The Diophantine Equation x² − 2y² = z² (Part IV)

A Method for Generating the Variables

Squares of Magic Squares Equation

Andrew Bremner's article on squares of squares included the 3x3 square:

The numbers in the right diagonal as the triple (205²,425²,565²) seem to have been derived elsewhere. This sequence as well as a host of others has been obtained from a number of web pages starting with PartIA-1 where the first number in the triple when added to a difference (Δ) gives the second square in the triple and when this same (Δ) is added to the second square produces a third square. In other words adding the two sums produces the Diophantine equation 2y² − x² = z². The method discovered therefore was useful in generating integers which could be used as the right diagonal entries of a magic square consisting primarily of squares. These integers, as well as a host of others, became entries of this particular Diophantine equation as was shown in Part I and Part II.

The Diophantine equation x² − 2y² = z² is well known and only a small number of values are available for this equation. This web page will show how to calculate a large number of triples which satisfy this equation using a simple novel method, completely different from the method used in Part I and Part II for the Diophantine equation 2y² − x² = z² but identical to the method used, with slight modifications, for the Diophantine equation x² + 2y² = z² covered in Part III. This section will construct the triples calculated without the attached imaginary symbol i to the first number in each triple as was done previously. In addition, the first and last numbers in the triples are reversed so as to correspond to the equation x² − 2y² = z².

Generation of Tables for odd and even numbers

The method uses tables of odd and even initial numbers starting first with Table I and Table II, where we use 1 for the first odd number and 2 for the first even number, then with Tables III to IX which covers other odd and even numbers. The triples are tabulated using the following titles for odd and even numbers:

where k is the increment starting at 0 and j equals the odd or even number chosen from the two sequences listed just below, a and c equals the equations just below them in the table above; and where b equals (c² − a²)/−2 in terms of k and where the δ₁ and δ₂ are the differences between the a and c columns, respectively. I must also mention that the square root of c² can take on positive or negative values according to the Diophantine equation. However, because of the equations in the header only one is allowed otherwise subtracting a from c will always give 0.

What are the initial numbers that may be used in the tables? By initial I mean the ns in the triple (n, 0, -n). The tables of initial even numbers are generated only from those numbers in the sequences generated from 2n², viz., 2,8,18,32,50,72,98,128,162,200... while the table of initial odd numbers are generated from (2n + 1)², viz., 1², 3², 5² 7², 9²... Only those numbers in these two sequences serve as the initial numbers as well as satisfy both Diophantine equations x² ± 2y² = z².

The next two tables III and IV, are shown below and use even number 8 and odd number 9 from the two sequences above where the light green triples of Table III are divisible by 4 and the light blue of Table IV by 9. In addition, the factored triples of III and IV are contained in previous tables.

Tables V, VI and VII tabulate the results of integers 18, 25 and 32, respectively.

Tables VIII and IX tabulate the results of integers 49 and 50, respectively.

Only nine integers were used in the calculation for the Diophantine equation x² − 2y² = z² to get an idea of its versatility. Note that the tables produced can be expanded downward indefinitely and that the entries in the two sequences, used as the initial numbers, are infinite in number, thereby, giving rise to an infinite number of triples for this Diophantine equation. In addition, one can determine the triple value of a particular odd/even number, chosen from any of the two sequences above, by employing any arbitrary k value and performing the calculation for a and c then calculating for b.

This concludes Part IV. Go back to Part III.

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