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

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 ym - zm = 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 + 2n- 2)2 - (pn - 2n- 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 yi 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 yi and one with two negative yi. 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, Fn = Fn- 1 + Fn- 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.

 S(7,2) £ 4 S*(7,2) £ 3 S(6,3) £ 7 S*(6,3) £ 3 S(7,3) £ 7 S*(7,3) £ 4 S(6,4) £ 7 S*(6,4) £ 5 S(4,5) £ 4 S*(4,5) £ 3 S(7,5) £ 7 S*(7,5) £ 4 S(4,6) £ 4 S*(4,6) £ 3 S(3,7) £ 4 S*(3,7) £ 3
 e(2,5,2) £ 2 e*(2,5,2) £ 1 e(2,7,2) £ 7 e*(2,7,2) £ 1 e(2,4,3) £ 5 e*(2,4,3) £ 1 e(2,5,3) £ 7 e*(2,5,3) £ 4 e(2,7,3) £ 9 e*(2,7,3) £ 2 e(2,3,4) £ 3 e*(2,3,4) £ 1 e(2,7,4) £ 86 e*(2,7,4) £ 33 e(2,4,5) £ 65 e*(2,4,5) £ 1 e(2,4,6) £ 23 e*(2,4,6) £ 1 e(2,5,6) £ 53 e*(2,5,6) £ 52 e(2,3,7) £ 2 e*(2,3,7) £ 1 e(2,5,7) £ 139 e*(2,5,7) £ 83 e(3,5,3) £ 3 e*(3,5,3) £ 1 e(3,7,3) £ 8 e*(3,7,3) £ 1 e(3,6,5) £ 51 e*(3,6,5) £ 9 e(3,7,5) £ 113 e*(3,7,5) £ 1 e(3,5,6) £ 41 e*(3,5,6) £ 20 e(3,7,6) £ 345 e*(3,7,6) £ 150 e(4,5,4) £ 3 e*(4,5,4) £ 1 e(4,6,4) £ 10 e*(4,6,4) £ 1 e(4,7,4) £ 16 e*(4,7,4) £ 5 e(4,6,5) £ 13 e*(4,6,5) £ 8 e(4,7,6) £ 217 e*(4,7,6) £ 22 e(4,5,7) £ 75 e*(4,5,7) £ 1

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.

 n\m 2 3 4 5 6 7 2 2 2 3 3 3 3 3 2 3 3 4 3 4 4 2 3 3 5 5 8 5 2 3 3 4 7 4 6 2 3 3 6 7 9 7 2 3 4 5 8 8

Table 5.1