II. Quartic Polynomials as kth Powers

1. ax⁴+by⁴ = cz²

2. ax⁴+bx²y²+cy⁴ = dz²

3. au⁴+bu²v²+cv⁴ = ax⁴+bx²y²+cy⁴

4. ax⁴+bx³y+cx²y²+dxy³+ey⁴ = z²

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,

a^k+b^k+c^k = d^k, and a^k+b^k = c^k+d^k

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

For a database of solns, see Jarek Wroblewski's results here. An infinite sequence of polynomial identities can be found using an elliptic "curve" discussed below, with the smallest having 7th deg. The beautiful version given by Gerardin has small coefficients, using only the integers {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

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. While there are identities for k = 6, none are known with such small coefficients. (Anyone can find one?)

J. Steggall

(a³+b)⁴ + (1+ad)⁴ = (a³+d)⁴ + (1+ab)⁴,

where d = 2ac-b and {a,b,c} must satisfy a quadratic in b, a⁴+3a²c-b²c+2abc²-2a²c³+1 = 0. Solving this in the rationals entails making its discriminant a square, and 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.

(Update, 2/22/10): Theorem: “The eqn a⁴+b⁴ = c⁶+d⁶ has an infinite number of unscaled, integral solns.” (Piezas)

Proof: The first known example by Giovanni Resta had the form,

(m²n)⁴(x⁴+y⁴) = (mn)⁶(u⁶+v⁶) (eq.1)

with primitive soln {m,n} = {5, 73}; and {u,v,x,y} = {3, 4, 18, 31}. Eq.1 can be simplified as,

m²(x⁴+y⁴) = n²(u⁶+v⁶)

Hence the situation is reduced to the easier problem of finding,

(x⁴+y⁴)(u⁶+v⁶) = w² (eq.2)

where gcd(x,y) = 1 and gcd(u,v) = 1. However, the second factor can also be treated as a constant, so this becomes,

25(193)(x⁴+y⁴) = w²

which is an elliptic curve. Starting with the known {x,y} = {18, 31}, an infinite number of other rational points can then be computed. Hence, another soln to eq.1 is:

{m,n} = {5, 228420709564570748140960777}; and {u,v,x,y} = {3, 4, 56328091105014, 7382523294847}

and so on, ad infinitum. It seems quite interesting that the Pythagorean triple {3, 4, 5} appears in this problem. Note: A computer search can probably find smaller co-prime solns to eq.2, though the relation u³y²-v³x² = 0 should be avoided as it is trivial. (End update.)

2. Form: pq(p²+q²) = rs(r²+s²)

It is easily seen that a⁴+b⁴ = c⁴+d⁴ and pq(p²+q²) = rs(r²+s²) are equivalent forms. 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: 1) equal products of a leg and hypotenuse, or 2) equal products of two legs,

(2pq)(p²+q²) = (2rs)(r²+s²) (eq.1)

(2pq)(p²-q²) = (2rs)(r²-s²) (eq.2)

is essentially 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 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 7^th 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 13^th 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 7^th 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 ff 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 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 (p+hq)⁴ + (r-hs)⁴ = (r+hs)⁴ + (p-hq)⁴ or equivalently, 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}. Thus if the RHS = 0, then so will be the LHS.

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”, (2nd ed, Diophantine Equations chap, p. 141.) R. Guy mentions it was not necessarily the case, as there is a d = 49, and Choudhry found a d = 25. Independently, 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 also 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-deg and 7-deg polynomials.

5. Form: x⁴+y⁴ = z⁴+nt²

Desboves

(a²+b²)⁴ + (a²-b²)⁴ = (2ab)⁴ + 2(a⁴-b⁴)²

Since the equation a²+b² = c^k is easily solved, this shows that x^4k+y⁴ = z⁴+2t² has an infinite no. of solns. (Or x⁴+y⁴ = z^4k+2t² since it is trivial to solve 2ab = c^k.)

E. Fauquembergue (also for x⁴+y⁴ = 1+z²)

(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 solves x⁴+y⁴ = z²+1. Since it was proven by Fermat that x⁴+y⁴ = z² has no non-trivial integral soln, then 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} < 10³. Why the asymmetry? To compare, there are roughly an equal number of positive integer solns to x³+y³ = z³±1 below a bound z.

(Update, Jan 3, 2011): By exhaustive computer search, Jim Cullen showed that there is no soln to x⁴+y⁴ = z²-1 with {x,y} < 10⁶. If any exists, by modular arguments it can be shown to have the form (10u)⁴+(10v)⁴ = z²-1. A table of solutions to the form (10x)⁴+(100y)² = z²-1 is also given in the link provided.

6. Form: x⁴+y⁴ = z⁴+nt⁴

N. Elkies

(192m⁸-24m⁴-1)⁴ + (192m⁷)⁴ = (192m⁸+24m⁴-1)⁴ + 192m⁴

or equivalently,

(192m⁸-24m⁴-1)⁴ + (192m⁷)⁴ = (192m⁸+24m⁴-1)⁴ + 12(2m)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 = p⁸+6np⁴q⁴+n²q⁸

See also Sophie Germain's Identity.

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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