Ed Pegg Jr., November 13, 2006

In 1772, Euler made a conjecture: If a sum of *n* positive *k*th
powers equals one
*k*th power, then *n* ≥ *k*.

In 1966, Lander, Parkin and
Selfridge proved Euler wrong with 144^{5 }= 133^{5}+110^{5}+84^{5}+27^{5}.
Let this solution be considered as (**5**.**1**.**4**).
All parts are **5**th powers, and
**1** of them is equal to **4** others, or (5.1.4). Lander,
Parkin and Selfridge further conjectured that for other
powers sums (*k,m,n*), that *m+n* ≥ *k*. So far, no counterexamples
have been found. Only seven forms are known with equality: 4.1.3, 4.2.2, 5.1.4,
5.2.3, 6.3.3, 8.3.5, and 8.4.4. The last of these
was discovered by Nuutti Kuosa on Nov 9, 2006.

**158 ^{4}+59^{4} = 134^{4}+133^{4} **(4.2.2)
infinite solutions

422481

144

14132

23

966

3113

Figure 1. Known solutions for (

Since March 1999, such results have been collected by EulerNet, headed by programmer Jean-Charles Meyrignac. Over 200 EulerNet contributors have put in almost 200 years of computer time looking for solutions, and improving search methods. Many of the solutions above are due to the EulerNet group. There are no known equalities for 6.1.5, 6.2.4, 7.1.6, 7.2.5, 7.3.4, 8.1.7, and 8.2.6, or for any higher powers. A plot of the excess for the best known solutions is below:

Figure 2. Excess of best known solutions. Only 4,5,6, and 8 have solutions with
0 excess.

Another power sum problem is the Fermat-Catalan
conjecture, which
claims there are a finite number of solutions to *x*^{p}+*y*^{q} = *z*^{r},
when 1/p + 1/q + 1/r < 1. There are ten known solutions, excluding trivial variations
of the first example.

**1 ^{7}+2^{3} = 3^{2}**

2^{5}+7^{2} = 3^{4}

7^{3}+13^{2} = 2^{9}

2^{7}+17^{3} = 71^{2}

3^{5}+11^{4} = 122^{2}

1414^{3}+2213459^{2} = 65^{7}

33^{8}+1549034^{2} = 15613^{3}

9262^{3}+15312283^{2} = 113^{7}

43^{8}+96222^{3} = 30042907^{2}

17^{7}+76271^{3} = 21063928^{2}

Figure 3. Known solutions for the Fermat-Catalan conjecture.

A third power problem is the *abc*-conjecture.
Consider a number that is a power, or which contains many powers. rad(*n*)
is the product of all distinct prime divisors of *n*. For example, if *n* =
3^{2}×5^{6}×7^{3}, then rad(*n*) is
3×5×7, or 105. For sums *a*+*b*=*c*, there are
only seven known examples where log(*c*)/log(rad(*abc*)) > 1.54.

**2 + 3 ^{10}×109 = 23^{5}**Figure 4. Known solutions for log(

11^{2 }+ 3^{2}×5^{6}×7^{3} = 2^{21}×23

19×1307 + 7×29^{2}×31^{8} = 2^{8}×3^{22}×5^{4}

283 + 5^{11}×13^{2} = 2^{8}×3^{8}×17^{3}

1 + 2×3^{7} = 5^{4}×7

7^{3 }+ 3^{10} = 2^{11}×29

7^{2}×41^{2}×311^{3} + 11^{16}×13^{2}×79 = 2×3^{3}×5^{23}×953

An alternate form of the problem is to consider log(*abc*)/log(rad(*abc*)).
Only ten examples are known where log(*abc*)/log(rad(*abc*))>4.2.

**13×19 ^{6 }+ 2^{30}×5 = 3^{13}×11^{2}×31**Figure 5. Known solutions for log(

2^{5}×11^{2}×19^{9 }+ 5^{15}×37^{2}×47 = 3^{7}×7^{11}×743

2^{19}×13×103 + 7^{11} = 3^{11}×5^{3}×11^{2}

19^{8}×43^{4}×149^{2 }+ 2^{15}×5^{23}×101 = 3^{13}×13×29^{2}×37^{6}×911

2^{35}×7^{2}×17^{2}×19 + 3^{27}×107^{2 }= 5^{15}×37^{2}×2311

3^{18}×23×2269 + 17^{3}×29×31^{8} = 2^{10}×5^{2}×7^{15 }

17^{4}×79^{3}×211 + 2^{29}×23×29^{2} = 5^{19}

5^{14}×19 + 2^{5}×3×7^{13} = 11^{7}×37^{2}×353

2^{7}×5^{4}×7^{22} + 19^{4}×37×47^{4}×53^{6} = 3^{14}×11×13^{9}×191×7829

3^{21} + 7^{2}×11^{6}×199 = 2×13^{8}×17

All of these problems are considered extremely difficult, so many
congratulations to Nuutti Kuosa for extending one
of these lists.

Richard Guy, *Unsolved Problems in Number Theory*, Springer, 2004.

Jean-Charles Meyrignac, *EulerNet*, "Computing Minimal Equal Sums Of Like
Powers" http://euler.free.fr/.

Abderrahmane Nitaj, "ABC Conjecture Home Page," http://www.math.unicaen.fr/~nitaj/abc.html.

Eric Weisstein, "Diophantine Equation, Fermat-Catalan Conjecture" From *MathWorld*--A
Wolfram Web Resource. http://mathworld.wolfram.com/DiophantineEquation.html

Comments are welcome. Please send comments to Ed Pegg Jr. at ed@mathpuzzle.com.

Ed Pegg Jr. is the webmaster of mathpuzzle.com.
He works at Wolfram Research, Inc. as an associate editor of *MathWorld*.
He is also a math consultant for the TV show *Numb3rs*.