Chapter 4
Generalisations and Approximations
The most obvious generalisation of equation 1.1 is to consider representations of one power in terms of a different power, such as
(4.1)
Definition: Let S(m,n) be the minimum value of k for which equation 4.1 has a nontrivial solution when both m and n are greater than 1. Thus s(m) = S(m,m). Without loss of generality, we shall assume that y_{1} ³ y_{2} ³ ...... ³ y_{k} and, when m = n, that x > y_{1}.
For equation 4.1, I have switched the sum to the left hand side of the equality, purely for convenience as it allows me to introduce an additional generalisation. Consider the Fermat Equation x^{n} + y^{n} = z^{n} for n > 1. We know that this has no solutions in positive integers apart from when n = 2. However, it is interesting to speculate on how close an approximation may be achieved. Specifically, let us consider the following extension of equation 4.1
(4.2)
Definition: Let e(k,m,n) be the smallest absolute value of d for which a nontrivial solution of equation 4.2 exists. In effect, this is the error margin between powers m and n for a particular length k. We can the redefine S(m,n) as the smallest value of k for which e(k,m,n) = 0.
In the case that gcd(m,n) = 1, it is appropriate to consider the case k = 1, which would not normally be of interest.
Before presenting known results, it is appropriate to clarify what we consider a nontrivial solution, including the cases where d = 0, and that is one which cannot be derived from any other solution (with the same or different powers m and n) by multiplication by a constant, or use of a simple parametric formula, or where d is the sum of a subset of the y’s, or where d would be smaller if one of the y’s was removed, giving a solution for a smaller value of k, trivial or otherwise. In the case k = 1, any (m*n)th power is both an mth power and and nth power and so any solution with gcd(m,n) > 1 is trivial. For completeness, we shall also consider solutions to be trivial if x = y_{1} or x = 1 or y_{1} = 1.
In the case d = 0, it is always possible to generate solutions whenever gcd(m,n) = 1 for any value of k. For example, consider m = 3, n = 7 and k = 2. Choose a random sum of two cubes, e.g. 3^{3} + 2^{3} = 27 + 8 = 35 = 5.7. All we then require is to find z such that z^{3}.(3^{3} + 2^{3}) is a 7th power. To do this we require z = 5^{a}.7^{b} so that z^{3}.5.7 = 5^{3a+1}.7^{3b+1} = 5^{7p}.7^{7q} = (5^{p}.7^{q})^{7}, i.e. that 7 divides both 3a+1 and 3b+1. We can choose a = 2, so that p = q = 1. We then have z = 5^{2}.7^{2} = 1225, so then (3z)^{3} + (2z)^{3} = (5.7)^{7}, i.e. the derived solution 3675^{3} + 2450^{3} = 35^{7}. In order to exclude such solutions, we therefore demand than when d = 0 in equation 4.2, the greatest common factor of x and all the y_{i} is greater than 1, otherwise the solution is derived and trivial.
Note that for k = 2, most searches will be to the stated limit, with y_{1} up to 3000 as preferred minimum. For k = 3, searches take y_{1} up to at least 250, k = 4 with y_{1} up to at least 100, k = 5 with y_{1} up to 50 and k = 6 with y_{1} up to 25. In the k = 1 case, limits will be well in excess of 10000. Also, for k = 1, we consider only m > n since e(1,m,n) = e(1,n,m) by subtracting d from each side of a solution equation.
The normal method of finding solutions of equation 4.2, in which we hope to find examples with d = 0 but must allow for nonzero errors, is by a brute force, exhaustive search that considers every possibility for each of the y_{i}. Once the overall picture is built up, we then try to improve bounds on S(m,n) directly by using a modified version of the decomposition algorithm from Chapter 1.
Consider n = 2.
For k = 1 and m = 3, the equation 2^{3} = 3^{2}  1 gives e(1,3,2) = 1.
For k = 1 and m = 5, the equation 1^{5} = 2^{2}  3 is considered trivial. After this, the closest known is 2^{5} = 6^{2}  4.
For k = 1 and m = 7, the equation 2^{7} = 11^{2} + 7 gives the best result.
For k = 2 and m = 2, there is an infinity of solutions for d = 0, i.e. the Pythagorean triples, and so S(2,2) = 2. For completeness, the smallest example is 3^{2} + 4^{2} = 5^{2}.
For k = 2 and m = 3, there are 27 solutions with d = 0 for y_{1} up to 10000, the smallest being 2^{3} + 1^{3} = 3^{2}, and the largest being 9265^{3} + 2184^{3} = 897623^{2}. Thus S(3,2) = 2. A variety of parametric solutions in 2 unknowns exist, but are not defined in such a way that sequential enumeration can be done.
For k = 2 and m = 4, a straightforward argument based on a method of Fermat shows that e(2,4,2) ¹ 0. However, there are 87 solutions with d = 1 for y_{1} up to 10000, the smallest being 7^{4} + 5^{4} = 55^{2} + 1 and the largest being 9901^{4} + 5333^{4} = 102072161^{2} + 1. Simple congruence rules imply that there are no solutions for d =  1.
For k = 2 and m = 5, and y_{1} up to 8120, there are two solutions with d = 2. These are 3^{5} + 3^{5} = 22^{2} + 2 and 111^{5} + 87^{5} = 147766^{2} + 2. Hence e(2,5,2) £ 2. Other solutions exist for d =  3, 4,  4,  5, 6 and  6.
For k = 2 and m = 6, we have 31^{6} + 29^{6} = 38501^{2} +1, giving e(2,6,2) £ 1. Any solution for m = 6 gives rise to a solution for m = 3 by squaring the y’s.
For k = 2 and m = 7, the best result at present is 4^{7} + 4^{7} = 181^{2} + 7.
For k = 3 and m = 3, there are many solutions with d = 0. The smallest is 3^{3} + 2^{3} + 1^{3} = 6^{2} and the smallest with none of the y’s equal to 1 is 14^{3} + 9^{3} + 2^{3} = 59^{2}. A particularly interesting solution is 25^{3} + 24^{3} + 23^{3} = 204^{2}, and in general it may be worth restricting the form of the y’s when looking for solutions in this case.
For k = 3 and m = 4, the equation 20^{4} + 15^{4} + 12^{4} = 481^{2} is the only solution for d = 0 with y_{1} up to 100. However, it does give us the result that S(4,2) = 3.
For k = 3 and m = 5, there are trivial parameterisable solutions with d = 1 and d =  1, as well as nontrivial solutions. However, there are 4 solutions with d = 0. These are 69^{5} + 49^{5} + 3^{5} = 42971^{2}, 81^{5} + 68^{5} + 20^{5} = 70313^{2}, 97^{5} + 80^{5} + 19^{5} = 108934^{2} and 97^{5} + 83^{5} + 16^{5} = 111926^{2}. Thus S(5,2) £ 3, with equality most likely.
For k = 3 and m = 6, there is a trivial parameterisable solution with d =  1. There are also nontrivial solutions with d = 1 and 2, e.g. 3^{6} + 3^{6} + 2^{6} = 39^{2} + 1, 5^{6} + 4^{6} + 3^{6} = 143^{2} + 1, 25^{6} + 25^{6} + 24^{6} = 26065^{2} + 1, 5^{6} + 5^{6} + 4^{6} = 188^{2} + 2 and 19^{6} + 11^{6} + 3^{6} = 6987^{2} + 2. However, we have S(6,2) £ 3 since 100^{6} + 81^{6} + 42^{6} = 1134865^{2}.
For k = 3 and m = 7, there are trivial parameterisable solutions with d = 1 and d =  1. The best nontrivial result available at present is e(3,7,2) £ 4, using 5^{7} + 4^{7} + 4^{7} = 333^{2} + 4.
For k = 4, there is a general parameterisable solution given by
for any positive integers x and y. Thus it could be said that S(m,2) £ 4 for all m. We have two choices: either ignore the k = 4 case for n = 2, or search for nonparameterised solutions. In the current situation, we are only interested in the m = 7 case since we have already proved that S(m,2) £ 3 for m £ 6. We will therefore push ahead searching for such solutions, as follows.
For k = 4 and m = 7, there are surprisingly many nontrivial solutions with small d, e.g. 27^{7} + 24^{7} + 21^{7} + 13^{7} = 130041^{2} + 4 and 40^{7} + 31^{7} + 27^{7} + 4^{7} = 449236^{2} + 2. However, the solutions 74^{7} + 59^{7} + 17^{7} + 4^{7} = 3826270^{2} and 78^{7} + 51^{7} + 42^{7} + 25^{7} = 4324306^{2} are found for y_{1} up to 100, and allow us to state that indeed S(7,2) £ 4.
Consider n = 3.
For k = 1 and m = 5, the equation 2^{5} = 3^{3} + 5 gives the best result.
For k = 1 and m = 7, the equation 2^{7} = 5^{3} + 3 gives the best result.
For k = 2 and m = 2, there are 80 solutions with d = 0 for y_{1} up to 10000, the smallest being 11^{2} + 2^{2} = 5^{3} and the largest being 9830^{2} + 1159^{2} = 461^{3}. Thus S(2,3) = 2.
For k = 2 and m = 3, Fermat’s Last Theorem implies that there is no solution for d = 0. The case for third powers only was first proved by Euler. However, there are 16 solutions with d = 1, the smallest being 10^{3} + 9^{3} = 12^{3} + 1 and 16 solutions with d =  1, the smallest being 8^{3} + 6^{3} = 9^{3}  1, with y_{1} up to 10000. Thus e(2,3,3) = 1. Of these, 5 solutions each of each sign result directly from the related parameteric solutions
(3^{2}b^{4})^{3} + (3^{2}b^{3} + 1)^{3} = (3^{2}b^{4} + 3b)^{3} + 1 and (3^{2}b^{4}  3b)^{3} + (3^{2}b^{3}  1)^{3} = (3^{2}b^{4})^{3}  1.
There is also a parameterisable solution for d =  2, (6p^{3}  1)^{3} + (6p^{2})^{3} = (6p^{3} + 1)^{3}  2.
For k = 2 and m = 4, the first nontrivial solution with small error is 2^{4} + 2^{4} = 3^{3} + 5.
For k = 2 and m = 5, the only nontrivial solution with small error is 5^{5} + 3^{5} = 15^{3}  7.
For k = 2 and m = 6, the only nontrivial solution with small error is 2^{6} + 2^{6} = 5^{3} + 3.
For k = 2 and m = 7, the only nontrivial solution with small error is 3^{7} + 1^{7} = 13^{3}  9, which is not trivial as the addition of a 1 decreases the error from a k = 1 solution. The solution 2^{7} + 1^{7} = 5^{3} + 4 is considered trivial as it is based on a k = 1 solution and the error is increased.
For k = 3 and m = 3, many nontrivial solutions exist with d = 0, some of which belong to parameterisable groups. Hence S(3,3) = 3. The smallest solution is 5^{3} + 4^{3} + 3^{3} = 6^{3}. Other notable solutions are 8^{3} + 6^{3} + 1^{3} = 9^{3} (related to the k = 2 solution), 22^{3} + 17^{3} + 4^{3} = 25^{3} and 74^{3} + 48^{3} + 25^{3} = 81^{3}, which all give solutions for n = 6 as well as n = 3.
For k = 3 and m = 4, there is one known solution with d = 0, i.e. 5^{4} + 5^{4} + 3^{4} = 11^{3}. Thus S(4,3) £ 3, with equality likely. Four solutions exist with d = 1 for y_{1} up to 100. These are 22^{4} + 16^{4} + 11^{4} = 68^{3} + 1, 78^{4} + 68^{4} + 11^{4} = 388^{3} + 1, 82^{4} + 53^{4} + 48^{4} = 388^{3} + 1 and 87^{4} + 82^{4} + 6^{4} = 468^{3} + 1.
For k = 3 and m = 5, there are several small solutions with d = 6 and d =  6, but the best result available is 81^{5} + 74^{5} + 15^{5} = 1787^{3}  3. Thus e(3,5,3) £ 3.
For k = 3 and m = 6, the solution 25^{6} + 23^{6} + 6^{6} = 732^{3} + 2 gives e(3,6,3) £ 2.
For k = 3 and m = 7, the solution 3^{7} + 1^{7} + 1^{7} = 13^{3}  8 gives the best result at present. As with the k = 2 case above, this is not considered a trivial solution. However, the best result not based on a previous solution is 15^{7} + 5^{7} + 4^{7} = 555^{3} + 9.
For k = 4 and m = 5, we have 4^{5} + 3^{5} + 2^{5} + 2^{5} = 11^{3}, and so s(5,3) £ 4. A larger example is 48^{5} + 8^{5} + 5^{5} + 3^{5} = 634^{3}. An interesting near miss is 6^{5} + 5^{5} + 4^{5} + 3^{5} = 23^{3} + 1. Solutions exist for most small d.
For k = 4 and m = 6, the earliest result is 3^{6} + 3^{6} + 3^{6} + 1^{6} = 13^{3}  9, which is far from ideal since the k = 3 case provides a solution for d = 2. Purer solutions include 8^{6} + 6^{6} + 5^{6} + 4^{6} = 69^{3} + 12 and 45^{6} + 44^{6} + 43^{6} + 19^{6} = 2799^{3} + 12, followed by 56^{6} + 33^{6} + 27^{6} + 12^{6} = 3192^{3} + 10. There is also a prevalence of solutions with d =  23. However, the current best results are for d = 3, e.g. 57^{6} + 19^{6} + 19^{6} + 10^{6} = 3252^{3} + 3 and 68^{6} + 53^{6} + 25^{6} + 15^{6} = 4950^{3} + 3.
For k = 4 and m = 7, the best result is 3^{7} + 1^{7} + 1^{7} + 1^{7} = 13^{3}  7, and the second best is 10^{7} + 6^{7} + 5^{7} + 3^{7} = 218^{3} + 16.
For k = 5 and m = 6, we first have solutions for d = 4, i.e. 5^{6} + 5^{6} + 3^{6} + 3^{6} + 2^{6} = 32^{3} + 4 and 23^{6} + 21^{6} + 15^{6} + 7^{6} + 4^{6} = 626^{3} + 4, followed by neighbouring solutions for d = 2, namely 28^{6} + 21^{6} + 15^{6} + 10^{6} + 6^{6} = 834^{3} + 2 and 28^{6} + 21^{6} + 16^{6} + 16^{6} + 3^{6} = 844^{3} + 2.
For k = 5 and m = 7, the best result is 3^{7} + 1^{7} + 1^{7} + 1^{7} + 1^{7} = 13^{3}  6.
For k = 6 and m = 6, we have 10^{6} + 9^{6} + 7^{6} + 7^{6} + 4^{6} + 3^{6} = 121^{3} + 3, and also 5 solutions for d =  3, the largest of which is 25^{6} + 22^{6} + 11^{6} + 7^{6} + 5^{6} + 2^{6} = 711^{3}  3.
For k = 6 and m = 7, we have 20^{7} + 7^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 1086^{3}  1. Thus e(6,7,3) £ 1 and we obviously also have that S(7,3) £ 7, adding 1 to each side.
Consider n = 4.
For k = 1 and m = 3, the best result is 3^{3} = 2^{4} + 11.
For k = 1 and m = 5, the best result is 3^{5} = 4^{4}  13.
For k = 1 and m = 7, the best result is 2^{7} = 3^{4} + 47.
For k = 2 and m = 2, there are many solutions with d = 0, which can be extracted from the lists of Pythagorean triples, e.g. 24^{2} + 7^{2} = 5^{4}, 120^{2} + 119^{2} = 13^{4} and 240^{2} + 161^{2} = 17^{4}. Thus S(2,4) = 2.
For k = 2 and m = 3, there are several solutions with small d, the best current result being 63^{3} + 31^{3} = 23^{4}  3. Thus e(2,3,4) £ 3. Another close call is 257^{4} + 147^{4} = 67^{3}  5.
For k = 2 and m = 4, the argument for n = 2 still holds and so e(2,4,4) ¹ 0. The closest that we get at present is 5^{4} + 5^{4} = 6^{4}  46.
For k = 2 and m = 5, the best result is 3^{5} + 1^{5} = 4^{4}  12, which is related to a k = 1 solution. Apart from this, the best is 2^{5} + 2^{5} = 3^{4}  17.
For k = 2 and m = 6, the best result is 2^{6} + 1^{6} = 3^{4}  16.
For k = 2 and m = 7, the best result is 3^{7} + 2^{7} = 7^{4}  86.
For k = 3 and m = 3, there are quite a few solutions with d = ± 1, but we also have the equations 20^{3} + 17^{3} + 12^{3} = 11^{4} and 74^{3} + 48^{3} + 25^{3} = 27^{4}, so that S(3,4) £ 3.
For k = 3 and m = 4, Elkies proved that e(3,4,4) = 0 and so S(4,4) = 3. The smallest solution with d = 0 is 414560^{4} + 217519^{4} + 95800^{4} = 422481^{4}, which was found using the very tight congruence restrictions that can be applied in this case. Without these restrictions, the closest we get to this for small x is 47^{4} + 22^{4} + 21^{4} = 48^{4} + 2.
For k = 3 and m = 5, the best result is 4^{5} + 3^{5} + 2^{5} = 6^{4} + 3.
For k = 3 and m = 6, the best result is 2^{6} + 1^{6} + 1^{6} = 3^{4}  15.
For k = 3 and m = 7, the best result is 4^{7} + 3^{7} + 3^{7} = 12^{4} + 22.
For k = 4 and m = 5, the earliest close calls are 8^{5} + 7^{5} + 4^{5} + 2^{5} = 15^{4} + 6 and 9^{5} + 5^{5} + 5^{5} + 3^{5} = 16^{4} + 6. However, the solution 104^{5} + 72^{5} + 52^{5} + 45^{5} = 348^{4}  3. provides the current best result.
For k = 4 and m = 6, the best result is 21^{6} + 18^{6} + 11^{6} + 3^{6} = 105^{4} + 10.
For k = 4 and m = 7, the best result is 9^{7} + 5^{7} + 4^{7} + 3^{7} = 47^{4}  16.
For k = 5 and m = 5, there are many solutions for small d, but the most important are 5^{5} + 3^{5} + 3^{5} + 3^{5} + 3^{5} = 8^{4} + 1,
22^{5} + 22^{5} + 15^{5} + 12^{5} + 4^{5} = 58^{4}  1 and 39^{5} + 22^{5} + 21^{5} + 14^{5} + 3^{5} = 100^{4}  1. Thus e(5,5,4) £ 1 and S(5,4) £ 6.
{N.B. Further studies provide the result 185^{5} + 169^{5} + 109^{5} + 46^{5} + 1^{5} = 780^{4}, so that, in fact, S(5,4) £ 5.}
For k = 5 and m = 6, the best result is 9^{6} + 8^{6} + 5^{6} + 3^{6} + 2^{6} = 30^{4} + 3. Another close solution is 31^{6} + 27^{6} + 26^{6} + 23^{6} + 5^{6} = 204^{4} + 4.
For k = 5 and m = 7, the best early results are for d = ± 15, given by 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 5^{4} + 15 and 9^{7} + 5^{7} + 4^{7} + 3^{7} + 1^{7} = 47^{4}  15, the latter based on the k = 4 solution. However, we have e(5,7,4) £ 7 since we have 31^{7} + 14^{7} + 13^{7} + 11^{7} + 10^{7} = 408^{4} + 7.
For k = 6 and m = 6, we have e(6,6,4) £ 1 since 9^{6} + 8^{6} + 7^{6} + 4^{6} + 4^{6} + 4^{6} = 31^{4} + 1.
For k = 6 and m = 7, the best result is 16^{7} + 13^{7} + 10^{7} + 7^{7} + 5^{7} + 4^{7} = 136^{4} + 9.
Consider n = 5.
For k = 1 and m = 7, the best result is 2^{7} = 3^{5}  115.
For k = 2 and m = 2, there are 6 solutions with d = 0, for y_{1} up to 9390. The smallest is 41^{2} + 38^{2} = 5^{5} and the largest is 6121^{2} + 5646^{2} = 37^{5}. Hence S(2,5) = 2.
For k = 2 and m = 3, for y_{1} up to 4000 we have e(2,3,5) £ 2 since 271^{3} + 239^{3} = 32^{5}  2.
For k = 2 and m = 4, the only nontrivial solution with d < 81 is 22^{4} + 11^{4} = 12^{5} + 65.
For k = 2 and m = 5, Fermat’s Last Theorem imples that e(2,5,5) ¹ 0, with the particular case for 5th powers proved independently by both Dirichlet and Legendre. The surprisingly good current best estimate is that e(2,5,5) £ 12 given by 16^{5} + 13^{5} = 17^{5} + 12, with a search range of y_{1} up to 4346.
For k = 2 and m = 6, the best result is 2^{6} + 2^{6} = 3^{5}  115.
For k = 2 and m = 7, the best result is 2^{7} + 2^{7} = 3^{5} + 13.
For k = 3 and m = 3, there are several solutions with d = ± 1. However there are two with d = 0. These are 77^{3} + 57^{3} + 49^{3} = 15^{5} = 86^{3} + 40^{3} + 39^{3}. Thus S(3,5) £ 3.
For k = 3 and m = 4, we have 8^{4} + 7^{4} + 6^{4} = 6^{5} + 17, then 10^{4} + 9^{4} + 4^{4} = 7^{5} + 10, then 19^{4} + 12^{4} + 10^{4} = 11^{5} + 6, but the best result so far known is e(3,4,5) £ 1 since we have 80^{4} + 66^{4} + 27^{4} = 36^{5} + 1. A proportionally very close solution is given by the equation 371^{4} + 369^{4} + 275^{4} = 134^{5} + 3.
For k = 3 and m = 5, we have 4^{5} + 4^{5} + 4^{5} = 5^{5}  53 and 13^{5} + 13^{5} + 7^{5} = 15^{5} + 18.
For k = 3 and m = 6, the best result is 2^{6} + 2^{6} + 2^{6} = 3^{5}  51.
For k = 3 and m = 7, the best result is 2^{7} + 1^{7} + 1^{7} = 3^{5}  113.
For k = 4 and m = 4, we have 33^{4} + 33^{4} + 17^{4} + 12^{4} = 19^{5} = 37^{4} + 26^{4} + 19^{4} + 11^{4} as a dual solution for d = 0 and so S(4,5) £ 4.
For k = 4 and m = 5, Lander and Parkin’s first known counterexample to Euler’s Conjecture, 133^{5} + 110^{5} + 84^{5} + 27^{5} = 144^{5}, gives us S(5,5) £ 4.
For k = 4 and m = 6, the best result is 2^{6} + 2^{6} + 2^{6} + 2^{6} = 3^{5} + 13.
For k = 4 and m = 7, the best result is 4^{7} + 2^{7} + 2^{7} + 2^{7} = 7^{5}  39.
For k = 5 and m = 6, the best result is 5^{6} + 5^{6} + 3^{6} + 3^{6} + 2^{6} = 8^{5} + 4.
For k = 5 and m = 7, the best result is 4^{7} + 2^{7} + 2^{7} + 2^{7} + 1^{7} = 7^{5}  38.
For k = 6 and m = 6, the best result is 8^{6} + 6^{6} + 5^{6} + 5^{6} + 5^{6} + 5^{6} = 13^{5} + 7.
For k = 6 and m = 7, the best result is 4^{7} + 2^{7} + 2^{7} + 2^{7} + 1^{7} + 1^{7} = 7^{5}  37.
Consider n = 6.
For k = 1 and m = 7, the best result is 4^{7} = 5^{6} + 759.
For k = 2 and m = 2, there are 3 solutions for d = 0 with y_{1} up to 10000. They are 117^{2} + 44^{2} = 5^{6} , 2035^{2} + 828^{2} = 13^{6} and 4888^{2} + 495^{2} = 17^{6}. These can be lifted directly from the solution list for n = 3. Thus S(2,6) = 2.
For k = 2 and m = 3, we use the n = 3 results to get e(2,3,6) = 1 using 8^{3} + 6^{3} = 3^{6}  1.
For k = 2 and m = 4, the best nontrivial solution is 5^{4} + 3^{4} = 3^{6}  23.
For k = 2 and m = 5, the best result is 5^{5} + 4^{5} = 4^{6} + 53.
For k = 2 and m = 6, the best approximation to Fermat’s equation is 8^{6} + 8^{6} = 9^{6}  7153 if we ignore the unsatisfactory 2^{6} + 2^{6} = 3^{6}  601.
For k = 2 and m = 7, the best result is 3^{7} + 3^{7} = 4^{6} + 278.
For k = 3 and m = 3, we use the n = 3 results to get S(3,6) = 3, using, for example 8^{3} + 6^{3} + 1^{3} = 3^{6} or 22^{3} + 17^{3} + 4^{3} = 5^{6}.
For k = 3 and m = 4, the best result is 5^{4} + 3^{4} + 2^{4} = 3^{6}  7.
For k = 3 and m = 5, the best result is 6^{5} + 6^{5} + 2^{5} = 5^{6}  41.
For k = 3 and m = 6, the best result is 5^{6} + 5^{6} + 5^{6} = 6^{6} + 219.
For k = 3 and m = 7, the best result is 2^{7} + 2^{7} + 2^{7} = 3^{6}  345.
For k = 4 and m = 4, we have several solutions for d = 3, 23^{4} + 21^{4} + 13^{4} + 13^{4} = 9^{6} + 3, 29^{4} + 20^{4} + 19^{4} + 7^{4} = 10^{6} + 3, 51^{4} + 27^{4} + 21^{4} + 14^{4} = 14^{6} + 3 and 111^{4} + 71^{4} + 61^{4} + 14^{4} = 24^{6} + 3. However, we have that S(4,6) £ 4 with the triple solution 61^{4} + 56^{4} + 26^{4} + 4^{4} = 62^{4} + 47^{4} + 46^{4} + 8^{4} = 67^{4} + 44^{4} + 22^{4} + 8^{4} = 17^{6}.
For k = 4 and m = 5, we have e(4,5,6) £ 2 since 19^{5} + 12^{5} + 11^{5} + 10^{5} = 12^{6}  2.
For k = 4 and m = 6, the best result is 12^{6} + 11^{6} + 11^{6} + 10^{6} = 14^{6}  430.
For k = 4 and m = 7, the best result is 2^{7} + 2^{7} + 2^{7} + 2^{7} = 3^{6}  217.
For k = 5 and m = 4, the situation is moot given the k = 4 solutions. However, for completeness, there are solutions with d = 3, 2 and 1, and the triple solution for d = 0 of
16^{4} + 14^{4} + 10^{4} + 7^{4} + 6^{4} = 7^{6}, 18^{4} + 10^{4} + 6^{4} + 6^{4} + 3^{4} = 7^{6}, 18^{4} + 10^{4} + 7^{4} + 4^{4} + 2^{4} = 7^{6}.
For k = 5 and m = 5, we have e(5,5,6) £ 1 since 5^{5} + 3^{5} + 3^{5} + 3^{5} + 3^{5} = 4^{6} + 1 and 19^{5} + 12^{5} + 11^{5} + 10^{5} + 1^{5} = 12^{6}  1. We also have 27^{5} + 17^{5} + 15^{5} + 12^{5} + 3^{5} = 16^{6}  2.
For k = 5 and m = 6, the best result is 12^{6} + 11^{6} + 11^{6} + 10^{6} + 3^{6} = 14^{6} + 299.
For k = 5 and m = 7, the best result is 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 3^{6}  89.
For k = 6 and m = 6, the n = 3 solution with d = 3 above gives an n = 6 solution.
For k = 6 and m = 7, 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 3^{6} + 39.
Consider n = 7.
For k = 2 and m = 2, we have 35^{2} + 31^{2} = 3^{7}  1, followed by the triple solution 92^{2} + 89^{2} = 103^{2} + 76^{2} = 127^{2} + 16^{2} = 4^{7} + 1. The smallest solution for d = 0 is 278^{2} + 29^{2} = 5^{7}, and so we have S(2,7) = 2.
For k = 2 and m = 3, the best result is 5^{3} + 1^{3} = 2^{7}  2.
For k = 2 and m = 4, the best result is 23^{4} + 3^{4} = 16^{7}  14.
For k = 2 and m = 5, the best result is 4^{5} + 4^{5} = 3^{7}  139.
For k = 2 and m = 6, the best result is 5^{6} + 3^{6} = 4^{7}  30.
For k = 2 and m = 7, we know from Fermat’s Last Theorem that e(2,7,7) ¹ 0, with the particular case for 7th powers proved by Lame. The best result is 5^{7} + 5^{7} = 6^{7}  123686 if we avoid the unsatisfactory 2^{7} + 2^{7} = 3^{7}  1931.
For k = 3 and m = 3, there are several solutions for small d. We have e(3,3,7) £ 1 given by the unsatisfactory 5^{3} + 1^{3} + 1^{3} = 2^{7}  1 and the satisfactory 11^{3} + 8^{3} + 7^{3} = 3^{7}  1 and 156^{3} + 80^{3} + 78^{3} = 9^{7}  1.
For k = 3 and m = 4, we have 14^{4} + 14^{4} + 6^{4} = 5^{7} + 3 and 23^{4} + 3^{4} + 2^{4} = 6^{7} + 2.
For k = 3 and m = 5, we have e(3,5,7) £ 7 with the solution 21^{5} + 14^{5} + 11^{5} = 9^{7} + 7.
For k = 3 and m = 6, the best result is 5^{6} + 3^{6} + 1^{6} = 4^{7}  29.
For k = 3 and m = 7, the best result is 3^{7} + 3^{7} + 3^{7} = 4^{7}  9823 if we avoid the unsatisfactory 2^{7} + 2^{7} + 2^{7} = 3^{7}  1803.
For k = 4 and m = 3, we have 5^{3} + 1^{3} + 1^{3} + 1^{3} = 2^{7} and 11^{3} + 8^{3} + 7^{3} + 1^{3} = 3^{7}. Hence we have S(3,7) £ 4. There are also solutions for all d = ± 1, ± 2, ± 3 and ± 4. The next d = 0 solution is 56^{3} + 46^{3} + 19^{3} + 5^{3} = 6^{7}. There are seven separate solutions for d = 0 at x = 7.
For k = 4 and m = 4, we have 3^{4} + 2^{4} + 2^{4} + 2^{4} = 2^{7} + 1 and 34^{4} + 25^{4} + 24^{4} + 14^{4} = 8^{7} + 1 suggesting that e(4,4,7) £ 1. Larger solutions for small d include 113^{4} + 81^{4} + 80^{4} + 68^{4} = 16^{7} + 2 and 146^{4} + 142^{4} + 125^{4} + 115^{4} = 20^{7} + 2. However, we have S(4,7) £ 4 with the equation 126^{4} + 111^{4} + 50^{4} + 22^{4} = 17^{7}.
For k = 4 and m = 5, the best result is 4^{5} + 4^{5} + 2^{5} + 2^{5} = 3^{7}  75.
For k = 4 and m = 6, the best result is 5^{6} + 3^{6} + 1^{6} + 1^{6} = 4^{7}  28.
For k = 4 and m = 7, the best result is 3^{7} + 3^{7} + 3^{7} + 3^{7} = 4^{7}  7636 if we avoid the unsatisfactory 2^{7} + 2^{7} + 2^{7} + 2^{7}= 3^{7}  1675.
For k = 5 and m = 4, given the k = 4 case, we would hope that solutions exist for d = 0. However, at present there are several solutions with d = 2, e.g. 21^{4} + 15^{4} + 12^{4} + 10^{4} + 8^{4} = 6^{7} + 2, 40^{4} + 40^{4} + 40^{4} + 39^{4} + 9^{4} = 10^{7} + 2 and 55^{4} + 26^{4} + 24^{4} + 15^{4} + 10^{4} = 10^{7} + 2, and the best current result of 60^{4} + 45^{4} + 35^{4} + 31^{4} + 7^{4} = 11^{7} + 1.
For k = 5 and m = 5, we have 9^{5} + 7^{5} + 4^{5} + 4^{5} + 3^{5} = 5^{7} + 22 followed by 20^{5} + 17^{5} + 11^{5} + 4^{5} + 4^{5} = 9^{7}  13 and then 25^{5} + 11^{5} + 8^{5} + 8^{5} + 6^{5} = 10^{7}  12. The solution 61^{5} + 33^{5} + 25^{5} + 13^{5} + 5^{5} = 19^{7}  2 gives e(5,5,7) £ 2.
{N.B. Further searches locate the solution 505^{5} + 315^{5} + 266^{5} + 212^{5} + 115^{5} = 87^{7}, giving S(5,7) £ 5.}
For k = 5 and m = 6, a good early result is 8^{6} + 5^{6} + 3^{6} + 3^{6} + 3^{6} = 6^{7} + 20. However, the current best is 44^{6} + 38^{6} + 24^{6} + 10^{6} + 10^{6} = 27^{7} + 13.
For k = 5 and m = 7, the best result is 9^{7} + 9^{7} + 6^{7} + 5^{7} + 5^{7} = 10^{7} + 2124 is we ignore the unsatisfactory 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 3^{7}  1547. The only other remotely close solution is 30^{7} + 26^{7} + 26^{7} + 23^{7} + 20^{7} = 33^{7} + 2822.
For k = 6 and m = 6, we have the impressive 24^{6} + 23^{6} + 19^{6} + 17^{6} + 5^{6} + 3^{6} = 17^{7}  4.
For k = 6 and m = 7, we have 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} + 2^{7} = 3^{7}  1419.
Note that there is a general solution in the case k = 2 and m = 2 for any n, as follows. Choose a pair of numbers p and q such that p is even and q is odd and gcd(p,q) = 1. Let a = p + iq. Calculate a^{n} = x + iy. Then x^{2} + y^{2} = z^{n} where z = p^{2} + q^{2}. For example, we may take p = 2 and q = 1, so that z = 5. Then, for instance, (2 + i)^{7} =  278 29i, and we have 278^{2} + 29^{2} = 5^{7}, as mentioned above. For n = 100, the values of x and y obtained in this way both have 35 digits, and for n = 257, they have 90 digits. As well as base 5, we can always generate nontrivial solutions for base 13 (= 2^{2} + 3^{2}), 17 (= 4^{2} + 1^{2}), etc. This result is obtained using the properties of the Euclidean field Q(i) in algebraic number theory (for which, see [17]).
If we consider (2+i)^{n} = x_{n} + iy_{n} , so that x_{n}^{2} + y_{n}^{2} = 5^{n}, we may use the recursive definition given by setting x_{0} = 1, y_{0} = 0, which makes sense since 1^{2} + 0^{2} = 5^{0}, and then the equations x_{n} = 2x_{n 1}  y_{n 1} and y_{n} = 2y_{n 1} + x_{n 1} . Similar recursive definitions are easily derived for other bases using the same starting parameters.
Alternatively, the standard series expansion (2+i)^{n} = 2^{n} + n2^{n 1}i + .... + i^{n} gives
real coefficient x_{n} =
imaginary coefficient y_{n} =
From this is is obvious that if n is even, x_{n} is odd and y_{n} is even, and when n is odd, x_{n} is even and y_{n} is odd.
The best estimates for S(m,n) based on the above data are presented in the following table. An asterisk indicates that equality has not been proved.
n\m 
2 
3 
4 
5 
6 
7 
2 
2 
2 
3 
3 * 
3 * 
4 * 
3 
2 
3 
3 * 
4 * 
9 * 
7 * 
4 
2 
3 * 
3 
6 * 


5 
2 
3 * 
4 * 
4 * 


6 
2 
3 
4 * 
6 * 
9 * 

7 
2 
4 * 
4 * 
7 * 
10 * 

Table 4.1
Additionally, for m and n with no d = 0 solution we give best error margins as a summary of what has been presented above.
e(2,4,2) = 1 




e(2,5,2) £ 2 




e(2,6,2) £ 1 




e(2,7,2) £ 7 
e(3,7,2) £ 4 



e(2,3,3) = 1 




e(2,4,3) £ 5 




e(2,5,3) £ 7 
e(3,5,3) £ 3 



e(2,6,3) £ 3 
e(3,6,3) £ 2 
e(4,6,3) £ 3 
e(5,6,3) £ 2 
e(6,6,3) £ 3 
e(2,7,3) £ 9 
e(3,7,3) £ 8 
e(4,7,3) £ 7 
e(5,7,3) £ 6 
e(6,7,3) £ 1 
e(2,3,4) £ 3 




e(2,4,4) £ 46 




e(2,5,4) £ 12 
e(3,5,4) £ 3 
e(4,5,4) £ 3 
e(5,5,4) £ 1 

e(2,6,4) £ 16 
e(3,6,4) £ 15 
e(4,6,4) £ 10 
e(5,6,4) £ 3 
e(6,6,4) £ 1 
e(2,7,4) £ 86 
e(3,7,4) £ 22 
e(4,7,4) £ 16 
e(5,7,4) £ 7 
e(6,7,4) £ 9 
e(2,3,5) £ 2 




e(2,4,5) £ 65 
e(3,4,5) £ 1 



e(2,5,5) £ 12 
e(3,5,5) £ 18 



e(2,6,5) £ 115 
e(3,6,5) £ 51 
e(4,6,5) £ 13 
e(5,6,5) £ 4 
e(6,6,5) £ 7 
e(2,7,5) £ 13 
e(3,7,5) £ 113 
e(4,7,5) £ 39 
e(5,7,5) £ 38 
e(6,7,5) £ 37 
e(2,3,6) = 1 




e(2,4,6) £ 23 
e(3,4,6) £ 7 



e(2,5,6) £ 53 
e(3,5,6) £ 41 
e(4,5,6) £ 2 
e(5,5,6) £ 1 

e(2,6,6) £ 601 
e(3,6,6) £ 219 
e(4,6,6) £ 430 
e(5,6,6) £ 299 
e(6,6,6) £ 3 
e(2,7,6) £ 278 
e(3,7,6) £ 345 
e(4,7,6) £ 217 
e(5,7,6) £ 89 
e(6,7,6) £ 39 
e(2,3,7) £ 2 
e(3,3,7) £ 1 



e(2,4,7) £ 14 
e(3,4,7) £ 2 

e(5,4,7) £ 1 

e(2,5,7) £ 139 
e(3,5,7) £ 7 
e(4,5,7) £ 75 
e(5,5,7) £ 2 

e(2,6,7) £ 30 
e(3,6,7) £ 29 
e(4,6,7) £ 28 
e(5,6,7) £ 13 
e(6,6,7) £ 4 
e(2,7,7) £ 1931 
e(3,7,7) £ 1803 
e(4,7,7) £ 1675 
e(5,7,7) £ 1547 
e(6,7,7) £ 1419 
For some combinations of m and n, it is obvious that the brute force method is inadequate for providing solutions when d = 0. In this case, we look for solutions of equation 4.2 using the decomposition algorithm. Thus we have
s(5,4) £ 5 
(22,22,15,12,4,1)^{5} = 58^{4} 


(34,29,22,14,4,3)^{5} = 92^{4} 


(39,22,21,14,3,1)^{5} = 100^{4} 


(40,17,10,10,8,6)^{5} = 101^{4} 


(40,34,33,32,15,12)^{5} = 122^{4} 


(47,38,29,28,21,18)^{5} = 137^{4} 


(54,29,10,6,4,3)^{5} = 148^{4} 


(54,28,26,20,17,6)^{5} = 149^{4} 


(55,33,22,12,10,9)^{5} = 153^{4} 


(54,36,33,27,21,4)^{5} = 155^{4} 


(61,24,19,4,2,1)^{5} = 171^{4} 


to x = 200 for k £ 6 


(185,169,109,46,1)^{5} = 780^{4} 


(207,126,103,59,6)^{5} = 807^{4} 


to x = 1045 for k £ 5 


to x = 1045 for k £ 4 

s(5,6) £ 6 
(19,12,11,10,1,1)^{5} = 12^{6} 


(119,81,46,41,24,14)^{5} = 55^{6} 


(156,140,120,86,74,9)^{5} = 75^{6} 


(181,157,128,122,47,31)^{5} = 84^{6} 


(217,198,172,144,71,48)^{5} = 100^{6} 


to x = 100 for k £ 6 


to x = 126 for k £ 5 

s(5,7) £ 5 
(15,15,14,8,6,2,1,1)^{5} = 8^{7} 
(k = 8) 

(20,16,13,11,4,4,1)^{5} = 9^{7} 
(k = 7) 

(35,24,18,13,5,2)^{5} = 13^{7} 
(k = 6) 

(107,106,103,80,50,31)^{5} = 33^{7} 
(k = 6) 

(130,123,101,76,30,26)^{5} = 36^{7} 
(k = 6) 

to x = 37 for k £ 6 


(505,315,266,212,115)^{5} = 87^{7} 
(k = 5) 

to x = 89 for k £ 5 

s(6,3) £ 7 
(13,11,10,5,5,5,2,2)^{6} = 197^{3} 
(k = 8) 

(30,24,9,9,3,3,1)^{6} = 973^{3} 
(k = 7) 

to x = 4460 for k £ 7 


to x = 9662 for k £ 6 

s(6,4) £ 7 
(13,13,7,5,5,5,5,3)^{6} = 56^{4} 
(k = 8) 

(18,16,9,8,8,8,6,6)^{6} = 85^{4} 
(k = 8) 

(92,66,57,50,48,22,8)^{6} = 931^{4} 
(k = 7) 

(84,84,62,56,52,32,9)^{6} = 949^{4} 
(k = 7) 

to x = 1084 for k £ 7 


to x = 1605 for k £ 6 

s(6,5) £ 7 
(17,17,14,11,11,10,7,3,1,1)^{6} = 36^{5} 
(k = 10) 

(30,26,23,22,18,16,14,1,1)^{6} = 67^{5} 
(k = 9) 

(29,29,26,25,19,13,13,13)^{6} = 71^{5} 
(k = 8) 

(90,85,68,68,57,54,19)^{6} = 259^{5} 
(k = 7) 

to x = 315 for k £ 7 


to x = 503 for k £ 6 

s(6,6) £ 7 
(1077,894,702,474,402,234,74)^{6} = 1141^{6} 
(Lander + Parkin) 
s(6,7) £ 8 
(19,19,18,17,14,14,11,11,5,1)^{6} = 15^{7} 
(k = 10) 

(29,16,7,7,7,7,7,5,5)^{6} = 18^{7} 
(k = 9) 

(34,32,27,27,15,5,5,5,3)^{6} = 23^{7} 
(k = 9) 

(66,62,62,56,54,52,23,13,13)^{6} = 43^{7} 
(k = 9) 

to x = 48 for k £ 9 


(112,95,81,73,20,8,7,7)^{6} = 61^{7} 
(k = 8) 

to x = 61 for k £ 8 


to x = 80 for k £ 7 

s(7,3) £ 7 
(5,4,3,2,2,2,2,2)^{7} = 46^{3} 
(k = 8) 

(8,4,4,4,2,2,2,1)^{7} = 129^{3} 
(k = 8) 

(16,16,11,9,6,5,4,2)^{7} = 825^{3} 
(k = 8) 

(23,22,16,16,11,5,4,1)^{7} = 1862^{3} 
(k = 8) 

to x = 2000 for k £ 8 


(20,7,2,2,2,2,1)^{7} = 1086^{3} 
(k = 7) 

(49,49,37,20,17,17,15)^{7} = 11328^{3} 
(k = 7) 

(49,49,39,36,25,21,4)^{7} = 11647^{3} 
(k = 7) 

to x = 12930 for k £ 7 


to x = 17600 for k £ 6 

s(7,4) £ 8 
(13,12,9,9,5,4,3,2,2,1)^{7} = 102^{4} 
(k = 10) 

(16,13,13,13,13,9,9,9,4,1)^{7} = 152^{4} 
(k = 10) 

to x = 400 for k £ 10 


(37,30,29,26,18,18,17,16,5)^{7} = 616^{4} 
(k = 9) 

(39,32,31,30,21,20,5,4,2)^{7} = 688^{4} 
(k = 9) 

(38,38,35,31,16,16,16,10,7)^{7} = 753^{4} 
(k = 9) 

(50,30,18,15,15,15,11,8,1)^{7} = 947^{4} 
(k = 9) 

to x = 1146 for k £ 9 


(64,57,48,36,25,12,11,3)^{7} = 1628^{4} 
(k = 8) 

(65,57,41,41,41,38,19,14)^{7} = 1658^{4} 
(k = 8) 

to x = 1706 for k £ 8 


to x = 2938 for k £ 7 

s(7,5) £ 7 
(14,12,12,12,11,11,8,7,1,1,1)^{7} = 48^{5} 
(k = 11) 

(23,18,16,13,12,12,10,9,8,6)^{7} = 85^{5} 
(k = 10) 

to x = 140 for k £ 10 


(46,45,32,23,12,7,6)^{7} = 243^{5} 
(k = 7) 

(93,88,79,44,12,12,3)^{7} = 655^{5} 
(k = 7) 

to x = 245 for k £ 9 


to x = 330 for k £ 8 


to x = 740 for k £ 7 


to x = 1210 for k £ 6 

s(7,6) £ 9 
(16,16,13,12,12,12,11,8,6,4,3,3,2,1,1)^{7} = 30^{6} 
(k = 15) 

(20,17,16,14,14,9,8,6,4,3,3,2,2,2)^{7} = 36^{6} 
(k = 14) 

(20,19,14,14,12,10,8,7,7,6,6,1,1)^{7} = 37^{6} 
(k = 13) 

(24,17,17,16,15,15,15,13,13,9,8,7)^{7} = 43^{6} 
(k = 12) 

(35,24,24,19,19,13,12,11,7,4,1)^{7} = 65^{6} 
(k = 11) 

(40,33,31,28,28,28,28,26,26,6)^{7} = 82^{6} 
(k = 10) 

(57,44,34,33,30,29,29,25,24,11)^{7} = 116^{6} 
(k = 10) 

(54,51,49,44,43,42,42,32,17,4)^{7} = 126^{6} 
(k = 10) 

to x = 134 for k £ 10 


(55,42,36,29,28,24,23,15,15)^{7} = 111^{6} 
(k = 9) 

(66,60,52,43,38,27,25,4,1)^{7} = 146^{6} 
(k = 9) 

to x = 170 for k £ 9 


to x = 244 for k £ 8 

s(7,7) £ 8 
(90,85,83,64,58,53,35,12)^{7} = 102^{7} 
(Lander + Parkin) 
We use these additional results to augment Table 4.1, as follows.
n\m 
2 
3 
4 
5 
6 
7 
2 
2 
2 
3 
3 * 
3 * 
4 * 
3 
2 
3 
3 * 
4 * 
7 * 
7 * 
4 
2 
3 * 
3 
5 * 
7 * 
8 * 
5 
2 
3 * 
4 * 
4 * 
7 * 
7 * 
6 
2 
3 
4 * 
6 * 
7 * 
9 * 
7 
2 
4 * 
4 * 
5 * 
8 * 
8 * 
Table 4.2
where, as well as filling in the gaps, upper limits on the values of s(5,4) and s(5,7) have been reduced.