Chapter 5

Extended Forms

As with the case for like powers, it is appropriate to consider the following extension of equation 3.2

(5.1)

For the purposes of this chapter, we will say that a positive y is one with a "+" (plus) sign preceding it and a negative y is one with a "- " (minus) sign preceding it in any solution of the above equation, although in fact all of the y_i are positive.

With the increased flexibility provided, there are many more ways of constructing solutions of equation 5.1 for a particular k value by manipulating a solution with smaller k. Given this, and other irregularities, in order to preserve some sort of consistency we shall enforce several conditions and also extend the concept of a trivial solution.

Since x is always positive (otherwise we simply reverse the signs of all the y_i), we may assume that at least one of the y_i is also positive. We shall let y₁ be the largest positive y, and if there are j positive y_i, then they are y₁ to y_j such that y₁ ³ y₂ ³ .... ³ y_j, and the negative y_i are y_j+1 to y_k such that y_j+1 ³ y_j+2 ³ .... ³ y_k. We continue to enforce the condition that y₁ ¹ x, but we do not require the same of y_j+1 since the signs differ. Without loss, we shall also assume that there is no common value between the positive and negative y_i.

Again, any solution that can be constructed from a solution with smaller k by inserting small powers and which makes d larger is trivial, but one which makes d smaller is acceptable as long as it does not violate any other condition. Also, any solution with x = 1 is considered trivial, since it is again obviously based on a previous solution for smaller k.

We extend the definitions also, and let e*(k,m,n) be the smallest absolute value of d for which a non-trivial solution of equation 5.1 exists, and let S*(m,n) be the smallest value of k for which e*(k,m,n) = 0. Even with the extra conditions applicable, we would expect, with the substantial overall increase in possible combinations, to see some tightening of the positive only results, and this is borne out in practice, as we shall show.

Since x is positive, the k = 1 case is identical to that for equation 4.2 and so is ignored. Also, since the case m = n is simply an alternative way of viewing the Extended Euler Conjecture, it too shall be ignored.

In the following, we shall only consider cases where results are at least as good as, if not better than, the positive only results. Search ranges are based on y₁, the largest positive y value. However, the largest possible negative y value may well exceed this depending on the circumstances, and as will be seen from some of the solutions given below.

The method of finding solutions in any of the following is by brute force. The only restrictions are based on the assumptions given above.

For k = 2 :

The only equation to be investigated is y^m - z^m = xⁿ. The minimum search limit on y₁ is 500.

For m = 3 and n = 2, there are 42 solutions for d = 0 with y₁ up to 10000, the first four of which are 8³ - 7³ = 13² ; 71³ - 23³ = 588² ; 74³ - 47³ = 549² ; 105³ - 104³ = 181² and the largest being 9632³ - 2743³ = 934331².

For m = 4 and n = 2, we find, as with the standard case, many solutions with d = ± 1. For y₁ up to 10000 there are 100 solutions, of which only 12 are for d = 1. The smallest and largest solutions this range are 3⁴ - 2⁴ = 8² + 1 and 6867⁴ - 4538⁴ = 42421328² + 1, 12⁴ - 8⁴ = 129² - 1 and 9961⁴ - 2873⁴ = 98877599² - 1.

For m = 5 and n = 2, we find 70⁵ - 56⁵ = 33615² - 1, giving the improved e*(2,5,2) £ 1.

For m = 6 and n = 2, we find 16⁶ - 14⁶ = 3041² - 1, 78⁶ - 23⁶ = 474396² - 1 and 242⁶ - 122⁶ = 14055679² - 1.

For m = 7 and n = 2, we find 64⁷ - 63⁷ = 677544² + 1, giving the improved e*(2,7,2) £ 1.

For m = 2 and n = 3, we know that every odd number can be represented as the difference of two consecutive squares, and so we have (2p + 1)³ = 2q + 1 = (q + 1)² - q² where q = 4p³ + 6p² + 3p. Also, since (2p)³ = 4.2p³ = 4.(q + 1) = (q + 2)² - q² where q = 2p³ - 1 we can also express every even third power as a difference of squares. Also, since (2p)³ = 8p³ = 8.(q + 2) = (q + 4)² - q² where q = p³ - 2, this provides an alternative way of expressing even third powers as a difference of squares. In this latter case, however, if p is even, then the solution is trivial as its greatest common divisor is not 1. Solutions that do not fall into any of the above categories include 43² - 11² = 12³, 76² - 49² = 15³ and 185² - 158² = 21³.

For m = 4 and n = 3, we have 3⁴ - 2⁴ = 4³ + 1, giving e*(2,4,3) £ 1.

For m = 5 and n = 3, we have 2⁵ - 1⁵ = 3³ + 4. The next best is 3⁵ - 2⁵ = 6³ - 5.

For m = 6 and n = 3, the best solution found is 4⁶ - 3⁶ = 15³ - 8, which is not as good as for the standard equation.

For m = 7 and n = 3, we have 2⁷ - 1⁷ = 5³ + 2, based on the standard k = 1 solution.

For m = 2 and n = 4, as with the n = 3 case, there are several similar parametric solutions for d = 0, e.g. (q + 1)² - q² = p⁴ where p odd and q = (p⁴ - 1) / 2, or (q + 2)² - q² = (2p)⁴ where q = 4p⁴ - 1. Solutions outwith these include 85² - 77² = 6⁴, 145² - 17² = 12⁴ and 353² - 272² = 15⁴.

For m = 3 and n = 4, we have 21³ - 19³ = 7⁴ + 1 and 37³ - 3³ = 15⁴ + 1, giving the improvement e*(2,3,4) £ 1.

For m = 5 and 6, and n = 4, there are no improvements.

For m = 7 and n = 4, we have 6⁷ - 2⁷ = 23⁴ - 33.

For m = 2 and n = 5, there are two obvious parametric solutions for d = 0 analogous to those for n = 4. The first solution not conforming to either of these is 251² - 235² = 6⁵. However, this gives a clue to another parameteric solution for all powers of even numbers, which becomes (pⁿ + 2^{n- 2})² - (pⁿ - 2^{n- 2})² = (2p)ⁿ. For n = 3, this coincides with the third of the parametric solutions for that power.

For m = 3 and n = 5, we can only match the standard result with 37³ - 35³ = 6⁵ + 2. However, this suggests a parameterisable solution that is true for all odd n, i.e.

and so, for instance, with n = 5 and p = 1 we get the above result. With n = 7 and p = 1 we get the equation 217³ - 215³ = 6⁷ + 2.

For m = 4 and n = 5, we have 37⁴ - 34⁴ = 14⁵ + 1, which is a substantial improvement over the standard result, and which also implies that S*(4,5) £ 3 if we introduce - 1⁴ to both sides.

For m = 6 and 7, and n = 5, there are no improvements.

For m = 2 and n = 6, the usual parametric solutions for d = 0 apply. However, these become very large very quickly, and so fall behind the many solutions for d = ± 1, some of which are no doubt also parameterisable.

For m = 3 and n = 6, we can match the standard result by looking at solutions of the standard equation for m = n = 3 to see if any one of the y_i is a square, e.g. we can convert 10³ + 9³ = 12³ + 1 into 12³ - 10³ = 3⁶ - 1.

For m = 4 and n = 6, we have the valid solution 3⁴ - 2⁴ = 2⁶ + 1.

For m = 5 and n = 6, the equation 4⁵ - 3⁵ = 3⁶ + 52 beats the standard current best by 1.

For m = 2 and n = 7, the usual applies.

For m = 3 and n = 7, we have 7³ - 6³ = 2⁷ - 1, an improvement over the standard.

For m = 5 and n = 7, we have 3⁵ - 2⁵ = 2⁷ + 83 and 5⁵ - 4⁵ = 3⁷ - 86, both of which are improvements over the standard.

For k = 3 :

There are two new equations to investigate, being one with a single negative y_i and one with two negative y_i. The minimum search limit on y₁ is 100.

For m = 5 and n = 2, we have, as expected, solutions with d = 0, e.g. 15⁵ + 5⁵ - 14⁵ = 474² and 16⁵ + 3⁵ - 15⁵ = 538². A larger example is 58⁵ + 21⁵ - 33⁵ = 24926².

For m = 6 and n = 2, there are lots of solutions with d = 0 and one negative, the smallest being 4⁶ + 3⁶ - 2⁶ = 69². However, there is only one known solution with two negatives, namely 57⁶ - 44⁶ - 28⁶ = 162967².

For m = 7 and n = 2, the solutions 16⁷ - 9⁷ - 7⁷ = 16212² and 64⁷ - 63⁷ - 1⁷ = 677544² give S*(7,2) £ 3. The latter is an extension of the k = 2 result above.

For m = 4 and n = 3, we have, as expected, solutions with d = 0, e.g. 3⁴ - 2⁴ - 1⁴ = 4³ and 9⁴ - 8⁴ - 7⁴ = 4³. A larger example is 31⁴ + 20⁴ - 17⁴ = 100³.

For m = 5 and n = 3, we have 92⁵ + 33⁵ - 74⁵ = 1640³ + 1 and 69⁵ - 67⁵ - 9⁵ = 598³ + 1.

For m = 6 and n = 3, we have 19⁶ - 18⁶ - 15⁶ = 118³ and so S*(6,3) £ 3, which is a substantial improvement over the standard case where we only have S(6,3) £ 7.

For m = 7 and n = 3, we have 2⁷ - 1⁷ - 1⁷ = 5³ + 1, based on the k = 2 current best.

For m = 3 and n = 4, there is a multitude of solutions for d = 0, e.g. 15³ + 9³ - 2³ = 8⁴, 17³ + 10³ - 18³ = 3⁴, 11³ - 3³ - 2³ = 6⁴, 21³ - 19³ - 1³ = 7⁴, 24³ + 17³ - 16³ = 11⁴.

For m = 5 and n = 4, we have 11⁵ - 4⁵ - 2⁵ = 20⁴ - 5 and 16⁵ + 13⁵ - 17⁵ = 2⁴ - 4 (which is a twisted version of the standard case solution for m = n = 5), but no improvement over the standard result.

For m = 6 and 7, and n = 4, there are no improvements.

For m = 3 and n = 5, we have the d = 0 solutions 17³ + 15³ - 8³ = 6⁵, 41³ - 33³ - 6³ = 8⁵ and 48³ - 21³ - 11³ = 10⁵.

For m = 4 and n = 5, we have 7⁴ - 6⁴ - 3⁴ = 4⁵ and 37⁴ - 34⁴ - 1⁴ = 14⁵, the latter from the k = 2 solution, giving S*(4,5) £ 3.

For m = 6 and n = 5, we have 31⁶ - 21⁶ - 17⁶ = 60⁵ - 9, a substantial improvement.

For m = 7 and n = 5, we have 19⁷ + 7⁷ - 18⁷ = 49⁵ + 1, so that e*(3,7,5) £ 1, and which clearly implies that S*(7,5) £ 4.

For m = 3 and n = 6, there are lots of solutions with d = 0, e.g. 12³ + 1³ - 10³ = 3⁶, 15³ + 9³ - 2³ = 4⁶, 6³ - 5³ - 3³ = 2⁶ and 25³ - 22³ - 17³ = 2⁶.

For m = 4 and n = 6, we have 31⁴ + 20⁴ - 17⁴ = 10⁶, and a number of d = 0 solutions with two negatives, the largest of which is 89⁴ - 73⁴ - 24⁴ = 18⁶.

For m = 5 and n = 6, we have 5⁵ + 4⁵ - 2⁵ = 4⁶ + 21 and 4⁵ - 3⁵ - 2⁵ = 3⁶ + 20.

For m = 7 and n = 6, the current best is 3⁷ + 3⁷ - 2⁷ = 4⁶ + 150.

For m = 3 and n = 7, we have 22³ + 22³ - 17³ = 4⁷ - 1. However, the double negative solutions 75³ - 52³ - 11³ = 6⁷ and 85³ - 77³ - 54³ = 2⁷ give S*(3,7) £ 3.

For m = 4, 5 and 6, and n = 7, there are no improvements.

For k = 4 :

There are three different equations to investigate, with one, two and three negatives respectively. The minimum search limit on y₁ is 35 in all cases.

For m = 4 and n = 3, there are many solutions with d = 0.

For m = 5 and n = 3, we know that d = 0 solutions exist, e.g. adding - 1⁵ to each side of the two k = 3 results above, but other solutions are found with smaller y values, namely 22⁵ + 19⁵ + 13⁵ - 4⁵ = 200³ and 25⁵ + 25⁵ + 13⁵ - 2⁵ = 271³.

For m = 6 and n = 3, there are many solutions with d = ± 1, including a parameterised solution using Fibonacci numbers, but none has been found with d = 0, which is unusual considering that a d = 0 solution exists for k = 3.

Let F₀ = 0, F₁ = 1, and, for n > 1, F_n = F_{n- 1} + F_{n- 2}. These are the Fibonacci Numbers. The parameterised solution mentioned above is

For m = 7 and n = 3, a near miss is given by 29⁷ + 23⁷ + 11⁷ - 15⁷ = 2737³ - 1. However, we have S*(7,3) £ 4 since we have 2⁷ - 1⁷ - 1⁷ - 1⁷ = 5³ (from the k = 3 solution).

For m = 5 and n = 4, we have 9⁵ + 5⁵ - 7⁵ - 7⁵ = 13⁴ - 1, 16⁵ + 13⁵ + 3⁵ - 17⁵ = 4⁴ - 1 and 34⁵ + 26⁵ + 20⁵ - 36⁵ = 15⁴ - 1.

For m = 6 and n = 4, we have 6⁶ + 4⁶ - 2⁶ - 2⁶ = 15⁴ - 1.

For m = 7 and n = 4, we find the improving results 8⁷ + 3⁷ + 3⁷ - 4⁷ = 38⁴ + 6 and 29⁷ + 10⁷ + 6⁷ - 25⁷ = 325⁴ - 5.

For m = 6 and n = 5, we have 31⁶ + 1⁶ - 21⁶ - 17⁶ = 60⁵ - 8.

For m = 7 and n = 5, as mentioned earlier, we have S*(7,5) £ 4 since we can obtain the solution 19⁷ + 7⁷ + 18⁷ - 1⁷ = 49⁵ from the k = 3 result.

For m = 5 and n = 6, there is no improvement over the standard equation.

For m = 7 and n = 6, we have the substantial improvement 3⁷ + 3⁷ - 2⁷ - 2⁷ = 4⁶ + 22.

For m = 4 and n = 7, we can match the standard result with 17⁴ - 14⁴ - 14⁴ - 9⁴ = 2⁷.

For m = 5 and n = 7, a great improvement is given by 28⁵ + 26⁵ + 11⁵ - 25⁵ = 11⁷ - 1.

For m = 6 and n = 7, there is no improvement.

To summarise the extended results, we identify every improvement over the standard.

Unimproved error bounds exist for :

e(2,6,3) ; e(2,5,4) ; e(2,6,4) ; e(2,3,5) ; e(2,6,5) ; e(2,7,6) ; e(2,4,7) ; e(2,6,7)

e(3,5,4) ; e(3,6,4) ; e(3,7,4) ; e(3,4,7) ; e(3,5,7) ; e(3,6,7)

e(4,5,6) ; e(4,6,7)

Finally, the following table summarises S*(m,n) analogously to Table 4.2.

Table 5.1

URL : www.glasgowg43.freeserve.co.uk/chapter5.htm