PART 7. Fourth PowersI. Sum / Sums of biquadrates
12.1 a+b = nc; d+e = nf 12.2 a+b ≠ c; d+e ≠ f 12.3 a+b±c = n(d+e±f) 12.4 na+b+c = d+e+nf 12.5 na+b = e+nf 12.6 a+d = n(c+f) 12.7 (a2-f2)c2 = -(b2-e2)d2
II. Quartic Polynomials as kth Powers
I. Sum / Sums of biquadrates
Fourth powers are rather interesting since quite a number of situations seem to stop at this degree. For example, it is already known that solving P(x) = 0 in the radicals is generally solvable only for degree k ≤ 4. Likewise, using Fermat’s method it seems that solving P(x) = y2 in the rationals and where P(x) has a square constant term is generally solvable only for k ≤ 4. Also, in the context of Diophantine equations, solns are known to,
ak+bk+ck = dk, and ak+bk = ck+dk
for k ≤ 4 and it is conjectured (a special case of Euler’s Extended Conjecture, or EEC) that these have no non-trivial integer solns for k > 4. Whether in fact this is the case remains to be seen.
1. Form: a4+b4 = c4+d4
J. Steggall
(a3+b)4 + (1+ad)4 = (a3+d)4 + (1+ab)4,
where d = 2ac-b and {a,b,c} must satisfy a4+3a2c-b2c+2abc2-2a2c3+1 = 0. Solving this quadratic in b in the rationals entails making its discriminant a square, and which involves the elliptic curve,
z2 = (1+a4)c+3a2c2-a2c4
If we do the substitution a = (n+1)/(n-1), one soln is c = 9n4/(2n8-2n6+n4-2n2+2) from which we can compute other rational points. Note that the resulting square, after reversing the substitution, has a factor in common with the square in Euler’s second alternative method to solve this eqn.
2. Form: pq(p2+q2) = rs(r2+s2)
Given two pairs of Pythagorean triples,
(p2-q2)2 + (2pq)2 = (p2+q2)2, and (r2-s2)2 + (2rs)2 = (r2+s2)2
which define two right triangles, the related problems of finding equal products of: 1) a leg and hypotenuse; 2) two legs,
(2pq)(p2+q2) = (2rs)(r2+s2), and, (2pq)(p2-q2) = (2rs)(r2-s2)
turn out to be equivalent to solving the equation,
(p+hq)4 + (r-hs)4 = (r+hs)4 + (p-hq)4
where h = 1 for the first, h = Ö-1 for the second and which can easily be proven by expanding this equation. Euler’s soln for the first case involves solving the simple elliptic curve y2 = (a3-b)(b3-a). To derive this, given,
pq(p2+q2) = rs(r2+s2) (eq.1)
let {p,q,r,s} = {x, ay, bx, y} to get
(a-b3)x2 + (a3-b)y2 = 0
or equivalently,
y2/x2 = (b3-a)/(a3-b)
1. First method: One can find a,b such that the above becomes a square. Euler, after a series of ingenious algebraic manipulation, gave,
b = a(1+v), where v = 3(1-a2)3/(1+10a2+a4+4a6)
which will give a 7th deg soln in the variables {p,q,r,s}.
2. Second method: Assume that x = (a3-b). One then has to find,
y2 = (a3-b)(b3-a)
The case a = b is trivial. But this curve has infinite number of rational points, another one given by,
b = a - 8a(-1+18a2-18a6+a8)/(1+100a2+190a4-44a6+9a8)
this yields a 13th deg poly soln to the problem and is related to Stegall’s. From this, other rational points can be found and parametrizations are known for degrees 7, 13, 19, etc. More will be said about this later. Other solns are,
A. Gerardin
{p,q,r,s} = {a7+a5-2a3+a, 3a2, a6-2a4+a2+1, 3a5}
equivalent to, but simpler than, Euler’s 7th deg. (Incidentally, this was also derived by Swinnerton-Dyer in his early papers at the tender age of 16.)
T. Hayashi
{p,q,r,s} = {2v3(u4+2v4), u4vw, 2uv6, uw(u4+2v4)}, if u4+3v4 = w2
The simplest non-trivial soln is {u,v,w} = {1,2,7} which yields {x1, x2, x3, x4} = {542, 103, 514, 359} so that x14+x24 = x34+x44. Note that given an initial soln to u4+3v4 = w2, subsequent ones can be found as,
{u,v,w} = {a4-3b4, 2abc, 12a4b4+c4}, if a4+3b4 = c2
which is just a special case of an identity by Lagrange. Q. Any polynomial soln to u4+3v4 = w2?
3. Form: pq(p2-q2) = rs(r2-s2)
Euler
If s = p, the equation reduces to solving p2 = q2-qr+r2, two solns of which are,
{p,q,r} = {u2+3v2, u2+2uv-3v2, 4uv} {p,q,r} = {u2+3v2, u2-2uv-3v2, u2+2uv-3v2}
The form x2+xy+y2, and its equivalent u2+3v2, appears quite often when dealing with third powers. Turns out it also plays a significant role for 4th powers.
L.J.Lander
{p,q,r,s} = {v5-2v, v5+v, -2v4+1, v4+1}
Using this initial soln, more can be derived using a general identity in the next section.
4. Form: pq(p2+hq2) = rs(r2+hs2)
To recall, via a substitution this is equivalent to the form,
a4 + b4 = c4 + d4
and it turns out this also non-trivially solves,
a4 + b4 + c4 = d4 + e4 + f4
J. Chernick
If pq(p2+hq2) = rs(r2+hs2), then,
(p2-hq2)k + (r2+hs2)k + (-h)(k/2)(2rs)k = (p2+hq2)k + (r2-hs2)k + (-h)(k/2)(2pq)k, for k = 2,4
though Chernick's identity was set at h = -1. For h = 1, this implies that the system,
a2 + b2 - c2 = d2 + e2 - f2 a4 + b4 + c4 = d4 + e4 + f4
is solvable, one of which is {a,b,c,d,e,f} = {7847, 21460, 3504, 21172, 10585, 7104} derived from the smallest soln, 594 + 1584 = 1334 + 1344. See also Form 18 on how this general form appears in the context of Descartes' Circle Theorem.
Piezas
Theorem: "Given one soln {p,q,r,s} to pq(p2+hq2) = rs(r2+hs2), then subsequent ones can be found."
Proof: (ab(a2+hb2)-cd(c2+hd2)) = (pq(p2+hq2)-rs(r2+hs2)) (m3q2-n3s2)4
a = m3pq2-3m2nqrs+2n3ps2 b = q(m3q2-n3s2) c = -2m3q2r+3mn2pqs-n3rs2 d = s(m3q2-n3s2)
where {m, n} = {3p2+hq2, 3r2+hs2}
The form x2+3y2 appears again. This identity is implicit in a method used by L.J. Lander. As was already mentioned, many polynomial solns for the case h = 1 were already known and Sinha observed that these were of degree d = 7, 13, 19, 31, 37, or d = 6n+1, and asked if it occurs only if d was prime. In “Unsolved Problems In Number Theory”, R. Guy mentions it was not necessarily the case, as there is a d = 49 though there was none known for d = 25. Turns out there is. This author found that using the above identity and initial soln,
{p,q,r,s} = {-2(1+10v2+v4+4v6), v-17v3-17v5+v7, -2v(4+v2+10v4+v6), 1-17v2-17v4+v6}
and after removing common factors, it yields a 25-deg poly. In fact, by judicious permutation of the {p,q,r,s}, one can find a 42-deg soln which is definitely not of form d = 6n+1 and doesn’t seem to be a composition of 6 and 7-deg polynomials.
5. Form: x4+y4 = z4+nt2
E. Fauquembergue (also for x4+y4 = z2+1)
(17p2-12pq-13q2)4 + (17p2+12pq-13q2)4 = (17p2-q2)4 + (289p4+14p2q2-239q4)2
As equal sums of two biquadrates, it remains to make the last term a square. With the trivial p = q to be avoided, one small soln is {p,q} = {11, 3}. Fauquembergue's identity depends on the unique integral soln of u2-2v4 = -1 with {u,v} > 1 and which is {u,v} = {239, 13}. Whether there are other identities similar in form to this remains to be seen. Incidentally, by making the third term as q2-17p2 = ±1, this also gives solns to x4+y4 = z2+1. It was proven by Fermat that x4+y4 = z2 has no non-trivial integral soln so this is the closest near-miss (just like there are integer solns to x3+y3 = z3±1 which are also near-misses of Fermat’s last theorm).
Note: This author checked x4+y4 = z2±1 and found a lot of solns for the positive case, with the smallest being {x,y,z} = {5, 7, 55}. In contrast, there were none at all for the negative case with {x,y} < 1000. Why the asymmetry? To compare, there are roughly an equal number of positive integer solns to x3+y3 = z3±1 below a bound z.
Desboves
(a2+b2)4 + (a2-b2)4 = (2ab)4 + 2(a4-b4)2
This shows that either x4k+y4 = z4+2t2, or x4+y4 = z4k+2t2 can have an infinite no. of solns.
6. Form: x4+y4 = z4+nt4
For n = 12, this has the soln,
N. Elkies
(192m8-24m4-1)4 + (192m7)4 = (192m8+24m4-1)4 + 12(4m)4
This was found while considering near-misses to the Fermat curve, x4+y4 = z4. Three of the terms are polynomials of high degree while the “excess” is only a quartic thus providing very good approximations to the curve. To illustrate, for m = 2, this gives,
245764 + 487674 ≈ (49535.0000000000063...)4
so the fourth root of the sum is very close to an integer.
7. Form: u4+nv4 = (p4+nq4)w2
R. Carmichael
Define {r,s} = {p4-nq4, p4+nq4}. Then,
(pm-2prs)4 + n(qm+2qrs)4 = (p4+nq4)(m2+16np4q4r2)2, with m = 4np4q4+s2
8. Form: u4+nv4 = x4+y4+nz4
A. Gerardin
Let r = p8-4n2,
(6pn2+pr)4 + n(3p4n-r)4 = (6pn2)4 + (pr)4 + n(3p4n+r)4
(Note that if n=0, the identity is trivial and doesn’t provide a counter-example to FLT for k = 4.)
9. Form: u4+v4 = x4+y4+nz4
R. Norrie
Define r = p8-q8. Then for any constant n,
((2n+r)p3q)4 + (2np4-q4r)4 = ((2n-r)p3q)4 + (2np4+q4r)4 + n(2pqr)4
for arbitrary p,q, so any number is the sum/difference of four rational fourth powers as was already mentioned in Part. Again, for n=0 the identity is trivial. It may be interesting to ask for what n is there a rational poly soln to x4+y4+z4 = t4+n?
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