# Do two pairs of distinct natural numbers exist such that AGM(A,B) equal to AGM(C,D)?

Here AGM is arithmetic-geometric mean.

Are there natural numbers A,B,C,D such that $$1\leq A 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?

• Isn't $A<AGM(A,B)<B?$ How could this be true? – saulspatz Jan 13 at 15:46
• 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. – user3482749 Jan 13 at 15:48
• Typo. Corrected – Stepan Jan 13 at 15:49