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 non-trivial solution when both m and n are greater than 1. Thus s(m) = S(m,m). Without loss of generality, we shall assume that y₁ ³ y₂ ³ ...... ³ y_k and, when m = n, that x > y₁.

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 xⁿ + yⁿ = zⁿ 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 non-trivial 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 non-trivial 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 = y₁ or x = 1 or y₁ = 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. 3³ + 2³ = 27 + 8 = 35 = 5.7. All we then require is to find z such that z³.(3³ + 2³) is a 7th power. To do this we require z = 5^a.7^b so that z³.5.7 = 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 = 5².7² = 1225, so then (3z)³ + (2z)³ = (5.7)⁷, i.e. the derived solution 3675³ + 2450³ = 35⁷. 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 y₁ up to 3000 as preferred minimum. For k = 3, searches take y₁ up to at least 250, k = 4 with y₁ up to at least 100, k = 5 with y₁ up to 50 and k = 6 with y₁ 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 non-zero 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 2³ = 3² - 1 gives e(1,3,2) = 1.

For k = 1 and m = 5, the equation 1⁵ = 2² - 3 is considered trivial. After this, the closest known is 2⁵ = 6² - 4.

For k = 1 and m = 7, the equation 2⁷ = 11² + 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 3² + 4² = 5².

For k = 2 and m = 3, there are 27 solutions with d = 0 for y₁ up to 10000, the smallest being 2³ + 1³ = 3², and the largest being 9265³ + 2184³ = 897623². 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 y₁ up to 10000, the smallest being 7⁴ + 5⁴ = 55² + 1 and the largest being 9901⁴ + 5333⁴ = 102072161² + 1. Simple congruence rules imply that there are no solutions for d = - 1.

For k = 2 and m = 5, and y₁ up to 8120, there are two solutions with d = 2. These are 3⁵ + 3⁵ = 22² + 2 and 111⁵ + 87⁵ = 147766² + 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 31⁶ + 29⁶ = 38501² +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 4⁷ + 4⁷ = 181² + 7.

For k = 3 and m = 3, there are many solutions with d = 0. The smallest is 3³ + 2³ + 1³ = 6² and the smallest with none of the y’s equal to 1 is 14³ + 9³ + 2³ = 59². A particularly interesting solution is 25³ + 24³ + 23³ = 204², 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 20⁴ + 15⁴ + 12⁴ = 481² is the only solution for d = 0 with y₁ 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 non-trivial solutions. However, there are 4 solutions with d = 0. These are 69⁵ + 49⁵ + 3⁵ = 42971², 81⁵ + 68⁵ + 20⁵ = 70313², 97⁵ + 80⁵ + 19⁵ = 108934² and 97⁵ + 83⁵ + 16⁵ = 111926². 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 non-trivial solutions with d = 1 and 2, e.g. 3⁶ + 3⁶ + 2⁶ = 39² + 1, 5⁶ + 4⁶ + 3⁶ = 143² + 1, 25⁶ + 25⁶ + 24⁶ = 26065² + 1, 5⁶ + 5⁶ + 4⁶ = 188² + 2 and 19⁶ + 11⁶ + 3⁶ = 6987² + 2. However, we have S(6,2) £ 3 since 100⁶ + 81⁶ + 42⁶ = 1134865².

For k = 3 and m = 7, there are trivial parameterisable solutions with d = 1 and d = - 1. The best non-trivial result available at present is e(3,7,2) £ 4, using 5⁷ + 4⁷ + 4⁷ = 333² + 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 non-parameterised 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 non-trivial solutions with small d, e.g. 27⁷ + 24⁷ + 21⁷ + 13⁷ = 130041² + 4 and 40⁷ + 31⁷ + 27⁷ + 4⁷ = 449236² + 2. However, the solutions 74⁷ + 59⁷ + 17⁷ + 4⁷ = 3826270² and 78⁷ + 51⁷ + 42⁷ + 25⁷ = 4324306² are found for y₁ 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 2⁵ = 3³ + 5 gives the best result.

For k = 1 and m = 7, the equation 2⁷ = 5³ + 3 gives the best result.

For k = 2 and m = 2, there are 80 solutions with d = 0 for y₁ up to 10000, the smallest being 11² + 2² = 5³ and the largest being 9830² + 1159² = 461³. 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 10³ + 9³ = 12³ + 1 and 16 solutions with d = - 1, the smallest being 8³ + 6³ = 9³ - 1, with y₁ up to 10000. Thus e(2,3,3) = 1. Of these, 5 solutions each of each sign result directly from the related parameteric solutions

(3²b⁴)³ + (3²b³ + 1)³ = (3²b⁴ + 3b)³ + 1 and (3²b⁴ - 3b)³ + (3²b³ - 1)³ = (3²b⁴)³ - 1.

There is also a parameterisable solution for d = - 2, (6p³ - 1)³ + (6p²)³ = (6p³ + 1)³ - 2.

For k = 2 and m = 4, the first non-trivial solution with small error is 2⁴ + 2⁴ = 3³ + 5.

For k = 2 and m = 5, the only non-trivial solution with small error is 5⁵ + 3⁵ = 15³ - 7.

For k = 2 and m = 6, the only non-trivial solution with small error is 2⁶ + 2⁶ = 5³ + 3.

For k = 2 and m = 7, the only non-trivial solution with small error is 3⁷ + 1⁷ = 13³ - 9, which is not trivial as the addition of a 1 decreases the error from a k = 1 solution. The solution 2⁷ + 1⁷ = 5³ + 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 non-trivial solutions exist with d = 0, some of which belong to parameterisable groups. Hence S(3,3) = 3. The smallest solution is 5³ + 4³ + 3³ = 6³. Other notable solutions are 8³ + 6³ + 1³ = 9³ (related to the k = 2 solution), 22³ + 17³ + 4³ = 25³ and 74³ + 48³ + 25³ = 81³, 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. 5⁴ + 5⁴ + 3⁴ = 11³. Thus S(4,3) £ 3, with equality likely. Four solutions exist with d = 1 for y₁ up to 100. These are 22⁴ + 16⁴ + 11⁴ = 68³ + 1, 78⁴ + 68⁴ + 11⁴ = 388³ + 1, 82⁴ + 53⁴ + 48⁴ = 388³ + 1 and 87⁴ + 82⁴ + 6⁴ = 468³ + 1.

For k = 3 and m = 5, there are several small solutions with d = 6 and d = - 6, but the best result available is 81⁵ + 74⁵ + 15⁵ = 1787³ - 3. Thus e(3,5,3) £ 3.

For k = 3 and m = 6, the solution 25⁶ + 23⁶ + 6⁶ = 732³ + 2 gives e(3,6,3) £ 2.

For k = 3 and m = 7, the solution 3⁷ + 1⁷ + 1⁷ = 13³ - 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 15⁷ + 5⁷ + 4⁷ = 555³ + 9.

For k = 4 and m = 5, we have 4⁵ + 3⁵ + 2⁵ + 2⁵ = 11³, and so s(5,3) £ 4. A larger example is 48⁵ + 8⁵ + 5⁵ + 3⁵ = 634³. An interesting near miss is 6⁵ + 5⁵ + 4⁵ + 3⁵ = 23³ + 1. Solutions exist for most small d.

For k = 4 and m = 6, the earliest result is 3⁶ + 3⁶ + 3⁶ + 1⁶ = 13³ - 9, which is far from ideal since the k = 3 case provides a solution for d = 2. Purer solutions include 8⁶ + 6⁶ + 5⁶ + 4⁶ = 69³ + 12 and 45⁶ + 44⁶ + 43⁶ + 19⁶ = 2799³ + 12, followed by 56⁶ + 33⁶ + 27⁶ + 12⁶ = 3192³ + 10. There is also a prevalence of solutions with d = - 23. However, the current best results are for d = 3, e.g. 57⁶ + 19⁶ + 19⁶ + 10⁶ = 3252³ + 3 and 68⁶ + 53⁶ + 25⁶ + 15⁶ = 4950³ + 3.

For k = 4 and m = 7, the best result is 3⁷ + 1⁷ + 1⁷ + 1⁷ = 13³ - 7, and the second best is 10⁷ + 6⁷ + 5⁷ + 3⁷ = 218³ + 16.

For k = 5 and m = 6, we first have solutions for d = 4, i.e. 5⁶ + 5⁶ + 3⁶ + 3⁶ + 2⁶ = 32³ + 4 and 23⁶ + 21⁶ + 15⁶ + 7⁶ + 4⁶ = 626³ + 4, followed by neighbouring solutions for d = 2, namely 28⁶ + 21⁶ + 15⁶ + 10⁶ + 6⁶ = 834³ + 2 and 28⁶ + 21⁶ + 16⁶ + 16⁶ + 3⁶ = 844³ + 2.

For k = 5 and m = 7, the best result is 3⁷ + 1⁷ + 1⁷ + 1⁷ + 1⁷ = 13³ - 6.

For k = 6 and m = 6, we have 10⁶ + 9⁶ + 7⁶ + 7⁶ + 4⁶ + 3⁶ = 121³ + 3, and also 5 solutions for d = - 3, the largest of which is 25⁶ + 22⁶ + 11⁶ + 7⁶ + 5⁶ + 2⁶ = 711³ - 3.

For k = 6 and m = 7, we have 20⁷ + 7⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 1086³ - 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 3³ = 2⁴ + 11.

For k = 1 and m = 5, the best result is 3⁵ = 4⁴ - 13.

For k = 1 and m = 7, the best result is 2⁷ = 3⁴ + 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. 24² + 7² = 5⁴, 120² + 119² = 13⁴ and 240² + 161² = 17⁴. Thus S(2,4) = 2.

For k = 2 and m = 3, there are several solutions with small d, the best current result being 63³ + 31³ = 23⁴ - 3. Thus e(2,3,4) £ 3. Another close call is 257⁴ + 147⁴ = 67³ - 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 5⁴ + 5⁴ = 6⁴ - 46.

For k = 2 and m = 5, the best result is 3⁵ + 1⁵ = 4⁴ - 12, which is related to a k = 1 solution. Apart from this, the best is 2⁵ + 2⁵ = 3⁴ - 17.

For k = 2 and m = 6, the best result is 2⁶ + 1⁶ = 3⁴ - 16.

For k = 2 and m = 7, the best result is 3⁷ + 2⁷ = 7⁴ - 86.

For k = 3 and m = 3, there are quite a few solutions with d = ± 1, but we also have the equations 20³ + 17³ + 12³ = 11⁴ and 74³ + 48³ + 25³ = 27⁴, 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 414560⁴ + 217519⁴ + 95800⁴ = 422481⁴, 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 47⁴ + 22⁴ + 21⁴ = 48⁴ + 2.

For k = 3 and m = 5, the best result is 4⁵ + 3⁵ + 2⁵ = 6⁴ + 3.

For k = 3 and m = 6, the best result is 2⁶ + 1⁶ + 1⁶ = 3⁴ - 15.

For k = 3 and m = 7, the best result is 4⁷ + 3⁷ + 3⁷ = 12⁴ + 22.

For k = 4 and m = 5, the earliest close calls are 8⁵ + 7⁵ + 4⁵ + 2⁵ = 15⁴ + 6 and 9⁵ + 5⁵ + 5⁵ + 3⁵ = 16⁴ + 6. However, the solution 104⁵ + 72⁵ + 52⁵ + 45⁵ = 348⁴ - 3. provides the current best result.

For k = 4 and m = 6, the best result is 21⁶ + 18⁶ + 11⁶ + 3⁶ = 105⁴ + 10.

For k = 4 and m = 7, the best result is 9⁷ + 5⁷ + 4⁷ + 3⁷ = 47⁴ - 16.

For k = 5 and m = 5, there are many solutions for small d, but the most important are 5⁵ + 3⁵ + 3⁵ + 3⁵ + 3⁵ = 8⁴ + 1,

22⁵ + 22⁵ + 15⁵ + 12⁵ + 4⁵ = 58⁴ - 1 and 39⁵ + 22⁵ + 21⁵ + 14⁵ + 3⁵ = 100⁴ - 1. Thus e(5,5,4) £ 1 and S(5,4) £ 6.

{N.B. Further studies provide the result 185⁵ + 169⁵ + 109⁵ + 46⁵ + 1⁵ = 780⁴, so that, in fact, S(5,4) £ 5.}

For k = 5 and m = 6, the best result is 9⁶ + 8⁶ + 5⁶ + 3⁶ + 2⁶ = 30⁴ + 3. Another close solution is 31⁶ + 27⁶ + 26⁶ + 23⁶ + 5⁶ = 204⁴ + 4.

For k = 5 and m = 7, the best early results are for d = ± 15, given by 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 5⁴ + 15 and 9⁷ + 5⁷ + 4⁷ + 3⁷ + 1⁷ = 47⁴ - 15, the latter based on the k = 4 solution. However, we have e(5,7,4) £ 7 since we have 31⁷ + 14⁷ + 13⁷ + 11⁷ + 10⁷ = 408⁴ + 7.

For k = 6 and m = 6, we have e(6,6,4) £ 1 since 9⁶ + 8⁶ + 7⁶ + 4⁶ + 4⁶ + 4⁶ = 31⁴ + 1.

For k = 6 and m = 7, the best result is 16⁷ + 13⁷ + 10⁷ + 7⁷ + 5⁷ + 4⁷ = 136⁴ + 9.

Consider n = 5.

For k = 1 and m = 7, the best result is 2⁷ = 3⁵ - 115.

For k = 2 and m = 2, there are 6 solutions with d = 0, for y₁ up to 9390. The smallest is 41² + 38² = 5⁵ and the largest is 6121² + 5646² = 37⁵. Hence S(2,5) = 2.

For k = 2 and m = 3, for y₁ up to 4000 we have e(2,3,5) £ 2 since 271³ + 239³ = 32⁵ - 2.

For k = 2 and m = 4, the only non-trivial solution with d < 81 is 22⁴ + 11⁴ = 12⁵ + 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 16⁵ + 13⁵ = 17⁵ + 12, with a search range of y₁ up to 4346.

For k = 2 and m = 6, the best result is 2⁶ + 2⁶ = 3⁵ - 115.

For k = 2 and m = 7, the best result is 2⁷ + 2⁷ = 3⁵ + 13.

For k = 3 and m = 3, there are several solutions with d = ± 1. However there are two with d = 0. These are 77³ + 57³ + 49³ = 15⁵ = 86³ + 40³ + 39³. Thus S(3,5) £ 3.

For k = 3 and m = 4, we have 8⁴ + 7⁴ + 6⁴ = 6⁵ + 17, then 10⁴ + 9⁴ + 4⁴ = 7⁵ + 10, then 19⁴ + 12⁴ + 10⁴ = 11⁵ + 6, but the best result so far known is e(3,4,5) £ 1 since we have 80⁴ + 66⁴ + 27⁴ = 36⁵ + 1. A proportionally very close solution is given by the equation 371⁴ + 369⁴ + 275⁴ = 134⁵ + 3.

For k = 3 and m = 5, we have 4⁵ + 4⁵ + 4⁵ = 5⁵ - 53 and 13⁵ + 13⁵ + 7⁵ = 15⁵ + 18.

For k = 3 and m = 6, the best result is 2⁶ + 2⁶ + 2⁶ = 3⁵ - 51.

For k = 3 and m = 7, the best result is 2⁷ + 1⁷ + 1⁷ = 3⁵ - 113.

For k = 4 and m = 4, we have 33⁴ + 33⁴ + 17⁴ + 12⁴ = 19⁵ = 37⁴ + 26⁴ + 19⁴ + 11⁴ 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, 133⁵ + 110⁵ + 84⁵ + 27⁵ = 144⁵, gives us S(5,5) £ 4.

For k = 4 and m = 6, the best result is 2⁶ + 2⁶ + 2⁶ + 2⁶ = 3⁵ + 13.

For k = 4 and m = 7, the best result is 4⁷ + 2⁷ + 2⁷ + 2⁷ = 7⁵ - 39.

For k = 5 and m = 6, the best result is 5⁶ + 5⁶ + 3⁶ + 3⁶ + 2⁶ = 8⁵ + 4.

For k = 5 and m = 7, the best result is 4⁷ + 2⁷ + 2⁷ + 2⁷ + 1⁷ = 7⁵ - 38.

For k = 6 and m = 6, the best result is 8⁶ + 6⁶ + 5⁶ + 5⁶ + 5⁶ + 5⁶ = 13⁵ + 7.

For k = 6 and m = 7, the best result is 4⁷ + 2⁷ + 2⁷ + 2⁷ + 1⁷ + 1⁷ = 7⁵ - 37.

Consider n = 6.

For k = 1 and m = 7, the best result is 4⁷ = 5⁶ + 759.

For k = 2 and m = 2, there are 3 solutions for d = 0 with y₁ up to 10000. They are 117² + 44² = 5⁶ , 2035² + 828² = 13⁶ and 4888² + 495² = 17⁶. 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 8³ + 6³ = 3⁶ - 1.

For k = 2 and m = 4, the best non-trivial solution is 5⁴ + 3⁴ = 3⁶ - 23.

For k = 2 and m = 5, the best result is 5⁵ + 4⁵ = 4⁶ + 53.

For k = 2 and m = 6, the best approximation to Fermat’s equation is 8⁶ + 8⁶ = 9⁶ - 7153 if we ignore the unsatisfactory 2⁶ + 2⁶ = 3⁶ - 601.

For k = 2 and m = 7, the best result is 3⁷ + 3⁷ = 4⁶ + 278.

For k = 3 and m = 3, we use the n = 3 results to get S(3,6) = 3, using, for example 8³ + 6³ + 1³ = 3⁶ or 22³ + 17³ + 4³ = 5⁶.

For k = 3 and m = 4, the best result is 5⁴ + 3⁴ + 2⁴ = 3⁶ - 7.

For k = 3 and m = 5, the best result is 6⁵ + 6⁵ + 2⁵ = 5⁶ - 41.

For k = 3 and m = 6, the best result is 5⁶ + 5⁶ + 5⁶ = 6⁶ + 219.

For k = 3 and m = 7, the best result is 2⁷ + 2⁷ + 2⁷ = 3⁶ - 345.

For k = 4 and m = 4, we have several solutions for d = 3, 23⁴ + 21⁴ + 13⁴ + 13⁴ = 9⁶ + 3, 29⁴ + 20⁴ + 19⁴ + 7⁴ = 10⁶ + 3, 51⁴ + 27⁴ + 21⁴ + 14⁴ = 14⁶ + 3 and 111⁴ + 71⁴ + 61⁴ + 14⁴ = 24⁶ + 3. However, we have that S(4,6) £ 4 with the triple solution 61⁴ + 56⁴ + 26⁴ + 4⁴ = 62⁴ + 47⁴ + 46⁴ + 8⁴ = 67⁴ + 44⁴ + 22⁴ + 8⁴ = 17⁶.

For k = 4 and m = 5, we have e(4,5,6) £ 2 since 19⁵ + 12⁵ + 11⁵ + 10⁵ = 12⁶ - 2.

For k = 4 and m = 6, the best result is 12⁶ + 11⁶ + 11⁶ + 10⁶ = 14⁶ - 430.

For k = 4 and m = 7, the best result is 2⁷ + 2⁷ + 2⁷ + 2⁷ = 3⁶ - 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

16⁴ + 14⁴ + 10⁴ + 7⁴ + 6⁴ = 7⁶, 18⁴ + 10⁴ + 6⁴ + 6⁴ + 3⁴ = 7⁶, 18⁴ + 10⁴ + 7⁴ + 4⁴ + 2⁴ = 7⁶.

For k = 5 and m = 5, we have e(5,5,6) £ 1 since 5⁵ + 3⁵ + 3⁵ + 3⁵ + 3⁵ = 4⁶ + 1 and 19⁵ + 12⁵ + 11⁵ + 10⁵ + 1⁵ = 12⁶ - 1. We also have 27⁵ + 17⁵ + 15⁵ + 12⁵ + 3⁵ = 16⁶ - 2.

For k = 5 and m = 6, the best result is 12⁶ + 11⁶ + 11⁶ + 10⁶ + 3⁶ = 14⁶ + 299.

For k = 5 and m = 7, the best result is 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 3⁶ - 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, 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 3⁶ + 39.

Consider n = 7.

For k = 2 and m = 2, we have 35² + 31² = 3⁷ - 1, followed by the triple solution 92² + 89² = 103² + 76² = 127² + 16² = 4⁷ + 1. The smallest solution for d = 0 is 278² + 29² = 5⁷, and so we have S(2,7) = 2.

For k = 2 and m = 3, the best result is 5³ + 1³ = 2⁷ - 2.

For k = 2 and m = 4, the best result is 23⁴ + 3⁴ = 16⁷ - 14.

For k = 2 and m = 5, the best result is 4⁵ + 4⁵ = 3⁷ - 139.

For k = 2 and m = 6, the best result is 5⁶ + 3⁶ = 4⁷ - 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 5⁷ + 5⁷ = 6⁷ - 123686 if we avoid the unsatisfactory 2⁷ + 2⁷ = 3⁷ - 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 5³ + 1³ + 1³ = 2⁷ - 1 and the satisfactory 11³ + 8³ + 7³ = 3⁷ - 1 and 156³ + 80³ + 78³ = 9⁷ - 1.

For k = 3 and m = 4, we have 14⁴ + 14⁴ + 6⁴ = 5⁷ + 3 and 23⁴ + 3⁴ + 2⁴ = 6⁷ + 2.

For k = 3 and m = 5, we have e(3,5,7) £ 7 with the solution 21⁵ + 14⁵ + 11⁵ = 9⁷ + 7.

For k = 3 and m = 6, the best result is 5⁶ + 3⁶ + 1⁶ = 4⁷ - 29.

For k = 3 and m = 7, the best result is 3⁷ + 3⁷ + 3⁷ = 4⁷ - 9823 if we avoid the unsatisfactory 2⁷ + 2⁷ + 2⁷ = 3⁷ - 1803.

For k = 4 and m = 3, we have 5³ + 1³ + 1³ + 1³ = 2⁷ and 11³ + 8³ + 7³ + 1³ = 3⁷. 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 56³ + 46³ + 19³ + 5³ = 6⁷. There are seven separate solutions for d = 0 at x = 7.

For k = 4 and m = 4, we have 3⁴ + 2⁴ + 2⁴ + 2⁴ = 2⁷ + 1 and 34⁴ + 25⁴ + 24⁴ + 14⁴ = 8⁷ + 1 suggesting that e(4,4,7) £ 1. Larger solutions for small d include 113⁴ + 81⁴ + 80⁴ + 68⁴ = 16⁷ + 2 and 146⁴ + 142⁴ + 125⁴ + 115⁴ = 20⁷ + 2. However, we have S(4,7) £ 4 with the equation 126⁴ + 111⁴ + 50⁴ + 22⁴ = 17⁷.

For k = 4 and m = 5, the best result is 4⁵ + 4⁵ + 2⁵ + 2⁵ = 3⁷ - 75.

For k = 4 and m = 6, the best result is 5⁶ + 3⁶ + 1⁶ + 1⁶ = 4⁷ - 28.

For k = 4 and m = 7, the best result is 3⁷ + 3⁷ + 3⁷ + 3⁷ = 4⁷ - 7636 if we avoid the unsatisfactory 2⁷ + 2⁷ + 2⁷ + 2⁷= 3⁷ - 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. 21⁴ + 15⁴ + 12⁴ + 10⁴ + 8⁴ = 6⁷ + 2, 40⁴ + 40⁴ + 40⁴ + 39⁴ + 9⁴ = 10⁷ + 2 and 55⁴ + 26⁴ + 24⁴ + 15⁴ + 10⁴ = 10⁷ + 2, and the best current result of 60⁴ + 45⁴ + 35⁴ + 31⁴ + 7⁴ = 11⁷ + 1.

For k = 5 and m = 5, we have 9⁵ + 7⁵ + 4⁵ + 4⁵ + 3⁵ = 5⁷ + 22 followed by 20⁵ + 17⁵ + 11⁵ + 4⁵ + 4⁵ = 9⁷ - 13 and then 25⁵ + 11⁵ + 8⁵ + 8⁵ + 6⁵ = 10⁷ - 12. The solution 61⁵ + 33⁵ + 25⁵ + 13⁵ + 5⁵ = 19⁷ - 2 gives e(5,5,7) £ 2.

{N.B. Further searches locate the solution 505⁵ + 315⁵ + 266⁵ + 212⁵ + 115⁵ = 87⁷, giving S(5,7) £ 5.}

For k = 5 and m = 6, a good early result is 8⁶ + 5⁶ + 3⁶ + 3⁶ + 3⁶ = 6⁷ + 20. However, the current best is 44⁶ + 38⁶ + 24⁶ + 10⁶ + 10⁶ = 27⁷ + 13.

For k = 5 and m = 7, the best result is 9⁷ + 9⁷ + 6⁷ + 5⁷ + 5⁷ = 10⁷ + 2124 is we ignore the unsatisfactory 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 3⁷ - 1547. The only other remotely close solution is 30⁷ + 26⁷ + 26⁷ + 23⁷ + 20⁷ = 33⁷ + 2822.

For k = 6 and m = 6, we have the impressive 24⁶ + 23⁶ + 19⁶ + 17⁶ + 5⁶ + 3⁶ = 17⁷ - 4.

For k = 6 and m = 7, we have 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ + 2⁷ = 3⁷ - 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 aⁿ = x + iy. Then |x|² + |y|² = zⁿ where z = p² + q². 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 278² + 29² = 5⁷, 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 non-trivial solutions for base 13 (= 2² + 3²), 17 (= 4² + 1²), 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 = 5ⁿ, we may use the recursive definition given by setting x₀ = 1, y₀ = 0, which makes sense since 1² + 0² = 5⁰, 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 = 2ⁿ + n2^{n- -1}i + .... + iⁿ 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.

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.

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

We use these additional results to augment Table 4.1, as follows.

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.

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