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Integer Solutions of the Equation x 2 + y 2 + z 2 +2 xyz = n
Author(s) -
Mordell L. J.
Publication year - 1957
Publication title -
journal of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.441
H-Index - 62
eISSN - 1469-7750
pISSN - 0024-6107
DOI - 10.1112/jlms/s1-32.3.383-s
Subject(s) - integer (computer science) , combinatorics , class (philosophy) , argument (complex analysis) , mathematics , physics , discrete mathematics , computer science , chemistry , biochemistry , artificial intelligence , programming language
In the preceding paper, Messrs. Schwartz and Muhly point out an oversight on page 507 of my paper “Integer solutions of the equation x 2 + y 2 + z 2 +2 xyz = n ”, Journal London Math. Soc. (28), 1953, 500‐510. It requires only a few lines to correct this, and I do so now. I state there that we can find out whether integer solutions of x 2 + y 2 + z 2 + axyz = b exist, and noted that the solutions may be arranged in three classes: (1) Those with xyz ⩽ 0; (2) Those with y 2 + z 2 ⩽ b 6, etc.; (3) The others and here we need only consider x > 0, y > 0, z > 0. Also that solutions in the first two classes are obvious on inspection. The remaining argument will be correct if we now add “we may suppose then that there are no solutions in the first two classes”. The argument on page 508 that when a = 2, there are no integer solutions when b = 4, 9, 49 except those in the first two classes typified by (±√ b ,0,0) needs only the slightest alteration. We may suppose that x, y, z is the minimum solution in the third class and that x ⩾ y ⩾ z > 0. Then either ayz − x ⩽ 0, i.e. y 2 + z 2 ⩽ b and the solution x, y, z is in the second class; or ayz − x > 0. Then by the minimum property ayz − x ⩾ x , and the argument at the end of page 507 and on page 508 is valid and shows again that the solution x, y, z cannot be in the third class.