“*Euler’s Extended Conjecture* and a^{k} +
b^{k} + c^{k} = d^{k} for *k*
> 4”

**by Titus Piezas
III**

“*Wir müssen
wissen. Wir werden wissen.*”

(“We must know. We shall know.”) – David Hilbert

Keywords: equal sums of like powers, diophantine equations, Fermat’s Last Theorem (FLT), Euler’s Extended Conjecture (EEC), cubic and quintic quadruples.

*Contents:*

I. Introduction

II. Historical Background

III. Derivation of a^{k} + b^{k} =
c^{k} + d^{k}, for *k* = 3 and 5

IV. Numerical Results

V. Conclusion

# I. Introduction

We’ll start this paper on *Euler’s Extended Conjecture ( EEC)*
by giving our main result, that is, a

**four-parameter**solution to the equation,

x_{1}^{5} + x_{2}^{5} +
x_{3}^{5} =
x_{4}^{5}
(eq.0)

which, if we disregard sign, is equivalent to the form
x_{1}^{5} + x_{2}^{5} =
x_{3}^{5} + x_{4}^{5}.
Let,

(Öp + Öq)^{5} + (Öp - Öq)^{5} = (Ör + Ös)^{5} + (Ör - Ös)^{5}

then,

p = 5(bc-ad)(c^{2}+10cd+5d^{2})^{2}

q = (a^{2}+10ab+5b^{2})^{3 }-
(ac+10bc+5bd)(c^{2}+10cd+5d^{2})^{2}

r = 5(bc-ad)(a^{2}+10ab+5b^{2})^{2}

s = -(c^{2}+10cd+5d^{2})^{3} +
(ac+10ad+5bd)(a^{2}+10ab+5b^{2})^{2}

for arbitrary *a,b,c,d*. As an
example, let a = 1, b = 9, c = 3, d = 3 and we have,

(-2272Ö22 +
288Ö30)^{5} +
(2272Ö22 + 288Ö30)^{5} = (-2528Ö6 + 992Ö30)^{5} + (2528Ö6 + 992Ö30)^{5}

The author is aware of only one other formula for (eq.0), a two-parameter
solution found by A. Desboves in 1880 and independently by N. Elkies in 1995 to
be given later. It may in fact be just a special case of the
new formula just given. The Desboves-Elkies solution is
*multigrade*, valid for multiple exponents *k* = 1,2,5.
The new formula, for *a,b,c,d* satisfying a certain condition, can
also be true for *k* = 1,2,5. It may be the complete
parametrization of these *quintic quadruples* since it was derived using
Euler’s method to find the complete one for cubic quadruples
x_{1}^{3} + x_{2}^{3} =
x_{3}^{3} + x_{4}^{3}. If so,
then to find a counter-example to EEC it suffices to find appropriate values
such that the expressions *p,q,r,s* are *all* squares.

So what exactly is EEC? Well, our story starts a long time ago, in a land far away…

# II. Historical Background

In their article “*Pythagoras’
Theorem In Babylonian Mathematics*”, J. O’Connor and E. Robertson mentions a
Babylonian tablet from **c.1900 BC**, which has the following problem,

*“4 is the length and 5 the diagonal. What is the
breadth?*

*
Its size is not known.*

*
4 times 4 is 16. 5 times 5 is 25.*

*
You take 16 from 25 and there remains 9.*

*
What times what shall I take in order to get 9?*

*
3 times 3 is 9. 3 is the breadth.”*

We can infer that the Babylonians knew there were many more examples of
two second powers equal to a second power and that 3^{2} + 4^{2}
= 5^{2} was not an isolated result. Another tablet, the famous Plimpton
322 also from the same time period and kept in Columbia University, contains
various paired values in sexigesimal which can be seen as part of a Pythagorean
triple. The largest pair, converted in decimal, is (18541,
12709) and a quick calculation shows that 18541^{2} – 12709^{2}
= 13500^{2}. Quick for us using a calculator, but the
size of this example shows the Babylonians *must* have known of a method to
generate solutions to a^{2} + b^{2} = c^{2} other than
by randomly scribbling values on the sand.

(Incidentally, most of the modern results in equal sums of like powers were found by computer search. And interestingly enough, silicon chips in computers get their silicon mostly from, what else, sand. After all these years, from Archimedes reckoning on it to scientists using computers, mathematics still gets help from sand.)

Jumping a few millennia to the early 1630’s, a certain lawyer and
“amateur”** ^{1}** mathematician by the name of Pierre de Fermat
(1601-1665) wrote on the margin of a copy of Diophantus’

*Arithmetica*,

*“…It is impossible for a cube
to be the sum of two cubes, a fourth power to be the sum of two fourth powers,
or in general for any number that is a power greater than the second to be the
sum of two like powers. I have discovered a truly marvelous
demonstration of this proposition that this margin is too narrow to
contain.”* (Nagell, T., Introduction to Number Theory, p.
251-253)

Fermat’s Last Theorem, or FLT, in
modern notation is that a^{k} + b^{k} = c^{k} has no
non-trivial integral solution for k > 2. This cryptic
remark** ^{2}** would tantalize generations of mathematicians and
amateurs alike. (And in some quarters, still is.) If indeed
he wrote this note when he first studied this book in the early 1630’s, then he
would be around 30 years old at the time. By Sept 1636 he
would mention the problem of two cubes whose sum is a cube in a letter to
Sainte-Croix.

The next figure is the great Swiss mathematician Leonhard Euler (1707-1783). In one of his letters he wrote,

*“…It has seemed to many
Geometers that this theorem [FLT] may be generalized. Just as
there do not exist two cubes whose sum is a cube, it is certain that it
is impossible to exhibit three biquadrates whose sum is a biquadrate, but that
at least four biquadrates are needed if their sum is to be a biquadrate,
although no one has been able up to the present to assign four such
biquadrates. In the same manner it would seem to be
impossible to exhibit four fifth powers whose sum is a fifth power, and
similarly for higher powers.”* (Dickson, L., History of
the Theory of Numbers, Vol 2, p. 648)

In a letter to Goldbach dated Aug
4, 1753, Euler claimed to have proven FLT for *k* = 3.
His conjecture above, now known as the **Euler’s Sum of Powers
Conjecture** or simply **ESC**, that** **it would take **at least**
*k* *k*th powers to sum to an *k*th power (other than the trivial
identity x^{k} = x^{k}), must be after this letter.
The first example of ESC for *k* > 3 turned up only after more
than 150 years, when R. Norrie finally found the four biquadrates in 1911,

30^{4} + 120^{4} +
272^{4} + 315^{4} = 353^{4}

One reason why ESC is not so
well-known is that the limelight was on FLT. Surprisingly, it
also turned out to be *false* and is one instance that Euler was wrong.
Lander, Parkin, and Selfridge in their seminal 1967 paper found that,

*“Since 27 ^{5} +
84^{5} + 110^{5} + 133^{5} = 144^{5}, then ESC
is false.”*

In their concluding remarks they also asked, given the diophantine equation (k,m,n),

x_{1}^{k} + x_{2}^{k} + … +
x_{m}^{k} = y_{1}^{k} +
y_{2}^{k} + … + y_{n}^{k}

where *x*_{i} and *y*_{i} are
assumed to be positive integers. Then,

[1] Is it true that (k,m,n) is never solvable when m+n < k?

[2] For which *k, m, n* such that m+n = k is
(k,m,n) solvable?

[3] Is (k,m,n) always solvable when m+n > k?

To quote further, *“…The results
presented in this paper tend to support an affirmative answer to
[1]. Question [2] appears to be especially
difficult. The only solvable cases with m+n = k known at
present are (4,2,2), (5,1,4), and (6,3,3).”* (Lander, L.,
Parkin, T., and Selfridge, J., “A Survey of Equal Sums of Like Powers”, Math. Of
Computation, Vol 21, 1967.)

This author made minor changes in
the numbering of questions but otherwise they are quoted verbatim.
Also, from this point onwards we will adopt the standard notation in the
literature of (k,m,n) with *k* as the power but with the modification that
*m is the number of terms on one side, and n on the other*.
Thus, the label (4,1,3) equally describes,

a^{4} + b^{4} + c^{4} = d^{4},
or
a^{4} = b^{4} + c^{4} + d^{4}

since mathematically they are equivalent anyway.

Some comments:
First, there is a related problem to the one discussed by Lander et al,
namely the *Prouhett-Tarry-Escott Problem* (*PTE*). However, this
seeks for a multigrade solution. Furthermore, it usually involves a so-called
“balanced equation”, with an equal number of terms on both sides.
We are more after a minimal number of terms for (k,m,n) and solutions
need not be multigrade. For more on PTE see Chen Shuwen’s
excellent site hosted at http://euler.free.fr/eslp/eslp.htm.

Second, three more cases of k =
m+n are now known, namely (4,1,3), (5,2,3), and (8,3,5).
Third, the remark “*…The results presented in this paper tend to
support an affirmative answer to [1]* is very cautiously worded, but if we
define a conjecture as “…a *proposition* which is consistent with known
data, but has neither been verified nor shown to be false” then while it does
not make a definite stand the authors couldn’t quite resist making a leading
statement. (Compare to Euler’s confident “*It is
certain that it is impossible…*” with only the data for (2,1,2) and
(3,1,3) known to him. Then again, he was Euler.)

So with the known result that ESC
was wrong plus question [1], Ekl in a follow-up “*New Results In Equal Sums Of
Like Powers*” (1998), then defined **Euler’s Extended Conjecture (EEC)**.
Just like FLT, this can be simply stated:

*“The diophantine equation (k,m,n),*

*x _{1}^{k} +
x_{2}^{k} + … + x_{m}^{k} =
y_{1}^{k} + y_{2}^{k} + … +
y_{n}^{k}*

*has no integral solution for k > m+n, other than the
trivial case when all x _{i} = y_{i}.”*

Since this is a very broad
conjecture, extending infinitely in two directions, namely the number of terms
and the exponent, it is convenient to label it as* cases* by the number of
terms *m+n*. The first two cases *m+n* = 1 and *m+n* = 2 are
trivial and it only the *third* that is the first non-trivial case,
*m+n* = 3 or EEC(k,1,2), and is the familiar claim that a^{k} +
b^{k} = c^{k} has no solutions for k > 3 which is true by
FLT. Hence, *FLT is just a special case of EEC, up to a
point*. A proof of the whole of EEC would automatically
imply FLT other than k = 3, which can be proven separately anyway as Euler
did. The fourth case is *m+n* = 4, with two versions,
EEC(k,1,3) and EEC(k,2,2),

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

which asserts that these have no solutions for k >
4. One can see the difference with FLT is that the greater
number of terms mean we can “move” them around in various versions and which
spell complications for proving the general conjecture. The
separation into *m* and *n* terms is obviously for the case of even
exponents since, if we drop the condition that *x*_{i} and
*y*_{i} are assumed to be positive integers (which, mathematically,
is rather arbitrary), then for *odd* exponents the various versions are
equivalent to the most symmetrical form,

*x _{1}^{k} + x_{2}^{k} + … +
x_{n}^{k} = 0*

and to prove the conjecture for a *particular* odd
exponent it suffices to prove it for one version. For the
equations (k,2,2) and (k,1,3) where k = 4, we have the first solutions,

59^{4} + 158^{4} = 133^{4} +
134^{4} (Euler, c.1750)

2,682,440^{4 }+ 15,365,639^{4} + 18,796,760^{4} =
20,615,673^{4} (Elkies, 1986)

where the latter is the second counter-example to ESC, but
the first for *k* = 4. A smaller one was later found by
J. Frye and more examples were subsequently found. However,
whether these two have solutions for k > 4 is definitely an **open**
question, as well as for the fifth *m+n* = 5 with versions EEC(k,1,4) and
EEC(k,2,3), and so all for an infinite number of cases.

The general objective of this
paper is to elaborate more on EEC. In particular, since the
first two cases are trivial and the third reduces to the known result for FLT,
then it will deal with the fourth, namely m+n = 4, the first unknown and
*only* its first exponent *k* = 5. This has two
versions but since we are dealing with an odd power and as was pointed out we
won’t distinguish between positive and negative solutions, then we will consider
our result as applicable for both. For convenience, we will
use the balanced equation (5,2,2). The next exponent and its two versions, as
well as all subsequent exponents, will be left for others.

It took about 200 years from the time Fermat sent the letter discussing
FLT for *k* = 3 to the complete solution of *k* = 5 by Germain,
Legendre, and Dirichlet in 1825. (Fermat himself proved the case for *k* =
4.) The next few decades saw more exponents *k* being
proven. Likewise, it took about 200 years from the time Euler
stated his original conjecture to the discovery of the counter-example
27^{5} + 84^{5} + 110^{5} + 133^{5} =
144^{5} in 1967. Let us hope that the next few decades will also spell
progress for EEC.

# III. Derivation of a^{k} + b^{k} = c^{k} +
d^{k}, for *k* = 3 and 5

Before we derive the formula given in the Introduction, the one found by Desboves and Elkies is, let,

*f*_{1} = x^{2} +(Ö2)xy-y^{2},
*f*_{2} = *i*x^{2} -(Ö2)xy+*i*y^{2},
*f*_{3} = -x^{2} +(Ö2)xy+y^{2},
*f*_{4} = -*i*x^{2} -(Ö2)xy-*i*y^{2},

Define *E*_{k} =
*f*_{1}^{k} + *f*_{2}^{k} +
*f*_{3}^{k} + *f*_{4}^{k}, then
*E*_{k} = 0 for k = 1,2,5.

Since *i* is the imaginary
unit Ö(-1), then we can’t
use this to find a counter-example to EEC. To find another
parametrization, we will first show how Euler found the complete one for cubic
quadruples as the quintic version was found by analogy. His
method in fact was related to his proof of FLT for *k* = 3 which used the
properties of the algebraic form p^{2} + 3q^{2}.
(We’ve come across this form before, as the related version
a^{2}+ab+b^{2} was prominent in the previous paper *“Ramanujan
and The Quartic Equation 2 ^{4} + 2^{4} + 3^{4} +
4^{4} + 4^{4} = 5^{4}”*.) The
method is as follows, given the quadruple,

x_{1}^{3} + x_{2}^{3} =
x_{3}^{3} + x_{4}^{3}

let x_{1} = p+q, x_{2} = p-q, x_{3} =
r+s, x_{4} = r-s, then,

p(p^{2} + 3q^{2}) = r(r^{2} +
3s^{2})
(eq.1)

Euler noted that the second
factors of each side have a common divisor of like form, call it u^{2} +
3v^{2}. The objective then is to find expressions
*p,q,r,s* such that each side factors. He then gave such
expressions as if plucked from thin air, assuming the gentle reader could
follow. The author knew that this step was crucial, since one
would have to find analogous expressions for the quintic case.
It turned out one simply had to set,

p^{2} + 3q^{2} = (a^{2} +
3b^{2})(u^{2} +
3v^{2})
(eq.2)

and *factor this over **Ö-3*. We then get two
equations,

(p + qÖ-3) = (a + bÖ-3)(u + vÖ-3) and, (p - qÖ-3) = (a - bÖ-3)(u - vÖ-3)

and solving for the two unknowns *p,q* should enable us
to express them in terms of *a,b,u,v*, given by,

p = au-3bv, q = bu+av

Similarly for r^{2} + 3s^{2}, we set,

r^{2} + 3s^{2} = (c^{2} +
3d^{2})(u^{2} +
3v^{2})
(eq.3)

and factoring over Ö-3 and solving for *r,s*
from the two resulting equations we find,

r = cu-3dv, s = du+cv

Substituting *p,r* and eq.2 and eq.3 into eq.1, we
get,

(au-3bv)(a^{2} + 3b^{2})(u^{2} + 3v^{2})
= (cu-3dv)(c^{2} + 3d^{2})(u^{2} + 3v^{2})

with the common divisor u^{2} + 3v^{2}, so
this is just linear in *u,v*, giving,

u(a^{3}+3ab^{2}-c^{3}-3cd^{2}) =
3v(a^{2}b+3b^{3}-c^{2}d-3d^{3})

So, u =
3(a^{2}b+3b^{3}-c^{2}d-3d^{3}) and v =
a^{3}+3ab^{2}-c^{3}-3cd^{2} (to eliminate
denominators). Substituting these two into the expressions
for *p,q,r,s*, we get,

(p+q)^{3} + (p-q)^{3} = (r+s)^{3} +
(r-s)^{3}

where,

p = 3(bc-ad)(c^{2}+3d^{2}),
q = (a^{2}+3b^{2})^{2} –
(ac+3bd)(c^{2}+3d^{2})

r =
3(bc-ad)(a^{2}+3b^{2}),
s = -(c^{2}+3d^{2})^{2} +
(ac+3bd)(a^{2}+3b^{2})

and we have the complete parametrization of cubic quadruples
with four free variables *a,b,c,d*! Most of this is
discussed in Dickson, p. 552-554. However, J. Binet gave a
version (p.555) of Euler’s formula with just two variables,

x_{1}^{3} + x_{2}^{3} =
x_{3}^{3} + x_{4}^{3}

where,

x_{1} = 1-(a-3b)(a^{2}+3b^{2}),
x_{2} = (a+3b)(a^{2}+3b^{2})-1

x_{3} = (a+3b)-(a^{2}+3b^{2})^{2},
x_{4} =
(a^{2}+3b^{2})^{2}-(a-3b)

and stated there was no loss of generality, though the justification for this assertion was not explicitly given by Dickson.

We now go into quintic quadruples by just following Euler’s footsteps,
*but with a small difference*. Given,

x_{1}^{5} + x_{2}^{5} =
x_{3}^{5} + x_{4}^{5}

let, x_{1} = Öp+Öq, x_{2} = Öp-Öq, x_{3} = Ör+Ös, x_{4} = Ör-Ös, then,

(Öp)(p^{2}+10pq+5q^{2})
= (Ör)(r^{2}+10rs+5s^{2})
(eq.4)

Like Euler, we consider the second factors as having a common divisor. We set,

p^{2}+10pq+5q^{2} =
(a^{2}+10ab+5b^{2})(u^{2}+10uv+5v^{2})
(eq.5)

and factor this over Ö5 to get two equations just
like before. We then solve for *p,q* which are given
by,

p = au-5bv, q = bu+av+10bv

Similarly,

r^{2}+10rs+5s^{2} =
(c^{2}+10cd+5d^{2})(u^{2}+10uv+5v^{2})
(eq.6)

so, r = cu-5dv, s = du+cv+10dv

Substituting *p,r* and (eq.5) and (eq.6) into the
*square* of (eq.4), we get,

(au-5bv)((a^{2}+10ab+5b^{2})(u^{2}+10uv+5v^{2}))^{2}
=
(cu-5dv)((c^{2}+10cd+5d^{2})(u^{2}+10uv+5v^{2}))^{2}

and find (after simplification) *u,v* as,

u = 5(b(a^{2}+10ab+5b^{2})^{2} –
d(c^{2}+10cd+5d^{2})^{2}),
v = a(a^{2}+10ab+5b^{2})^{2} –
c(c^{2}+10cd+5d^{2})^{2}

Substituting these two into the expressions *p,q,r,s*,
we get,

(Öp + Öq)^{5} + (Öp - Öq)^{5} = (Ör + Ös)^{5} + (Ör - Ös)^{5}

where,

*p* =
5(bc-ad)(c^{2}+10cd+5d^{2})^{2},
*q* = (a^{2}+10ab+5b^{2})^{3
}- (ac+10bc+5bd)(c^{2}+10cd+5d^{2})^{2}

*r *=
5(bc-ad)(a^{2}+10ab+5b^{2})^{2},
*s* =
-(c^{2}+10cd+5d^{2})^{33} +
(ac+10ad+5bd)(a^{2}+10ab+5b^{2})^{2}

as given in the Introduction. Note that
this is vaguely reminiscent of the formula for cubics. The
big question is then: *Is this the general parametrization for quintic
quadruples?* After all, we essentially used the same
steps. *If* it is, and *if* it can be proven that
*p,q,r,s* can never be simultaneously squares (or if they can be but only
for trivial *x*_{i} = *y*_{i}), then we can prove
EEC(5,2,2)! A small step, but one has to start
somewhere. Note that *p,r* can easily be made into
squares by setting 5(bc-ad) = v^{2} and solving for any of the free
variables, say, *d*. The problem is that by substituting
this value for *d* into *q,s* we end up with two polynomials of degree
10 and 12 in four variables (a,b,c,v^{2}) which simultaneously must be
made into squares! Whether one can find such values is
uncertain.

Earlier it was pointed out that just like the Desboves-Elkies formula, we
can make our parametrization solve the multi-grade x_{1}^{k} +
x_{2}^{k} = x_{3}^{k} +
x_{4}^{k} for k = 1,2,5. To illustrate this,
define,

*R*_{k} = (Öp + Öq)^{k} + (Öp - Öq)^{k} + (-Ör - Ös)^{k} + (-Ör + Ös)^{k}

then for k = 1,2,5,

*R*_{1} = 2(Öp - Ör),
*R*_{2} =
2(p+q+r+s),
*R*_{5} = 2(p^{2}+10pq+5q^{2})Öp-2(r^{2}+10rs+5s^{2})Ör

Using the formulas established for *p,q,r,s*, we find
that,

*R*_{1} =
2(a^{2}+10ab+5b^{2}-c^{2}-10cd-5d^{2})Ö(bc-ad),

*R*_{2} =
2(a^{2}+10ab+5b^{2}-c^{2}-10cd-5d^{2}) P(z)

where P(z) is a polynomial of degree 4.
Obviously, *R*_{5} = 0. Thus to set
*R*_{1} = *R*_{2} = *R*_{5} = 0, one
simply has to find values *a,b,c,d* such that,

a^{2}+10ab+5b^{2 }= c^{2}+10cd+5d^{2}

which can easily be found.

Before we go to the next section, some interesting incidental facts about
the parametrizations for cubic and quintic quadruples can be
mentioned. **First**, **Q**Ö(-3) is the smallest
*imaginary *quadratic field with class number 1 while **Q**Ö5 is the smallest *real*
quadratic field also with class number 1. **Second**, to
recall Euler used the algebraic properties of the form *F*_{1} =
p^{2} + 3q^{2} to find the formula for k = 3.
With a small linear substitution p = (-a+b)/2 and q = (a+b)/2, we get
*F*_{1} = a^{2}+ab+b^{2}. For k
= 5, we used the form *F*_{2} =
p^{2}+10pq+5q^{2}. Letting p = (a-2b)/4 and q
= (a+2b)/4, we get *F*_{2} = a^{2}+ab-b^{2} and
which differs from the previous *only by a sign*, yet it makes all the
difference as the discriminant *d* of the former is *d* = -3 while the
latter is *d* = 5. **Third**, the algebraic form
*F*_{1}, much discussed in a previous paper, also appears in a
geometric context (perhaps not surprisingly) in the formula for the volume of
the *triangular and square pyramidal frustrums*. The
general pyramidal frustrum is simply a pyramid with a base of *n* sides and
the top chopped off. For *n* = 3, 4, the volume formulas
are,

V_{3} =
(a^{2}+ab+b^{2})(h/12)Ö3

V_{4} = (a^{2}+ab+b^{2})(h/3)

where *a* is the top side length, *b* is the base
side length, and *h* is the height (with the assumption that top and base
are regular *n*-gons). It should be interesting to know
if the form *F*_{2} appears somewhere as a formula, perhaps
involving pentagons.
**Fourth**, mirroring the coefficients of *F*_{2} =
p^{2}+10pq+5q^{2}, the equation,

x^{4}-10x^{2}+5 = 0

is connected to the *tangent function*, having the
solutions tan(p*n*/10) for *n* =
1,2,3,4 given explicitly by,

x_{1} = (1/5)Ö(25-10Ö5),
x_{2} = Ö(5-2Ö5),
x_{3} = (1/5)Ö(25+10Ö5),
x_{4} = Ö(5+2Ö5).

**Finally**, note that
the Pythagorean formula is essentially the *distance* formula between two
points (x_{1}, y_{1}) and (x_{2}, y_{2}) in the
Euclidean plane. This can be generalized to *n*-space
and for ** n = 3**, the analogous one for two points (x

_{2}, y

_{2}, z

_{2}) and (x

_{1}, y

_{1}, z

_{1}) is,

d^{2} = (x_{2}-x_{1})^{2} +
(y_{2}-y_{1})^{2} +
(z_{2}-z_{1})^{2}

so perhaps it may not be
surprising that Euler’s parametrization of x_{1}^{k} +
x_{2}^{k} + x_{3}^{k} =
x_{4}^{k} for *k* = 3 depended on the algebraic properties
of a^{2}+ab+b^{2} which was connected to the *volume*
formula of certain 3-dimensional objects.

# IV. Numerical Results

For this section, solutions for diophantine equations (k,m,n) will be
given. Since fortunately very extensive results are now
known, we will restrict our list into three categories corresponding to the
questions asked in Lander et al but with further restrictions, for: *Type
I* (k>m+n), *Type II* (k=m+n), and *Type III* (k<m+n).

**A. Type I (k>m+n)**

(No examples known)

**B. Type II (k=m+n)**

For the sake of conciseness, this will be limited only to first results
and *k* up to 10.

(3,1,2):
*(No solutions by Fermat’s Last Theorem)*

(4,1,3):
2,682,440^{4 }+ 15,365,639^{4} + 18,796,760^{4} =
20,615,673^{4} (Elkies, 1986)

(4,2,2):
59^{4} + 158^{4} = 133^{4} +
134^{4} (Euler, c.1750)

(5,1,4):
27^{5} + 84^{5} + 110^{5} + 133^{5} =
144^{5} (Lander, Parkin, 1967)

(5,2,3): 5027^{5} +
6237^{5} + 14068^{5} = 220^{5} + 14132^{5}
(Scher, Seidl, 1996)

(6,1,5):

(6,2,4):

(6,3,3): 3^{6} +
19^{6} + 22^{6} = 10^{6} + 15^{6} +
23^{6} (Rao, 1934)

(7,1,6):

(7,2,5):

(7,3,4):

(8,1,7):

(8,2,6):

(8,3,5): 81^{8} +
539^{8} + 966^{8} = 1588 + 310^{8} + 481^{8} +
725^{8} + 954^{8} (Chase, 2000)

(8,4,4):

(9,1,8):

(9,2,7):

(9,3,6):

(9,4,5):

(10,1,9):

(10,2,8):

(10,3,7):

(10,4,6):

(10,5,5):

As one can see, much work still needs to be done even for this partial list though at least we now know more than the previous generations.

**C. Type III (k<m+n)**

As this has the most extensive results, to make this list manageable it
will be limited to the minimum possible difference, just a step above k=m+n,
namely, k+1=m+n. These then would be the most difficult to
find of this type but there are still many results known. To
keep numbers down, we can just focus on *k* up to 11 and on two sub-types:
the *Eulerian *(k,1,k) and the balanced equation (2n-1,n,n) for odd powers.
(For even powers, (2n,n,n) is Type II.)

## Eulerian solutions (k,1,k):

(2,1,2):
3^{2} + 4^{2} = 5^{2}

(3,1,3):
3^{3} + 4^{3} + 5^{3} = 6^{3}

(4,1,4):
30^{4} + 120^{4} + 272^{4} + 315^{4} =
353^{4} (Norrie, 1911)

(5,1,5):
7^{5} + 43^{5} + 57^{5} + 80^{5} +
100^{5} = 107^{5} (Sastry, 1934)

(6,1,6):

(7,1,7):
127^{7} +258^{7} + 266^{7} + 413^{7} +
430^{7} + 439^{7} + 525^{7} =
568^{7} (Dodrill, 1999)

(8,1,8):
90^{8} + 223^{8} + 478^{8} + 524^{8} +
748^{8} + 1088^{8} + 1190^{8} + 1324^{8} =
1409^{8} (Chase, 2000)

(9,1,9):

(10,1,10):

(11,1,11):

The case (6,1,6) is a bit odd. For *k* *k*th
powers equal to a *k*th power, call this sum *z*^{k}, one
might assume that perhaps the larger the *k*, the larger would be
*z*. However, Lander et al actually searched for
this (*k* = 6) below a certain bound and failed to find a
solution. But this bound was z = 38300, much larger than
Chase’s z = 1409 for *k* = 8! Either there was something
wrong with the congruences they used (unlikely) or a programming or computer
error (who knows) or it is really the case that if (6,1,6) does have a solution,
then it will have z > 38300. Meyrignac’s *Eulernet*
at http://euler.free.fr/ prominently has
this equation as most wanted on its main page and they still have no solution
(and the project’s been running since 1999) so they may have pushed this bound
even higher.

## Balanced equations (2n-1,n,n):

(3,2,2):
1^{3} + 12^{3} = 9^{3} + 10^{3} (Frenicle
de Bessy, 1657; Ramanujan, 1919)

(5,3,3):
24^{5} + 28^{5} + 67^{5} = 3^{5} +
54^{5} + 62^{5} (Moessner, 1939)

(7,4,4):
10^{7} + 14^{7} + 123^{7} + 149^{7} =
15^{7} + 90^{7} +129^{7} + 146^{7} (Ekl,
1996)

(9,5,5):
26^{9} + 30^{9} + 91^{9} + 101^{9} +
192^{9} = 12^{9} + 17^{9} + 116^{9} +
175^{9} + 180^{9} (Ekl, 1998)

(11,6,6):

These type of equations are also interesting since for the equation pairs (3,2,2), (4,2,2) and (5,3,3), (6,3,3) parametric solutions are known. The question obviously is: does this generalize, or do (2n-1,n,n) and (2n,n,n) always have parametric solutions? The cases (7,4,4) and (9,5,5) were found by numerical search and the former in fact already has many examples, some of which found by A. Choudhry (2000) are even multigrade for k = 1,3,7. Whether these would turn out to be members of a parametric family is unknown.

In summary, there are 39 diophantine equations (k,m,n) in this admittedly
arbitrary list. Out of the 39, less than half are known to
have solutions. Some will undoubtedly fall in the next few
years with the advent of faster computers, the Internet, and more people working
on the field (most likely would be (8,4,4) and (10,5,5)). *Moore’s Law* in
particular, which states that computing power doubles roughly every two years,
makes the feasibility of a numerical search coupled with congruence restraints
very reasonable for some (k,m,n). The author will maintain
this list and update it as necessary. For more details, see
the separate article “*Timeline of Euler’s Extended Conjecture
(EEC)”*.

# V. Conclusion

Almost 4000 years have passed since the ancient Babylonians knew that two
second powers could sum to a second power and the time S. Chase found by
computer the first example of eight eighth powers whose sum is an eight
power. We now have many examples of *k* *k*th
powers equal to a *k*th power, equations of such a high degree perhaps
inconceivable to the Babylonians and definitely out of reach even if they could
conceive of the possibility. While now we routinely discuss
equations of any degree *k*, the solution of Eulerian (k,1,k) for higher
powers (as “high” as *k*=6) still eludes us, and for
anyone interested in diophantine equations this not knowing is as irritating as
a pebble in the shoe. *We must
know*.^{3}^{}

However, we also now know that
*three* fourth powers can sum to a fourth power, as well as that the sum of
four fifth powers can be a fifth power. But similar equations
for the sixth? Or seventh, and so on?

Euler’s extended conjecture (EEC)
generalizes these questions and asserts that *m* *k*th powers equal to
*n*th *k*th powers have no non-trivial solution for k > m+n and it
is hoped this paper gave a clear introduction to the topic.
After all, it was its objective, as well as a small contribution to the
fourth case m+n = 4 for *k* = 5 by giving a new
parametrization. Some questions, basically versions of the
ones asked in Lander et al, are worth pointing out again, namely:

[1] Is k = m+n always solvable?

[2] Is (k,1,k) always solvable?

[3] Are (2n-1,n,n) and (2n,n,n) always parametrically solvable?

as well as a new one focused on this paper,

[4] Is the new solution for (5,2,2) the complete parametrization?

If indeed EEC is true, then why? Why can’t, for example, be three
12^{th} powers equal to three 12^{th} powers, all
distinct? There must be a reason forbidding such
solutions. For comparison, Fermat’s investigations on
generalizing a^{k} + b^{k} = c^{k} for k > 2
eventually lead, via the Taniyama-Shimura conjecture, to an interesting
connection between modular forms and elliptic curves. So
where will EEC lead us? Who knows, but it may also be to
previously unknown and unexpected connections between different and seemingly
unrelated mathematical objects.

---

*Footnotes*:

- Fermat is sometimes considered as an amateur
mathematician. However, the word “amateur” is derived from
the Latin “
*amare*”, which means “to love”, hence words like amorous, amicable, amigo, etc. The mathematics amateur then, etymologically speaking, is a lover of mathematics in a similar sense that a philosopher is a lover of wisdom (the Greek “*philein*” also means “to love”). Thus, professional mathematicians who love the field can truly be called, paradoxically, as _*amateurs*_.

- Hardy almost left a message as tantalizing as
Fermat’s. This was about the famous Riemann
Hypothesis. To quote de Sautoy,
*“…On a rough sea crossing fearing for his life, he sent a joke telegram saying that he had found a wonderful proof of the hypothesis. The ship, however, did not sink.”*Since Hardy did prove there were an infinite number of zeros on the critical line, imagine the “Did he really?” questions if the ship did sink. (See de Sautoy’s article link below.)

- For Hilbert, what he
*really*wanted to know was the proof of the Riemann hypothesis. He allegedly said that if he could wake from a thousand year sleep, his first question would be to ask whether anyone had proved it.

--End--

© 2005

Titus Piezas III

Oct. 13, 2005

http://www.oocities.com/titus_piezas/ramanujan.html ® Click here for an index

tpiezasIII@uap.edu.ph ® **Remove** “III” for
email

References:

- Choudhry, A., “Equal Sums of Seventh Powers”, Rocky Mountain Journal of Mathematics, Vol 30, 2000.
- Dickson, L., History of the Theory of Numbers, Vol 2, Carnegie Institute of Washington, 1919. Reprinted by Chelsea Publishing, NY, 1971.
- du Sautoy, M., “The Music of The Primes”, Science Spectra (11), 1998, http://www.maths.ex.ac.uk/~mwatkins/zeta/dusautoy.htm
- Ekl, R., “New Results In Equal Sums Of Like Powers”, Math. Of Computation, Vol 67, 1998
- Joyce, D., “Plimpton 322”, http://aleph0.clarku.edu/~djoyce/mathhist/plimpnote.html
- Lander, L., Parkin, T., and Selfridge, J., “A Survey of Equal Sums of Like Powers”, Math. Of Computation, Vol 21, 1967.
- O’Connor, J. and Robertson, E., “Fermat’s Last Theorem”, http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html
- O’Connor, J. and Robertson, E., “Pythagoras’ Theorem in Babylonian Mathematics”, http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Babylonian_Pythagoras.html
- Merignac, Jean-Charles, Computing Minimal Equal Sums of Like Powers, http://euler.free.fr/
- Nagell, T., Introduction to Number Theory, New York, 1951.
- Piezas, T., “Ramanujan and The Quartic Equation
2
^{4}+ 2^{4}+ 3^{4}+ 4^{4}+ 4^{4}= 5^{4}”, http://www.oocities.com/titus_piezas/ramanujan_page10.html - Reznick, B., “Patterns of Dependence Among Powers of Polynomials”.
- Shuwen, Chen, “Equal Sums of Like Powers”, http://euler.free.fr/eslp/eslp.htm
- Weisstein, Eric et al, “Diophantine Equations”, http://mathworld.wolfram.com/
- et al.