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Here AGM is arithmetic-geometric mean.

Are there natural numbers A,B,C,D such that $1\leq A<C<D<B$ and arithmetic-geometric mean AGM(A,B)=AGM(C,D) ?

In other words, is AGM a homomorphism of an unordered pair of natural numbers on a set of real numbers?

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  • $\begingroup$ Isn't $A<AGM(A,B)<B?$ How could this be true? $\endgroup$ – saulspatz Jan 13 at 15:46
  • $\begingroup$ The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) \leq B < C \leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that. $\endgroup$ – user3482749 Jan 13 at 15:48
  • $\begingroup$ Typo. Corrected $\endgroup$ – Stepan Jan 13 at 15:49

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