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 y_{1} be the largest positive y, and if there are j positive y_{i}, then they are y_{1} to y_{j} such that y_{1} ³ y_{2} 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 y_{1} 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 y_{1}, 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} = x^{n}. The minimum search limit on y_{1} is 500.

For m = 3 and n = 2, there are 42 solutions for d = 0 with y_{1} up to 10000, the first four of which are 8^{3} - 7^{3} = 13^{2} ; 71^{3} - 23^{3} = 588^{2} ; 74^{3} - 47^{3} = 549^{2} ; 105^{3} - 104^{3} = 181^{2} and the largest being 9632^{3} - 2743^{3} = 934331^{2}.

For m = 4 and n = 2, we find, as with the standard case, many solutions with d = ± 1. For y_{1} up to 10000 there are 100 solutions, of which only 12 are for d = 1. The smallest and largest solutions this range are 3^{4} - 2^{4} = 8^{2} + 1 and 6867^{4} - 4538^{4} = 42421328^{2} + 1, 12^{4} - 8^{4} = 129^{2} - 1 and 9961^{4} - 2873^{4} = 98877599^{2} - 1.

For m = 5 and n = 2, we find 70^{5} - 56^{5} = 33615^{2} - 1, giving the improved e*(2,5,2) £ 1.

For m = 6 and n = 2, we find 16^{6} - 14^{6} = 3041^{2} - 1, 78^{6} - 23^{6} = 474396^{2} - 1 and 242^{6} - 122^{6} = 14055679^{2} - 1.

For m = 7 and n = 2, we find 64^{7} - 63^{7} = 677544^{2} + 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} - q^{2} where q = 4p^{3} + 6p^{2} + 3p. Also, since (2p)^{3} = 4.2p^{3} = 4.(q + 1) = (q + 2)^{2} - q^{2} where q = 2p^{3} - 1 we can also express every even third power as a difference of squares. Also, since (2p)^{3} = 8p^{3} = 8.(q + 2) = (q + 4)^{2} - q^{2} where q = p^{3} - 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 43^{2} - 11^{2} = 12^{3}, 76^{2} - 49^{2} = 15^{3} and 185^{2} - 158^{2} = 21^{3}.

For m = 4 and n = 3, we have 3^{4} - 2^{4} = 4^{3} + 1, giving e*(2,4,3) £ 1.

For m = 5 and n = 3, we have 2^{5} - 1^{5} = 3^{3} + 4. The next best is 3^{5} - 2^{5} = 6^{3} - 5.

For m = 6 and n = 3, the best solution found is 4^{6} - 3^{6} = 15^{3} - 8, which is not as good as for the standard equation.

For m = 7 and n = 3, we have 2^{7} - 1^{7} = 5^{3} + 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} - q^{2} = p^{4} where p odd and q = (p^{4} - 1) / 2, or (q + 2)^{2} - q^{2} = (2p)^{4} where q = 4p^{4} - 1. Solutions outwith these include 85^{2} - 77^{2} = 6^{4}, 145^{2} - 17^{2} = 12^{4} and 353^{2} - 272^{2} = 15^{4}.

For m = 3 and n = 4, we have 21^{3} - 19^{3} = 7^{4} + 1 and 37^{3} - 3^{3} = 15^{4} + 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 6^{7} - 2^{7} = 23^{4} - 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 251^{2} - 235^{2} = 6^{5}. However, this gives a clue to another parameteric solution for all powers of even numbers, which becomes (p^{n} + 2^{n- 2})^{2} - (p^{n} - 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 37^{3} - 35^{3} = 6^{5} + 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 217^{3} - 215^{3} = 6^{7} + 2.

For m = 4 and n = 5, we have 37^{4} - 34^{4} = 14^{5} + 1, which is a substantial improvement over the standard result, and which also implies that S*(4,5) £ 3 if we introduce - 1^{4} 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 10^{3} + 9^{3} = 12^{3} + 1 into 12^{3} - 10^{3} = 3^{6} - 1.

For m = 4 and n = 6, we have the valid solution 3^{4} - 2^{4} = 2^{6} + 1.

For m = 5 and n = 6, the equation 4^{5} - 3^{5} = 3^{6} + 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 7^{3} - 6^{3} = 2^{7} - 1, an improvement over the standard.

For m = 5 and n = 7, we have 3^{5} - 2^{5} = 2^{7} + 83 and 5^{5} - 4^{5} = 3^{7} - 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 y_{1} is 100.

For m = 5 and n = 2, we have, as expected, solutions with d = 0, e.g. 15^{5} + 5^{5} - 14^{5} = 474^{2} and 16^{5} + 3^{5} - 15^{5} = 538^{2}. A larger example is 58^{5} + 21^{5} - 33^{5} = 24926^{2}.

For m = 6 and n = 2, there are lots of solutions with d = 0 and one negative, the smallest being 4^{6} + 3^{6} - 2^{6} = 69^{2}. However, there is only one known solution with two negatives, namely 57^{6} - 44^{6} - 28^{6} = 162967^{2}.

For m = 7 and n = 2, the solutions 16^{7} - 9^{7} - 7^{7} = 16212^{2} and 64^{7} - 63^{7} - 1^{7} = 677544^{2} 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. 3^{4} - 2^{4} - 1^{4} = 4^{3} and 9^{4} - 8^{4} - 7^{4} = 4^{3}. A larger example is 31^{4} + 20^{4} - 17^{4} = 100^{3}.

For m = 5 and n = 3, we have 92^{5} + 33^{5} - 74^{5} = 1640^{3} + 1 and 69^{5} - 67^{5} - 9^{5} = 598^{3} + 1.

For m = 6 and n = 3, we have 19^{6} - 18^{6} - 15^{6} = 118^{3} 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 2^{7} - 1^{7} - 1^{7} = 5^{3} + 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. 15^{3} + 9^{3} - 2^{3} = 8^{4}, 17^{3} + 10^{3} - 18^{3} = 3^{4}, 11^{3} - 3^{3} - 2^{3} = 6^{4}, 21^{3} - 19^{3} - 1^{3} = 7^{4}, 24^{3} + 17^{3} - 16^{3} = 11^{4}.

For m = 5 and n = 4, we have 11^{5} - 4^{5} - 2^{5} = 20^{4} - 5 and 16^{5} + 13^{5} - 17^{5} = 2^{4} - 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 17^{3} + 15^{3} - 8^{3} = 6^{5}, 41^{3} - 33^{3} - 6^{3} = 8^{5} and 48^{3} - 21^{3} - 11^{3} = 10^{5}.

For m = 4 and n = 5, we have 7^{4} - 6^{4} - 3^{4} = 4^{5} and 37^{4} - 34^{4} - 1^{4} = 14^{5}, the latter from the k = 2 solution, giving S*(4,5) £ 3.

For m = 6 and n = 5, we have 31^{6} - 21^{6} - 17^{6} = 60^{5} - 9, a substantial improvement.

For m = 7 and n = 5, we have 19^{7} + 7^{7} - 18^{7} = 49^{5} + 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. 12^{3} + 1^{3} - 10^{3} = 3^{6}, 15^{3} + 9^{3} - 2^{3} = 4^{6}, 6^{3} - 5^{3} - 3^{3} = 2^{6} and 25^{3} - 22^{3} - 17^{3} = 2^{6}.

For m = 4 and n = 6, we have 31^{4} + 20^{4} - 17^{4} = 10^{6}, and a number of d = 0 solutions with two negatives, the largest of which is 89^{4} - 73^{4} - 24^{4} = 18^{6}.

For m = 5 and n = 6, we have 5^{5} + 4^{5} - 2^{5} = 4^{6} + 21 and 4^{5} - 3^{5} - 2^{5} = 3^{6} + 20.

For m = 7 and n = 6, the current best is 3^{7} + 3^{7} - 2^{7} = 4^{6} + 150.

For m = 3 and n = 7, we have 22^{3} + 22^{3} - 17^{3} = 4^{7} - 1. However, the double negative solutions 75^{3} - 52^{3} - 11^{3} = 6^{7} and 85^{3} - 77^{3} - 54^{3} = 2^{7} 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 y_{1} 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 - 1^{5} to each side of the two k = 3 results above, but other solutions are found with smaller y values, namely 22^{5} + 19^{5} + 13^{5} - 4^{5} = 200^{3} and 25^{5} + 25^{5} + 13^{5} - 2^{5} = 271^{3}.

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 F_{0} = 0, F_{1} = 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 29^{7} + 23^{7} + 11^{7} - 15^{7} = 2737^{3} - 1. However, we have S*(7,3) £ 4 since we have 2^{7} - 1^{7} - 1^{7} - 1^{7} = 5^{3} (from the k = 3 solution).

For m = 5 and n = 4, we have 9^{5} + 5^{5} - 7^{5} - 7^{5} = 13^{4} - 1, 16^{5} + 13^{5} + 3^{5} - 17^{5} = 4^{4} - 1 and 34^{5} + 26^{5} + 20^{5} - 36^{5} = 15^{4} - 1.

For m = 6 and n = 4, we have 6^{6} + 4^{6} - 2^{6} - 2^{6} = 15^{4} - 1.

For m = 7 and n = 4, we find the improving results 8^{7} + 3^{7} + 3^{7} - 4^{7} = 38^{4} + 6 and 29^{7} + 10^{7} + 6^{7} - 25^{7} = 325^{4} - 5.

For m = 6 and n = 5, we have 31^{6} + 1^{6} - 21^{6} - 17^{6} = 60^{5} - 8.

For m = 7 and n = 5, as mentioned earlier, we have S*(7,5) £ 4 since we can obtain the solution 19^{7} + 7^{7} + 18^{7} - 1^{7} = 49^{5} 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 3^{7} + 3^{7} - 2^{7} - 2^{7} = 4^{6} + 22.

For m = 4 and n = 7, we can match the standard result with 17^{4} - 14^{4} - 14^{4} - 9^{4} = 2^{7}.

For m = 5 and n = 7, a great improvement is given by 28^{5} + 26^{5} + 11^{5} - 25^{5} = 11^{7} - 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