A Variant of Fermat’s Diophantine Equation

A variant of Fermat’s last Diophantine equation is proposed by adjusting the number of terms in accord with the power of terms and a theorem describing the solubility conditions is stated. Numerically obtained primitive solutions are presented for several cases with number of terms equal to or greater than powers. Further, geometric representations of solutions for the second and third power equations are devised by recasting the general equation in a form with rational solutions less than unity. Finally, it is suggested to consider negative and complex integers in seeking solutions to Diophantine forms in general.


Introduction
Diophantus (200-284), famed as the father of algebra, is known for his works on quadratic equations and puzzle-like algebraic problems such as finding numbers satisfying the condition that difference of the cubes of two numbers is equal to the sum of the cubes of two other numbers [1]. Fermat (1601-1665), quite familiar with the works of Diophantus, posed several Diophantine problems, among them the special case of the last theorem that the sum of two cubes cannot be a cube. Concerning the last theorem Fermat's note in the margin of his now lost copy of Diophantus's Arithmetica that "To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it" has unquestionably been the most firing remark giving rise to hopes for a relatively simple proof [1]. Thus, regardless of the complete rigorous proof of Wiles [2], works aiming at a simple proof of the theorem continue to be reported constantly as reviewed by Schorer [3]. There are several reasons to this unceasing interest; some obvious while others slightly hidden. First of all, the theorem has such charming qualities of being simple, elegant, and manageable appearance that anyone half interested in mathematics cannot but feel like being capable of toying with it to some good end. The fact that the complete proof had not come for centuries and when it did it came in hundreds of pages has not disheartened the initiated ones at all. A very recent outcome of such steadfast efforts is due to Nag [4] [5] who reported a neat and short proof of Fermat's last theorem in a manner quite befitting to the general character of Diophantine equations. Of course, the proof is yet to stand against probable objections.
On the other hand, this attractive and amusing challenge has not been unanimously praised. Gauss (1777-1855), the most distinguished opponent, replied to Olbers in 1816 [1] that "I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of." Likewise, Hilbert (1862-1943) was not keen on working Fermat's theorem as he explained why he was unwilling to do so [1]: "Before beginning I should put in three years of intensive study, and I haven't that much time to squander on a probable failure." The present work in some manner sides with these reserved views and rather than tackling with the original equation suggests first a generalized form and then presents numerically obtained solutions to this variant of Fermat's last Diophantine equation. Thus, the quest to prove insolubility is reversed to find the forms and conditions that provide solutions. Accordingly, cubes are partitioned into three cubes; fourth powers into five different fourth powers, etc. In some cases geometric representations of solutions are offered as well as some conjectures concerning solubility of the general form for definite powers and terms.

A Generalization of Fermat's Last Theorem
Fermat's well-known last theorem states that the Diophantine equation The above theorem together with the statement concerning 6 ≥ n is a con- . Accordingly, it is challenged to find rationals whose summation of the nth powers equals exactly to unity provided that ≥ m n . The most important advantage of Equation (3) is that it provides visual observations of the solutions for 2 = = n m and 3 = = n m as demonstrated in §3 and §4, respectively.

Second Power n = 2 with Two Terms m = 2
We begin with the case for which the solutions are possible; namely, 2 = = n m in (2) so that For this case the solutions can be written as = + z k l , which represent all the primitive integer solutions or Pythagorean integer triples. Here, k and l are relatively prime and . It is obvious that infinitely many solutions can be produced from each primitive solution by multiplying that particular solution by different integers. Table 1 gives ten different primitive triples for 2  Figure 1 is therefore symmetric about the line drawn at 45˚ to both axes.

Third Power n = 3 with Three Terms m = 3
in Equation (2) gives A simple FORTRAN program given in the Appendix for 3 = = n m is employed to seek integers satisfying Equation (5). The first ten primitive solutions obtained from a search covering integers in the range 1 -100 are listed in Table  2. Thus, while it is not possible to express the cube of a whole number as a summation of two cubes, it can be expressed as a summation of three or more cubes. The corresponding rational solutions satisfying 3

Fourth Power n = 4 with Four m = 4 and Five Terms m = 5
in Equation (2) Running the program for 4 = = n m yields no primitive integer solutions for the range 1 -150. The range could not be increased further due to restricted machine capability of operating large numbers. Nevertheless, from this particular and other computations we make a tentative inference that if no solution is found in the range 1 -100 it is unlikely to be any solution at all. Accordingly we now increase the number of terms to 1 for which the solutions are obtained by using the second program given in the Appendix. Table 3 lists ten primitive solutions of Equation (7) The search program for 5 = = n m gives no primitive integer solutions for the range 1 -50. Again, the range could not be increased more because of machine limits. On the other hand, increasing the number of terms to 6 = m and thus considering + + + + + = z z z z z z z (9) gives just two primitive solutions shown in Table 4 for the 50 integers that could

Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.