Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

First of all, notice that one of the numbers is $0$, then we have a solution, no matter what the other number is, so let's assume that they're both non-$0$. Another thing worth noticing is that the condition is equivalent to $a\{nb\}=b\{na\},\ \forall n\in\mathbb N^*$. Clearly, from this condition we get $a\in\mathbb Q\iff b\in\mathbb Q$. Let's consider the case when both our numbers are irrational. We have $\frac ba=\frac{\{nb\}}{\{na\}}$. $\{na\}$ is dense in $(0,1)$, so if we choose a sequence $n_k$ s.t. $\{n_ka\}\to 1$, and then we get $\{n_kb\}\to 1$, so $\frac{\{n_kb\}}{\{n_ka\}}\to 1$, meaning that $\frac ba=1\iff a=b$, and we can see that, conversely, all pairs $(a,b)=(a,a)$ satisfy the equation. Assume now one of the numbers is an integer. We quickly find the other one to be an integer too, so this is another type of solution: $(a,b)\in\mathbb Z^2$. Finally, the last case: the numbers are rational but not integral. Let $a=\frac{a'}q,\ b=\frac {b'}{q'}$, where $q,q'\in\mathbb N^*,a',b'\in\mathbb Z$, and $(a',q)=(a',q')=1$. By making $n=q$, and then $n=q'$, we find $q=q'$. We can use the relation $\frac ba=\frac{\{nb\}}{\{na\}}$. We can find $n$ s.t. $\{nb\}=\frac{q-1}q$, so $\frac ba\ge 1$. At the same time, we can show in exactly the same way that $\frac ab\ge 1$, so $a=b$. Combining everything, the solutions are either of the form $(a,a)$ with $a$ being any real number, or $(a,b),\ a,b\in\mathbb Z$, or $(0,b),(a,0)$.

Lemma.$[nx]=[x+\frac{1}{n}]+[x+\frac{2}{n}]+\dots+[x+\frac{n-1}{n}]$ if $a\not =b$from this lemma we have $a[2b]=b[2a]\rightarrow a[b]+a[b+\frac 12]=b[a]+b[a+\frac 12]\rightarrow a[b+\frac 12]=b[a+\frac 12]$from this we can reach to$\{a\},\{b\}<\frac 12$ now try for $n=4,8,\dots,2^k$ we will recieve to $\{a\},\{b\}<\frac{1}{2^k} \ for\ every\ k\in\ \mathbb{N}$ then the solutions are: $a=b\in \mathbb{R}$ $a,b\in \mathbb {Z}$

Can anyone tell me if this solution is correct? Note that $a\lfloor bn \rfloor=b\lfloor an \rfloor \Longrightarrow a(bn - \{bn\})=b(an - \{an\}) \Longrightarrow a\{bn\}=b\{an\}$ $(*)$ This means $a,b \in \mathbb Q$ or $a,b \notin \mathbb Q$. Moreover, if one of them is an integer, then also the other one is. Clearly, equality holds in this case. So we may assume neither $a$ nor $b$ is an integer. It follows that we may also assume $a=(X,a_1a_2\cdots)_2$ and $b=(Y,b_1b_2\cdots)_2$ where $a_i, b_i \in \{0,1\}$ but not all $0$. If $a, b\in \mathbb Q$, sequences $a_i$ and $b_i$ are finite. In the equality $(*)$, take $n=2^k \Longrightarrow a\cdot(0,b_kb_{k+1}\cdots)_2=b\cdot(0,a_ka_{k+1}\cdots)_2$. Multiply both sides by $2 \Longrightarrow a(b_k,b_{k+1}\cdots)_2=b(a_k,a_{k+1}\cdots)_2$ Take $n=2^{k+1} \Longrightarrow a\cdot (0,b_{k+1}b_{k+2}\cdots)_2=b\cdot (0,a_{k+1}a_{k+2}\cdots)_2$ Subtracting these two equalities, we get $a\cdot (b_k)_2 = b\cdot (a_k)_2 \Longrightarrow a\cdot b_k = b\cdot a_k$ Since there is an $a_m\neq 0$, taking $k=m$ gives $a=b$ if $b_m=1$ and $a=0$ if $b_m=0$. Doing the same for $b_n$, we obtain all the solutions are $a=b, a=0, b=0$ or $a, b \in \mathbb Z$.

Storage (essentially Kronecker/Eray's proof) 1998 N2 wrote: Determine all pairs $(a,b)$ of real numbers such that $a \lfloor bn \rfloor =b \lfloor an \rfloor $ for all positive integers $n$. (Note that $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.) Firstly, if one of them is an integer, so is the other, and all such pairs work. If one of them is zero, that's also clear. If the two are equal, then that works. Plugging $n=q$ will give ${an}=0$, forcing $a$ to be rational. So either both $a, b$ are rational or neither of them is. If both are rational, then their denominators are equal (plug $n$ to be the smaller one). Say $a=p/q$ and $b=r/q$ with $q>0$ and $p>r$. Plug $n$ such that $rn \equiv -1 \pmod{q}$ to get a contradiction. Thus, if both non-integer rationals, then both are equal. Otherwise, either one can finish by Kronecker, or do the following: Let $a=u+.u_1u_2\dots$ and $b=v+.v_1v_2\dots$ in binary, with $u, v$ integers, and $u_i, v_i \in \{0, 1\}$ for all $i$ with not all of them as 0 (or 1). Plugging $n=2^k$ we get that $\frac{.u_k\dots}{.v_k\dots}=\frac{.u_{k+1}\dots}{.v_{k+1}\dots}=\frac{a}{b}$ hence $bu_k=av_k$ for all $k \ge 1$ and $bu=av$. Pick $k$ such that $u_k=1$ hence $b=av_k$ so either $b=0$ or $a=b$. The result follows. Remark. It is somewhat annoying that the above argument does not work so nicely with other bases (I spent 15mins doodling base 10 )

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Clearly, $a=b$ is a valid solution. If $a$ were an integer, then $a \lfloor b \rfloor =b \lfloor a \rfloor $ implies $b$ is an integer. Note that $a,b$ both integers is also a valid solution. Also, if $a=0$ or $b=0$ the equation holds. Now, suppose distinct $a,b \neq 0$ are non-integer. Then, \[\frac{b}{a}=\frac{\lfloor b \rfloor}{\lfloor a \rfloor}=\frac{\{b\}}{\{a\}}\]where $\{x\}=x-\lfloor x \rfloor$ is the fractional part of $x.$ Clearly, if $\frac{\lfloor b \rfloor}{\lfloor a \rfloor}=1$ then this reduces to $a=b$ so WLOG, assume $\lfloor a\rfloor >\lfloor b\rfloor.$ Now, we have \[\frac{b}{a}=\frac{\{bn\}}{\{an\}}=\frac{\{b\}}{\{a\}}\]for sufficiently large $n$ this pattern no longer holds, so the only answers are when two are equal out of $(a,b,0)$ or $a,b$ integers.

Suppose that one of $a,b$ is equal to $0$. Clearly this works, so this is a solution. Now, suppose that $a,b\neq 0$. Then we have $a\{bn\}=b\{an\}$. If $a=b$ we are done, which gives another solution. Next, note that if $a$ is an integer, then $b$ must also be an integer. This works, and gives a third solution set. Finally, I claim that no other solutions exist. Note that $a,b$ must have the same sign, so WLOG let $0<a<b$. Then if $a$ is irrational then the RHS can become arbitrarily close to $b$, a contradiction. Thus $a$ is rational, so the RHS is periodic. Then $b$ is also rational. Since both the LHS and RHS have a common period, we can set $a=\frac{x}{d}$ and $b=\frac{y}{d}$. But now the maximum values of both sides are the same, so that $\frac{x}{d}\cdot \frac{d-1}{d}=\frac{y}{d}\cdot \frac{d-1}{d}\implies x=y$, contradiction. We are done. $\blacksquare$

ISL Marabot solve Clearly, $(a,b) = (0,x)$ or $(x,0)$ works where $x\in \mathbb R$. Also, $a,b \in \mathbb Z$ works. Now, let's assume that $a,b\notin \mathbb Z$. And we will prove that $a=b$. Let, $a_i, b_i$ be the $i$th digits of $a,b$ when written in base-$2$. And here, we are considering the representation where there is no infinite tail of $1$s. [If there is $\dots 0111\dots$ then we can replace that with $\dots1000\dots$] Here, $a \lfloor bn \rfloor =b \lfloor an \rfloor \implies \dfrac a b = \dfrac{\lfloor an \rfloor}{\lfloor bn \rfloor}$. Now, $b\notin \mathbb Z$ so, we have $k$ such that $b_k =1$. But, $$\dfrac a b = \dfrac{\lfloor 2^{k-1}a \rfloor}{\lfloor 2^{k-1}b \rfloor} = \dfrac{\lfloor 2^ka \rfloor}{\lfloor 2^kb \rfloor} = \dfrac{2\lfloor 2^{k-1}a \rfloor+a_k}{2\lfloor 2^{k-1}b \rfloor+b_k}$$$$\implies \dfrac a b = \dfrac{a_k}{b_k}$$Therefore, $a=b$ as, $a \neq 0$ and we are done. So, the solution are, $a,b \in \mathbb Z$ or $(a,b) = (x,0)$ or $(0,x)$ or $(x,x)$ where $x \in \mathbb R. \ \ \ \ \blacksquare$

The condition can be rewritten as $$a\{bn\}=b\{an\}.$$ Note that $a$ and $b$ are either both rational or both irrational. Case 1: $a$ and $b$ are both irrational. If $a=b$ then obviously it works. Otherwise WLOG $a>b.$ Then, pick $n$ such that $\{bn\}$ is as close to 1 as needed (possible due to irrationality). Then, $a\{bn\}>b\{an\},$ contradiction. Case 2: $a$ and $b$ are both rational. Then, let $a=x/y$ and $b=z/w$ such that $\gcd(x,y)=\gcd(z,w)=1.$ We then have $$\frac{x}{y}\{\frac{z}{w}n\}=\frac{z}{w}\{\frac{x}{y}n\}.$$Letting $n=y$, we have that $$\frac{x}{y}\{\frac{z}{w}y\}=0,$$so either $x=0$ (or just $a=0$) or $w|y.$ Similarly, plugging in $n=w$ gives either $b=0$ or $y|w$. Therefore, unless either of $a$ or $b$ is 0 (giving solutions $(0,b),(a,0)$), $y=w.$ If $y=w$, then we have $$\frac{x}{y}\{\frac{z}{y}n\}=\frac{z}{y}\{\frac{x}{y}n\}.$$Multiplying by $y^2$ gives $$x(zn\pmod y)=z(xn\pmod y).$$Note that $z$ and $x$ are both relatively prime to $y$ by definition, so if we sum over $0\leq n\leq y-1$, each residue occurs exactly once, so we have $$x(\frac{y(y-1)}{2})=z(\frac{y(y-1)}{2}),$$which means that either $y=1$ (which means $a$ and $b$ are both integers) or $x=z$ (which means $a=b$ since $y=w$). Therefore, we have either $(a,b)=(a,0),(0,b)$, or $a,b$ are both integers, or $a=b$. Putting this all together, the solution set is: (1) when $a=b$, (2) when $a$ and $b$ are both integers, (3) when at least one of $a$ and $b$ is 0.

The answer is $(a,a)$, $(a,0)$, $(0,a)$, and $(p,q)$, where $a \in \mathbb R$ and $p,q \in \mathbb Z$. Assume that $a \neq b$ and $ab \neq 0$ and $a$ and $b$ aren't both integers. Rewrite the condition as $a \{ bn \} = b \{ an \}$. Taking $n=1$ implies $\text{sgn} a = \text{sgn} b$. Assume $|a|>|b|$. Next note that taking $n=1$ gives $\frac{a}{b} = \frac{\lfloor a \rfloor}{\lfloor b \rfloor}$ is rational. If $a$ and $b$ are irrational, note that $\{ bn \}$ is dense in $[0,1]$ by the irrationality of $b$. Choose $n$ so that $\{ bn \} \in \left( \frac{b}{a},1 \right)$. Then $\{an\} = \frac{a\{bn\}}{b} > 1$ and we're done with this case. If $a$ and $b$ are both rational, note that $an \in \mathbb Z \equiv \{an\} = 0 \equiv \{ bn \} = 0 \equiv b\in \mathbb Z$. Thus $a=\frac{x}{z}$ and $b=\frac{y}{z}$ where $|x|>|y|$ and $\gcd(x,z)=\gcd(y,z)=1$. By assumption, $z>1$. Now take $n$ so that $xn \equiv 1 \pmod z$ which exists by the $\gcd$ condition. Then $\{ bn \} = \frac{b \{ an \} } {a} = \frac{b}{az} < \frac{1}{z}$. Thus $\{ bn \} = 0$ and $z \mid n$, an obvious contradiction. This concludes the proof.

bruh this could actually be a real analysis problem The answer is all pairs of integers, $a=b$, $a=0$, and $b=0$. These obviously work. Let $p=\lfloor a\rfloor$, $q=\lfloor b\rfloor$, $a=p+r$, and $b=q+s$. Note that if $a$ or $b$ is a nonzero integer, obviously the other is an integer as well, so assume $r,s\ne 0$. Claim. $r=s$. Proof. Suppose not. Note that \[(p+r)\lfloor qn+sn\rfloor = (q+s)\lfloor pn+rn\rfloor\rightarrow (rq-sp)n=(q+s)\lfloor rn\rfloor-(p+r)\lfloor sn\rfloor.\]Let the LHS be $x_n$ and the RHS be $y_n$. $x_n$ is clearly an arithmetic sequence. But consider some $k$, which gives $y_{k+1}-y_k=rq-sp$. This common difference can't be the common difference every time though, since there obviously exists $n_1$ and $n_2$ such that \[\lfloor r(n_2+1)\rfloor-\lfloor rn_2\rfloor=\lfloor r(n_1+1)\rfloor-\lfloor rn_1\rfloor\]but \[\lfloor s(n_2+1)\rfloor-\lfloor sn_2\rfloor\ne \lfloor s(n_1+1)\rfloor-\lfloor sn_1\rfloor\]since all four of these (LHS or RHS) are binary sequences, and them always changing at the same time would imply that $\lfloor rn\rfloor=\lfloor sn\rfloor$ for all positive integers $n$, so $|r-s|<\varepsilon$ for all $\varepsilon>0$, so $r=s$. Now $n=1$ gives $(p+r)q=(q+r)p$, so $r=0$ or $p=q$ as desired. $\blacksquare$

Note that trivially all integers work, and so does when $a = b$. Furthermore, when $a = 0$ or $b = 0$, then it also works. So assume $a \neq b$ and $a, b \neq 0$. I will show all other cases fail. First, if $a$ is irrational and $b$ is not, obviously this fails. Second, if $a$ and $b$ are irrational and $a/b$ is irrational, this also fails. But if $a/b$ is rational, then $a/b = p/q$ for relatively prime integers $p$ and $q$ and we must have \[ \frac{p}{q} = \frac{\lfloor prn \rfloor}{\lfloor qrn \rfloor} \]for some irrational $r$. But then we can choose $n$ such that $p \nmid \lfloor prn \rfloor$, which is bad. Thus, we just need to resolve the case when both are rationals $p/q$ and $r/s$ where $(p, q) = 1$ and $(r, s) = 1$. Note we can assume that $(q, s) = 1$ because if $\gcd(q, s) \nmid n$ then the floor function makes those cases irrelevant. This means that we only need to look at $n \pmod {qs}$ because after that it'll just repeat. Thus, we must have \[ \frac pq \lfloor \frac rs n \rfloor = \frac rs \lfloor \frac pq n \rfloor \implies ps \lfloor \frac rs n \rfloor = rq \lfloor \frac pq n \rfloor \implies ps\{\frac{rn}{s}\} = rq\{\frac{pn}{q}\} \]But from the assumption that $(q, p) = 1$ and $(q, s) = 1$ then $q \mid \{ \frac{rn}{s}\}$. This means that $\frac{rn}{s}$ must always be an integer, so $s \mid r$, contradiction. $\blacksquare$

Nice Problem! We claim the answer is $(a,b)=(a,0);(0,a);(a,a),(a,b) \quad a,b \in \mathbb{Z}$ Note that the given condition is equivalent to $\dfrac{\{an\}}{\{bn\}}=\frac{a}{b}$ Assume $a,b \neq 0$ also if either one is integer then the other is forced to be integer which works. If $a$ and $b$ are rational, let $a = \frac{c}{d}$, $b = \frac{e}{f}$, then $d=f$ by putting $n = d$, then choose $\{cn\} = \frac{d-1}{d}$ to get $a=b$ If $a,b$ are irrational then, choosing ${an} \rightarrow 1$ makes RHS arbitarily close to $b$, contradiction!.

I claim the answer is only the pairs such that at least one of the following holds: $a,b$ are integers or $a,b$ are equal or one of $a,b$ is zero. It is trivial to see that all of these pairs work. We show no other pairs work. Assume $a,b$ are unequal and nonzero. If exactly one of $a,b$ is an integer, without loss of generality let it be $a$, then we would be forced to have $a \lfloor bn \rfloor = abn$, forcing $\lfloor bn \rfloor = bn$, forcing $bn$ is an integer for all $n$, forcing $b$ to be an integer, which is a contradiction. So assume neither $a$ nor $b$ is an integer. Let $k$ be the smallest value of $n$ that satisfies at least one of $\lfloor an \rfloor \neq n \lfloor a \rfloor$ or $\lfloor bn \rfloor \neq n \lfloor b \rfloor$. At this value of $n$, $\lfloor an \rfloor$ deviates by at most $1$ from $n \lfloor a \rfloor$ and likewise for $b$. Thus we have $\frac ab = \frac{\lfloor a \rfloor}{\lfloor b \rfloor} = \frac{k\lfloor a \rfloor + c}{k \lfloor b \rfloor + d}$, where $c,d$ have absolute value at most $1$ and at least one is nonzero. Without loss of generality, let $a > b$. It is then easy to compare $\frac{k \lfloor a \rfloor + 1}{k \lfloor b \rfloor} , \frac{k \lfloor a \rfloor + 1}{k \lfloor b \rfloor - 1}, \frac{k \lfloor a \rfloor - 1}{k \lfloor b \rfloor - 1}, \frac{k \lfloor a \rfloor}{k \lfloor b \rfloor - 1}> \frac{k \lfloor a \rfloor }{k \lfloor b \rfloor} = \frac{\lfloor a \rfloor }{\lfloor b \rfloor } > \frac{k \lfloor a \rfloor - 1}{k \lfloor b \rfloor} , \frac{k \lfloor a \rfloor - 1}{k \lfloor b \rfloor + 1}, \frac{k \lfloor a \rfloor + 1}{k \lfloor b \rfloor + 1}, \frac{k \lfloor a \rfloor}{k \lfloor b \rfloor + 1} $ , so no values of $c,d$ can achieve the desired equality, done.