Timeline for Representation of a number as a product of $\sqrt{n^2 + 1} + n$
Current License: CC BY-SA 4.0
6 events
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Sep 13, 2023 at 15:57 | comment | added | Pavel Gubkin | One can show that for all pairs $(x, y)$ one has $x^2 - y^2 = \pm 1$ (described operation comes from $SU(1,1)$ matrix multiplication.). Hence the answer is no greater than $4$. Your proof implies that the pairs $(x,y)$ and $(y,x)$ cannot be on the blackboard simultaneously (these two pairs give two factorizations of $x + y$) hence the answer is not greated then $2$ which is achieved for $x = 3$. The problem becomes easier if initially one has pairs $(\sqrt{n^2 + 1}, n)$. In this case for all pairs $(x,y)$ the inequality $x\ge y$ holds and the bound $2$ for the answer is straightforward. | |
Sep 13, 2023 at 15:53 | vote | accept | Pavel Gubkin | ||
Sep 13, 2023 at 15:47 | comment | added | Pavel Gubkin | Oh, I see, thanks! Here is the original problem I came up with and was trying to solve. Pairs of numbers $(n, \sqrt{n^2 + 1})$ for all $n\ge 1$ are written on the blackboard. If pairs $(a,b)$ and $(c,d)$ are written one can also write $(ac + bd, ad + bc)$. Show that after a finite number of operations there will be no more than two different pairs with some fixed $x\in \mathbb{R}$. | |
Sep 13, 2023 at 14:48 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
added 370 characters in body
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Sep 13, 2023 at 14:24 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
deleted 6 characters in body
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Sep 13, 2023 at 14:10 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |