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

Question

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?

Answer 1

$\operatorname{AGM}(A,B) = \operatorname{AGM}(\sqrt{AB}, \frac{A+B}{2})$. If $A<B$, we have $A < \sqrt{AB} <\operatorname{AGM}(A, B) < \frac{A+B}{2} < B$.