorme solved, x^{4} + y^{4} + z^{4} + t^{4} = (x^{2}+y^{2}+z^{2}t^{2})^{2} If we let a^{2}+b^{2} = c^{2} and, {x, y, z, t} = {(a^{2}b^{2})c^{4}, 2a^{2}bc^{3}, 2ab^{2}c^{3}, 2ab(a^{4}+b^{4})} then this is equivalent to Fauquembergue's. Delorme also gave a new solution as, {x, y, z, t} = {2a^{2}b^{2}c(a^{4}+b^{4}), a(a^{8}b^{8}), b(a^{8}b^{8}), abc^{5}(a^{2}b^{2})} and pointed out that if {x, y, z, t} is a solution, then so is {1/x, 1/y, 1/z, 1/t}.
(Update, 2/9/10): Lee Jacobi, Daniel Madden
The complete soln to a^{4}+b^{4} = c^{4}+d^{4} and a^{4}+b^{4}+c^{4} = d^{4} can be reduced to solving an elliptic curve. It turns out that for a special case of four 4th powers equal to a 4th power namely,
a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4} (eq.0)
with variables as ±, then one can do so as well, a small example of which is,
955^{4}+1770^{4}+(2634)^{4}+5400^{4} = (955+17702634+5400)^{4} = 5491^{4}
Jacobi's and Madden’s clever soln depended on the identity,
a^{4}+b^{4}+(a+b)^{4} = 2(a^{2}+ab+b^{2})^{2}
an identity also useful for solving a^{4}+b^{4}+c^{4}+d^{4}+e^{4} = f^{4} as quadratic forms. (See Form 21 below.) Their method starts by adding (a+b)^{4} + (c+d)^{4} to both sides of eq.0,
a^{4}+b^{4}+(a+b)^{4}+c^{4}+d^{4}+(c+d)^{4} = (a+b)^{4}+(c+d)^{4}+(a+b+c+d)^{4}
and, using the identity, it is seen that eq.0 is equivalent to the special Pythagorean triple,
(a^{2}+ab+b^{2})^{2} + (c^{2}+cd+d^{2})^{2} = ((a+b)^{2} + (a+b)(c+d) + (c+d)^{2})^{2}
Sparing the reader the intermediate algebraic manipulations, using the transformation, {a,b,c,d} = {p+r, pr, q+s, qs}, the eqn,
(p+r)^{4}+(pr)^{4}+(q+s)^{4}+(qs)^{4} = (2p+2q)^{4}
can be solved if {p,q,r,s} satisfy two quadratics to be made squares,
(m^{2}7)p^{2}+4(m^{2}1)pq+4(m^{2}1)q^{2} = (m^{2}+1)r^{2} (eq.1) 8mp^{2}+8mpq(3m^{2}8m+3)q^{2} = (m^{2}+1)s^{2} (eq.2)
for some constant m. This is easily proven by solving {r,s} in radicals and substituting it into the eqn to see that it holds. These two quadratic conditions define an elliptic curve. One can then try to find a suitable m and find rational {p,q,r,s}. Or, given a known {p,q,r,s}, a value m can be derived as,
m = (3q^{2}+s^{2}) / ((p+2q)^{2}r^{2})
Example: Given {a,b,c,d} = {5400, 1770, 2634, 955}, then {p,q,r,s} = {3585, 1679/2, 1815, 3589/2}, and m = 961/61. This initial soln set can be used to find an infinite more. Since eq.1 and eq.2 are homogeneous, we can assume q = 1 without loss of generality. Starting with a soln to eq.1 as p_{1} = 2(3585)/1679, a complete parameterization is,
p = (30/1679) (4291863+112196282u+110805419u^{2})/(448737463621u^{2})
which makes eq.1 a square for any u. But substituting this to eq.2, and still with q = 1, produces a rational 4th deg polynomial in u that still has to be made a square. To find an initial rational point u, equating pp_{1} = 0 yields the factor 55770003+56098141u = 0, hence this is one suitable value. Treating the quartic polynomial as an elliptic curve, an infinite more rational u can be computed, thus proving that the eqn, a^{4}+b^{4}+c^{4}+d^{4} = (a+b+c+d)^{4}
has an infinite number of distinct, rational solns. (End update.)
17. Form: v^{k}+x^{k}+y^{k}+z^{k} = a^{k}+b^{k}+c^{k}+d^{k}, k = 2,4
Crussol
For the special form,
[0] (p+u)^{k} + (p+u)^{k} + (q+v)^{k} + (q+v)^{k} = (r+u)^{k} + (r+u)^{k} + (s+v)^{k} + (s+v)^{k}, k = 1, 2, 4
one must satisfy two conditions,
p^{2}+q^{2} = r^{2}+s^{2}
p^{2}+3u^{2} = s^{2}+3v^{2}
The system is also equivalent to the two simultaneous equations,
[1] p^{2}+3u^{2} = s^{2}+3v^{2}
[2] r^{2}+3u^{2} = q^{2}+3v^{2}
which is quite easily solved, though the trivial case p^{2}3v^{2} = q^{2}3u^{2} and others that merely permute the x_{i}, y_{i }should be avoided. If [0] is to be valid for k = 6 as well, then a third condition must be met,
[3] p^{2}+q^{2} = 5(u^{2}+v^{2})
See also the section on Crussol's approach.
Piezas
Theorem: If a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, for k = 2,4, and abcd = efgh. Define n as a^{4}+b^{4}+c^{4}+d^{4} = n(a^{2}+b^{2}+c^{2}+d^{2})^{2} then,
32(a^{6}+b^{6}+c^{6}+d^{6}e^{6}f^{6}g^{6}h^{6})(a^{10}+b^{10}+c^{10}+d^{10}e^{10}f^{10}g^{10}h^{10}) = 15(n+1)(a^{8}+b^{8}+c^{8}+d^{8}e^{8}f^{8}g^{8}h^{8})^{2}
Note if d = h = 0, the condition abcd = efgh disappears and this reduces to the theorem given earlier of which Ramanujan’s 6108 Identity is a special case. Proof: Given,
a^{2}+b^{2}+c^{2}+d^{2} = e^{2}+f^{2}+g^{2}+h^{2} (eq.1) a^{4}+b^{4}+c^{4}+d^{4} = e^{4}+f^{4}+g^{4}+h^{4} (eq.2)
Eliminate h between the two to get a quadratic in {a,b,c,d} and solve for any, say d. Solve for h in eq.1 and we get the complete radical soln of eq.1 and eq.2. Substitute these two values into the 6108 identity and it has a factor, call this P, which is the same polynomial that results when the values are substituted into abcdefgh = 0. Thus if the latter is satisfied so is the former. (End proof.) A solution will be given below.
Theorem. If a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, for k = 2,4,6, where abcd = efgh, and a+b ≠ ±(c+d); e+f ≠ ±(g+h), (call the entire system V_{1}) then,
(a+b+cd)^{k} + (a+bc+d)^{k} + (ab+c+d)^{k} + (a+b+c+d)^{k} + (2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k} = (e+f+gh)^{k} + (e+fg+h)^{k} + (ef+g+h)^{k} + (e+f+g+h)^{k} + (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k}
for k = 1,2,4,6,8,10.
Again, for d = h = 0, the condition abcd = efgh disappears so reduces to Birck’s Theorem (given later). This is then an extension to tenth powers and the proof will be discussed in that section. Even for nonzero variables it is possible one or two of the addends will vanish, as we shall see later. This author hasn’t been able to find a parametric soln to V_{1} but was able to do so for the simpler case when valid only for k = 2,4. Given,
x_{1}^{k }+ x_{2}^{k }+ x_{3}^{k }+ x_{4}^{k} = y_{1}^{k }+ y_{2}^{k }+ y_{3}^{k }+ y_{4}^{k}, k = 2,4
assume it to have the form, call it F_{1},
a^{k}(p+q)^{k} + b^{k}(rs)^{k} + c^{k}(r+s)^{k} + d^{k}(pq)^{k} = a^{k}(pq)^{k} + b^{k}(r+s)^{k} + c^{k}(rs)^{k} + d^{k}(p+q)^{k}
which has x_{1}x_{4} = y_{1}y_{4} and x_{2}x_{3} = y_{2}y_{3}. This sufficient but not necessary constraint is completely parameterized by F_{1}. To see this, use the Mathematica command,
Solve[{a(p+q), d(pq), a(pq)} = {x_{1}, x_{4}, y_{1}}, {a,p,d}]
to express {a,d,p} in terms of {x_{1}, x_{4}, y_{1}}. Substituting these into d(p+q) = y_{4}, we get,
x_{1}x_{4}/y_{1} = y_{4}
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 x_{1}+x_{2}+x_{3}+x_{4} = y_{1}+y_{2}+y_{3}+y_{4} = 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. F_{1} takes care of the necessary requirement that x_{1}x_{2}x_{3}x_{4} = y_{1}y_{2}y_{3}y_{4}. To simplify matters, assume further that x_{1}+x_{2} = y_{1}+y_{2} (not obeyed by the example above). After some algebra, we get the soln,
{p,q,r,s} = {ab^{2}ac^{2}, be, a^{2}bbd^{2}, ae}, where a^{2}(3b^{2}c^{2}) + e^{2} = (b^{2}+c^{2})d^{2}.
This conditional eqn is easily solved parametrically. First, let n = 3b^{2}c^{2} and assume LHS as a^{2}n+e^{2} = y^{2}. One can easily find {a,e}. Then assume RHS as b^{2}+c^{2} = z^{2} and find {b,c}, to get the form y^{2} = z^{2}d^{2} which then gives d. For example, let {b,c} = {3,4} entails solving the conditional equation,
11a^{2}+e^{2} = (5d)^{2}
which is easily done.
(Note: If the particular form F_{1} is extended to k = 6, it yields only trivial solns so another form has to be used.)
18. Form: 2(v^{4}+x^{4}+y^{4}+z^{4}) = (v^{2}+x^{2}+y^{2}+z^{2})^{2}
The equation,
2(a^{4}+b^{4}+c^{4}) = (a^{2}+b^{2}+c^{2})^{2}
has the simple soln a+b = c, which is central to Ramanujan's 6108 Identity. Turns out when there are four terms,
2(v^{4}+x^{4}+y^{4}+z^{4}) = (v^{2}+x^{2}+y^{2}+z^{2})^{2} (eq.1)
it has a parametric soln as well. Note that this is a special case of Descartes' Circle Theorem,
2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}) = (x_{1}+x_{2}+x_{3}+x_{4})^{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} = (v^{2}+x^{2}y^{2}z^{2})^{2} (2vy)^{2} + (2xz)^{2} = (v^{2}x^{2}+y^{2}z^{2})^{2} (2vz)^{2} + (2xy)^{2} = (v^{2}x^{2}y^{2}+z^{2})^{2}
so is a special case of three simultaneous Pythagorean triples. More generally, for any d, we are to solve,
2(d^{2}v^{4}+x^{4}+y^{4}+z^{4}) = (dv^{2}+x^{2}+y^{2}+z^{2})^{2} (eq.2)
and Euler's soln can be given as {v,x,y,z} = {2pqrs(a^{2}t^{2}), t(a^{2}t^{2}), t(abrscpqt), a(abrscpqt)}, where {a,b,c} = {pq(r^{2}ds^{2}), p^{2}+dq^{2}, r^{2}+ds^{2}}, and {p,q,r,s,t} satisfies the conditional eqn,
pqrs(p^{2}dq^{2})(r^{2}ds^{2}) = t^{2}
For d = 1, one soln is {p,q,r,s} = {2m^{2}n^{2}, m^{2}+n^{2}, m^{2}+n^{2}, m^{2}2n^{2}}. 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(r^{2}ds^{2}) and the conditional eqn becomes the symmetric,
pq(p^{2}dq^{2}) = rs(r^{2}ds^{2})
This has appeared already in Section 013 and has many solns and is connected to the two distinct eqns,
a^{4}+b^{4} = c^{4}+d^{4} a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4
Thus, we can also solve 2(n^{2}v^{4}+x^{4}+y^{4}+z^{4}) = (nv^{2}+x^{2}+y^{2}+z^{2})^{2} as:
I. When n = 1:
{v,x,y,z} = {2(abcd)(ab+cd), (a^{2}+b^{2}+c^{2}+d^{2})(a^{2}b^{2}+c^{2}d^{2}), 2(acbd)(a^{2}+c^{2}), 2(acbd)(b^{2}+d^{2})}, where a^{4}+b^{4} = c^{4}+d^{4}.
II. When n = 1:
Let a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,4, where a^{2}+b^{2} = f^{2}, d^{2}+e^{2} = c^{2}, and ab = de, then,
{v,x,y,z} = {aebd, (a+d)(ad), a(c+f), d(c+f)}
A soln given by Gloden is {a,b,c,d,e,f} = {p^{2}q^{2}, p^{2}+r^{2}, 2pq, p^{2}+q^{2}, p^{2}r^{2}, 2pr}, where p^{2} = q^{2}+qr+r^{2}. This can be solved by {p,q,r} = {u^{2}+uv+v^{2}, u^{2}v^{2}, 2uv+v^{2}}. (End update.)
19. Form: x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}, k = 1,2,3,4
The system of eqns,
x_{1}^{k}+x_{1}^{k}+…+ x_{n}^{k} = y_{1}^{k}+y_{1}^{k}+…+ y_{n}^{k}, k = 1,2,3,4
has nontrivial 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},
for the special case when h = a+b+c, then,
{d,e,f,g} = {(p+u)/(2q), (pu)/(2q), (p+v)/(2q), (pv)/(2q)} where p = (a+b)(a+c)(b+c), q = (ab+ac+bc), and {a,b,c} satisfies the two quadratic conditions, p(p+4abc) = u^{2} p(p+4abc)  4q^{3} = v^{2}
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: x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}, k = 1,2,3,4
The complete soln to this system, call this E_{4}, is unknown. Use the form,
(a+bj)^{k}+(c+dj)^{k}+(e+fj)^{k}+(g+hj)^{k}+(i+j)^{k} = (abj)^{k}+(cdj)^{k}+(efj)^{k}+(ghj)^{k}+(ij)^{k}
and let i = (ab+cd+ef+gh), h = (1+b+d+f) to make it identically true for k = 1,2. I. Case x_{2}+x_{3} = y_{2}+y_{3}: 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)j^{2}+(Poly20) = 0 (eq.1) (Poly11)j^{4}+(Poly21)j^{2}+(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 = y^{2}, one uses a quadratic form e = Q(x) for some variable x. After a rational g is found, to find j, solve,
(Poly10)(Poly20) = z^{2}
which is a quadratic in e. Substituting the quadratic form e = Q(x) into this, it becomes,
Poly(x) = z^{2} (eq.3)
which is a quartic in x, hence any soln to E_{4} with the constraint x_{2}+x_{3} = y_{2}+y_{3} must satisfy eq.3, a quartic polynomial to be made a square. II. Case x_{1}+x_{2} = y_{1}+y_{2}; x_{2}+x_{3} = y_{2}+y_{3}: 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 multivariable polynomial parametrization to E_{4} .
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} = (ab)^{k}+(cd)^{k}+(ef)^{k}+(gh)^{k}+(ij)^{k}
for example, is messier.
21. Form: x_{1}^{4}+x_{2}^{4}+…x_{n}^{4}, n > 4
E. Fauquembergue
(4x^{4}y^{4})^{4} + 2(4x^{3}y)^{4} + 2(2xy^{3})^{4} = (4x^{4}+y^{4})^{4}
C.Haldeman
(2x^{2}+12xy6y^{2})^{4} + (2x^{2}12xy6y^{2})^{4} + (4x^{2}12y^{2})^{4} + (3x^{2}+9y^{2})^{4} + (4x^{2}+12y^{2})^{4} = 5^{4}(x^{2}+3y^{2})^{4}
This gives as a first instance 2^{4}+2^{4}+4^{4}+3^{4}+4^{4 }= 5^{4}. Similar identities were found by A. Martin and Ramanujan. A generalization has been found by this author.
Piezas
(ax^{2}+2v_{1}xy3ay^{2})^{k} + (bx^{2}2v_{2}xy3by^{2})^{k} + (cx^{2}2v_{3}xy3cy^{2})^{k} = (a^{k}+b^{k}+c^{k})(x^{2}+3y^{2})^{k}
where {v_{1}, v_{2}, v_{3}, c} = {a+2b, 2a+b, ab, a+b}, for k = 2,4
Thus, it suffices to decompose a sum which starts as a^{4}+b^{4}+(a+b)^{4}+… into a sum and difference of any number of biquadrates and apply it to the identity. For example, using the eqn 50^{4}+50^{4}+100^{4}+4^{4}+15^{4} = 103^{4}, this yields,
(50x^{2}+300xy150y^{2})^{4} + (50x^{2}300xy150y^{2})^{4} + (100x^{2}300y^{2})^{4} + (4x^{2}+12y^{2})^{4} + (15x^{2}+45y^{2})^{4} = 103^{4}(x^{2}+3y^{2})^{4}
R. Norrie
(x+y)^{4} + (x+y)^{4} + (2y)^{4} + (u^{2}v^{2})^{4} + (2uv)^{4} = (u^{2}+v^{2})^{4}, if 2uv(u^{2}v^{2}) = x^{2}+3y^{2}
Note that {u^{2}v^{2}, 2uv} are the legs of a Pythagorean triple. Thus, if the product of the legs can be expressed as x^{2}+3y^{2}, 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
x^{4}+4y^{4} = ((x+y)^{2}+y^{2}) ((xy)^{2}+y^{2}) = (x^{2}+2xy+2y^{2})(x^{2}2xy+2y^{2})
The quartic form can be generalized as,
x^{4}+(a^{2}+2b)x^{2}y^{2}+b^{2}y^{4} = (x^{2}+axy+by^{2})(x^{2}axy+by^{2})
where Germain's was the case {a,b} = {2,2}. Another interesting case is {a,b} = {2,1},
x^{4}6x^{2}y^{2}+y^{4} = (x^{2}+2xyy^{2})(x^{2}2xyy^{2}) = (x^{2}y^{2})^{2}  (2xy)^{2}
which sometimes appear when dealing with Pythagorean triples.
(Update, 2/3/10): Choudhry
Given the eqn,
a^{4}+4b^{4} = c^{4}+4d^{4} (eq.1)
use the transformation,
{a,b,c,d} = {(x^{2}+2x2)z+xy, 2xz+y, (x^{2}+2x+2)z+xy, y}
so that, substituted into eq.1, it has a linear factor in y. This first soln can be used to compute an infinite sequence of polynomial solns, as eq.1 is an elliptic curve in disguise. The transformations, {a,b,c,d} = {p+q, rs, pq, r+s}, then {p,q,r,s} = {u, mv, nu, v} reduce eq.1 to the form,
(m4n^{3})u^{2}+(m^{3}4n)v^{2} = 0
so one is to solve,
(m4n^{3})(m^{3}4n) = z^{2}
One soln is,
{p,q,r,s} = {8x+2x^{3}, 4+2x^{2}, 4+4x^{2}, 2x+x^{3}}
from which others can then be computed. Source: The Diophantine Equation A^{4}+4B^{4} = C^{4}+4D^{4}, Indian Journal of Pure and Applied Mathematics, Vol. 29, 1998. (End update.)
R.Carmichael
(a^{4}2b^{4})^{4} + (2a^{3}b)^{4} + 4(2ab^{3})^{4} = (a^{4}+2b^{4})^{4}
Note that the form (a^{4}+2b^{4})^{4} can also be expressed as a sum of both squares and biquadrates,
E. Fauquembergue
(a^{4}2b^{4})^{4} + (2a^{3}b)^{4} + (8a^{2}b^{6})^{2} = (a^{4}+2b^{4})^{4}
(2a^{2}b^{2})^{4} + (2a^{3}b)^{4} + (a^{8}4a^{4}b^{4}4b^{8})^{2} = (a^{4}+2b^{4})^{4}
