Divide all sides by $ n.$ We will have: \[ \frac{4\lfloor an\rfloor}{n}=1+\frac{\lfloor a\lfloor an\rfloor \rfloor}{n}\] \[ \lfloor an\rfloor = an - \{an\}\] \[ \lfloor a\lfloor an\rfloor \rfloor = a^2 n - a\{an\} - \{\lfloor a\lfloor an\rfloor \rfloor \}\] We know, that $ 0\leq\{x\}<1$ always. If we take $ n \to \infty$ we will get equation with $ a$ to solve it: \[ 4a=1+a^2\] The answer is $ a=2+\sqrt 3$ Am i right?

Yuriy Solovyov wrote: The answer is $ a = 2 + \sqrt 3$ Am i right? I don't think so $ \lfloor 2 + \sqrt {3} \rfloor =3$ Using $ n = 1$ equality doesn't hold.

SimonM wrote: Yuriy Solovyov wrote: The answer is $ a = 2 + \sqrt 3$ Am i right? I don't think so $ \lfloor 2 + \sqrt {3} \rfloor = 3$ Using $ n = 1$ equality doesn't hold. $ 2 + \sqrt 3 = 3.732... etc$ $ \lfloor 2 + \sqrt {3} \rfloor = 3$ If $ n = 1$ we get: $ 4\cdot 3 = 1 + \lfloor (2 + \sqrt 3)\cdot 3 \rfloor$ $ 5 = \lfloor 3\sqrt 3 \rfloor$ $ 3\sqrt 3 = 5.169...etc$ Last is correct.

a=2+sqrt(3),prove that 4[an]=n+[a[an]] .

To prove that 4[an]=n+[a[an]], where a^2+1=4a ,a=2+3^(1/2), It is enough to prove that 4[an]-n<=a[an]<4[an]-n+1 , To prove the left part : [an](4-a)<=an(4-a)=n(4a-a^2)=n. The right part:4[an]-a[an]=(4-a)[an]>(4-a)(na-1)=n(4a-a^2)-(4-a)=n-(4-a)>n-1<=>a>3 ,which is true because a=2+3^(1/2)>3 .

I came up with the following nice generalization. Perhaps further generalizations are possible. Theorem. Let $p,q \in \mathbb{N}$ be such that $1 \le q < p-1$. Then $a =\frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$ is the only real number $a$ such that $p \lfloor an \rfloor = qn + \lfloor a \lfloor an \rfloor \rfloor$ for all $n \in \mathbb{N}$. Proof. To prove this, first note that we always have $pan \ge p \lfloor an \rfloor > pan - p$. On the other hand, it's fairly clear that $a < p$, otherwise the right-hand side of our equation would always be bigger. This in turn implies that we also have bounds on the right-hand side: $qn + a^2n \ge qn + \lfloor a \lfloor an \rfloor \rfloor > qn + a\lfloor an \rfloor - 1 > qn + a(an - 1) - 1 > qn + a^2n - (p+1)$. We claim that $a$ must be such that $a^2 - pa + q = 0$. If not, then there is an $n$ such that $|a^2n - pan + qn| > p+1$. So then either $pan > qn + a^2n + p+1$, in which case $p \lfloor an \rfloor > pan - p > qn + a^2n + 1 > qn + \lfloor a \lfloor an \rfloor \rfloor$; contradiction. Or $pan < qn + a^2n - (p+1)$, in which case $p \lfloor an \rfloor \le pan < qn + a^2n - (p+1) < qn + \lfloor a \lfloor an \rfloor \rfloor$; contradiction. Conclusion: $a$ must be such that $a^2 - pa + q = 0$. The two solutions to that quadratic equation are $a_1 = \frac{1}{2} \left(p - \sqrt{p^2 - 4q}\right)$ and $a_2 = \frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$. Since $p > \sqrt{p^2 - 4q} > \sqrt{p^2 - 4p + 4} = p-2$, we get $0 < a_1 < 1$ and $p-1 < a_2 < p$. The first option can be ruled out by checking $n = 1$. From now on assume $a = a_2 = \frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$. We will prove that it works. Define $\epsilon_n$ to be $\epsilon_n = an - \lfloor an \rfloor = \{an\}$ and define $\delta_n$ to be $\delta_n = a \lfloor an \rfloor - \lfloor a \lfloor an \rfloor \rfloor = \{a \lfloor an \rfloor\}$. Then: \begin{align*} \delta_n &= \{a \lfloor an \rfloor\} \\ &= \{a (an - \epsilon_n)\} \\ &= \{a^2n - a\epsilon_n\} \\ &= \{(pa - q)n - a\epsilon_n\} \\ &= \{pan - qn - a\epsilon_n\} \\ &= \{p \lfloor an \rfloor + p\epsilon_n - qn - a\epsilon_n \} \\ &= \{(p - a)\epsilon_n\} \\ &= (p - a)\epsilon_n \end{align*} With this in mind we can finish the proof. \begin{align*} qn + \lfloor a \lfloor an \rfloor \rfloor &= qn + a \lfloor an \rfloor - \delta_n \\ &= qn + a (an - \epsilon_n) - \delta_n \\ &= qn + a^2n - a\epsilon_n - (p-a)\epsilon_n \\ &= qn + a^2n - p \epsilon_n \\ &= pan - p\epsilon_n \\ &= p \lfloor an \rfloor \end{align*}