Peter wrote: Determine all pairs $ (a, b)$ of real numbers such that $ a\lfloor bn\rfloor = b\lfloor an\rfloor$ for all positive integer $ n$. http://www.kalva.demon.co.uk/short/soln/sh98n2.html

The above link doesn´t work anymore.. It should be: http://web.archive.org/web/20040821052526/http://www.kalva.demon.co.uk/short/soln/sh98n2.html My proof (note that I denote real numbers by Greek letters): Since it is clear that integral values of $\alpha$ and $\beta$ work and that it is impossible for exactly 1 of the numbers to be integral, I'll asumme $\alpha$ and $\beta$ are non-integral (and by implication non-zero) and show that they must be equal. So, assume $\alpha \left\lfloor \beta n \right\rfloor = \beta \left\lfloor \alpha n \right\rfloor$ for all positive integers n and some non-integral real numbers $\alpha$, $\beta$. Let $\alpha = a + \gamma$ and $\beta = b + \delta$ with $0 < \gamma, \delta < 1$ and $a, b \in Z$. We now have: $\alpha \left\lfloor \beta n \right\rfloor = \beta \left\lfloor \alpha n \right\rfloor \\ (a + \gamma) \left\lfloor (b + \delta) n \right\rfloor = (b + \delta) \left\lfloor (a + \gamma) n \right\rfloor \\ (a + \gamma)(bn + \left\lfloor \delta n \right\rfloor) = (b + \delta)(an + \left\lfloor \gamma n \right\rfloor) \\ (a + \gamma) \left\lfloor \delta n \right \rfloor + \gamma bn = (b + \delta) \left \lfloor \gamma n \right \rfloor + \delta an \\ \alpha \left\lfloor \delta n \right \rfloor + \gamma bn \&= \beta \left \lfloor \gamma n \right \rfloor + \delta an$ (1) If $n = 1$, this simplifies to: $\gamma b = \delta a$ Which implies that (1) simplifies to: $\alpha \left\lfloor \delta n \right \rfloor = \beta \left \lfloor \gamma n \right \rfloor$ (2) Now, let $n$ be the smallest positive integer such that $\max(\gamma n, \delta n) \ge 1$. Then $\min(\gamma n, \delta n) \ge 1$ as well, because otherwise we would have $0$ on one side of the equation while the other side is non-zero. For this $n$, (2) simplifies to: $\alpha = \beta$ So the only real numbers $\alpha$, $\beta$ for which $\alpha \left\lfloor \beta n \right\rfloor$ equals $\beta \left\lfloor \alpha n \right\rfloor$ for all $n \in \aleph$ (and is not always $0$), are those real numbers which are either the same or both integral.

Another, rather cute, proof I found today: Assume $\alpha$ and $\beta$ are distinct non-integral real numbers and assume $\alpha [\beta n] = \beta [\alpha n]$ for all $n \in \aleph$. Then, after dividing by $\beta[\beta n]$ (which is allowed for large enough $n$) we get that $\dfrac{\alpha}{\beta} = \dfrac{[\alpha n]}{[\beta n]}$ is rational, say $\dfrac{p}{q}$. Since it is clear that $\alpha = -\beta$ is not a solution, we may assume $q \geq 2$. Assume, wlog, furthermore that $p$ and $q$ are coprime. Substituting $\dfrac{\beta p}{q}$ for $\alpha$, dividing by $\beta$ and multiplying by $q$ and we obtain: $p[\beta n] = q[\beta n \dfrac{p}{q}]$ Since the RHS is divisible by $q$, LHS must be a multiple of $q$ too. And since $\gcd(p, q) = 1$, we have, in the particular case that $n = q$, that $q$ divides $[\beta q]$. Now, put $\beta = b + \delta$ with $b$ integral and $0 < \delta < 1$. Then we have: $\delta \in (0, \dfrac{1}{q})$. Assume otherwise (thus, assume $\delta \in (\dfrac{1}{q}, 1)$): $[\beta q] = [(b+\delta) q] = [bq + \delta q] = bq + [\delta q] \geq bq + [\dfrac{1}{q} q] = bq + 1$ $[\beta q] = [(b+\delta) q] = [bq + \delta q] = bq + [\delta q] < bq + [q] = bq + q$ And this is clearly not possible if $q$ has to be a divisor of $[\beta q]$. Now, assume that we have showed $\delta \in (0, \dfrac{1}{mq})$ for some integral $m \geq 1$. Either $\delta \in (0, \dfrac{1}{(m+1)q})$ or $\delta \in (\dfrac{1}{(m+1)q}, \dfrac{1}{mq})$. Assume the latter. Let $n = q(m+1)$. We then have: $[\beta q(m+1)] = [(b+\delta)q(m+1)] = [bq(m+1) + \delta q(m+1)] = bq(m+1) + [\delta q(m+1)] = bq(m+1) + 1$ And since this is not a multiple of $q$, $\delta \in (\dfrac{1}{(m+1)q}, \dfrac{1}{mq})$ cannot occur. By induction we now have proved: $\delta \in (0, \dfrac{1}{mq})$ for all $m \in \aleph$. This is nonsense. So the assumption that there are distinct, non-integral reals $\alpha$ and $\beta$ for which $\alpha[\beta n] = \beta[\alpha n]$ holds for all positive integers $n$ must be wrong.

Still another variation (where I quote the first part of the last post and then take another road): Assume $\alpha$ and $\beta$ are distinct non-integral real numbers and assume $\alpha [\beta n] = \beta [\alpha n]$ for all $n \in \aleph$. Then, after dividing by $\beta[\beta n]$ (which is allowed for large enough $n$) we get that $\dfrac{\alpha}{\beta} = \dfrac{[\alpha n]}{[\beta n]}$ is rational, say $\dfrac{p}{q}$. Since it is clear that $\alpha = -\beta$ is not a solution, we may assume $q \geq 2$. Assume, wlog, furthermore that $p$ and $q$ are coprime. Substituting $\dfrac{\beta p}{q}$ for $\alpha$, dividing by $\beta$ and multiplying by $q$ and we obtain: $p[\beta n] = q[\beta n \dfrac{p}{q}]$ Since the RHS is divisible by $q$, LHS must be a multiple of $q$ too. And since $\gcd(p, q) = 1$, we have, in the particular case that $n = 1$, that $q$ divides $[\beta]$, say $mq = [\beta]$. Assume $kmq = [\beta k] < \beta k < kmq + 1$ for some $m \ge 1$. Then $[\beta (k+1)] < \beta (k+1) < (k+1)mq + \dfrac{k+1}{k}$. So $[\beta (k+1)]$ is either $(k+1)mq$ or $(k+1)mq + 1$. But since the last one is not divisible by $q$, we have, for all $k \in \aleph$: $[\beta k] = kmq$ Now, put $\beta = b + \delta$ with $b$ integral and $0 < \delta < 1$ and consider the smallest element $N$ such that $\delta N > 1$. For this $N$ we have: $Nmq = [\beta N] = [\beta (N -1)] + [\beta] + 1 = Nmq + 1$ A contradiction.