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Statement
Setting (pp. 237--238). The equation (1) is in positive integers. A solution is trivial when or . The index of a solution is . Mills's Theorem 1, recalled on p. 237, gives no non-trivial solution with , that is , and his Theorem 2 gives exactly Ko's family (2) with , that is . For the remaining non-trivial solutions, those with , the paper assumes by symmetry (its (3)), so that is a rational number with .
Theorem 4 (p. 238). If with rational and , then the equation has no non-trivial solutions of index .
Equivalently, with in lowest terms, no non-trivial solution has an index for which is a perfect square (p. 241).
Proof pointer
§ 4, p. 241, in the notation of § 1:
Notation of § 1 (pp. 238--239). For a non-trivial solution put , , , with , so that , and , a positive odd integer with . With , , , one has , with an integer , and ; with , , one has (from Schinzel) and in lowest terms with .
With one has and ; this quadratic form splits into integer linear factors exactly when is a square, that is, when is rational. The case is excluded directly. For or the paper writes with positive integers ; both factors are positive and of the same parity, so the first is at least 2, which forces and contradicts .
Read depth
Claims checked: the statement and the proof on p. 241 were read clause by clause on the page images of the print. The facts of § 1 taken from Mills and Schinzel are cited, not proved, in the paper and were not read. Nothing here is independently reviewed.
Dependencies
None in the corpus. External inputs named by the paper: Mills's 1959 report and Schinzel (1958) for the facts of § 1.
Source. S. Uchiyama, On the Diophantine equation , Trudy Mat. Inst. Steklov. 163 (1984), 237--243; the edition read is named on the source card.
Bears on
- Problem 674: the theorem does not touch the problem's question, which the family (2) that the paper recalls from Ko already answers. It excludes non-trivial solutions with for every index of the form with rational, .