In number theory, Euler's conjecture is a disproved conjecture related to Fermat's Last Theorem. It was presented by Leonhard Euler in 1778 to the Academy of Sciences of St. Petersburg. It states that for all integers n and k greater than 1, if the sum of n many kth powers of positive integers is itself a kth power, then n is greater than or equal to k:
a
1
k
+
a
2
k
+
⋯
+
a
n
k
=
b
k
⟹
n
≥
k
{\displaystyle a_{1}^{k}+a_{2}^{k}+\dots +a_{n}^{k}=b^{k}\implies n\geq k}
The conjecture represents an attempt to generalize Fermat's Last Theorem, which is the special case n = 2: if
a
1
k
+
a
2
k
=
b
k
,
{\displaystyle a_{1}^{k}+a_{2}^{k}=b^{k},}
then 2 ≥ k.
Although the conjecture holds for the case k = 3 (which follows from Fermat's Last Theorem for the third powers), it was disproved for k = 4 and k = 5. It is unknown whether the conjecture fails or holds for any value k ≥ 6.
Contents
Background
Euler was aware of the equality 594 + 1584 = 1334 + 1344 involving sums of four fourth powers; this, however, is not a counterexample because no term is isolated on one side of the equation. He also provided a complete solution to the four cubes problem as in Plato's number 33 + 43 + 53 = 63 or the taxicab number 1729. The general solution of the equation
x
1
3
+
x
2
3
=
x
3
3
+
x
4
3
{\displaystyle x_{1}^{3}+x_{2}^{3}=x_{3}^{3}+x_{4}^{3}}
is
x
1
=
λ
(
1
−
(
a
−
3
b
)
(
Counterexamples
Euler's conjecture was disproven by L. J. Lander and T. R. Parkin in 1966 when, through a direct computer search on a CDC 6600, they found a counterexample for k = 5. This was published in a paper comprising just two sentences. A total of four primitive (that is, in which the summands do not all have a common factor) counterexamples are known:
144
5
=
27
5
+
84
5
+
110
5
+
133
5
14132
5
=
(
−
220
)
5
+
5027
5
+
6237
5
+
14068
5
85359
Generalizations
In 1967, L. J. Lander, T. R. Parkin, and John Selfridge conjectured that if
∑
i
=
1
n
a
i
k
=
∑
j
=
1
m
b
j
k
{\displaystyle \sum _{i=1}^{n}a_{i}^{k}=\sum _{j=1}^{m}b_{j}^{k}}
,
where ai ≠ bj are positive integers for all 1 ≤ i ≤ n and 1 ≤ j ≤ m, then m + n ≥ k. In the special case m = 1, the conjecture states that if
∑
i
=
1
n
a
i
k
=
b
k
k = 3
From Fermat's Last Theorem, we know that there can't be a solution to
a
3
+
b
3
=
c
3
{\displaystyle a^{3}+b^{3}=c^{3}}
.
(The minimum positive value of a sum of third powers is
9
3
−
8
3
−
6
3
=
1
{\displaystyle 9^{3}-8^{3}-6^{3}=1}
, which provides a solution to the equation (a = (1, 6, 8), b = 9), where however the smallest member isn't larger than 1.)
The smallest solution with terms > 1 is
3
3
+
4
3
+
k = 4
422481
4
=
95800
4
+
217519
4
+
414560
4
353
4
=
30
4
+
120
4
+
272
4
+
315
4
{\displaystyle {\begin{aligned}422481^{4}&=95800^{4}+217519^{4}+414560^{4}\\[4pt]353^{4}&=30^{4}+120^{4}+272^{4}+315^{4}\end{aligned}}}
(R. Frye, 1988); (R. Norrie, smallest, 1911).
k = 5
144
5
=
27
5
+
84
5
+
110
5
+
133
5
72
5
=
19
5
+
43
5
+
46
5
+
47
5
+
67
5
94
5
=
21
5
+
23
5
k = 6
It has been known since 2002 that there are no solutions for k = 6 whose final term is ≤ 730000.
k = 7
568
7
=
127
7
+
258
7
+
266
7
+
413
7
+
430
7
+
439
7
+
525
7
{\displaystyle 568^{7}=127^{7}+258^{7}+266^{7}+413^{7}+430^{7}+439^{7}+525^{7}}
(M. Dodrill, 1999).
k = 8
1409
8
=
90
8
+
223
8
+
478
8
+
524
8
+
748
8
+
1088
8
+
1190
8
+
1324
8
{\displaystyle 1409^{8}=90^{8}+223^{8}+478^{8}+524^{8}+748^{8}+1088^{8}+1190^{8}+1324^{8}}
(S. Chase, 2000).