Answer: no. There are many constructions, and I invite you to share your own. Define a sequence by $x_0 = 1$ and \begin{align*} 2x_1 &= x_0 \\ 2x_2 &= x_1 + 1 \\ 2x_3 &= x_2 \\ 2x_4 &= x_3 + 1 \\ 2x_5 &= x_4 \\ 2x_6 &= x_5 + 1 \\ &\phantom=\vdots \end{align*} Set $f(2^{-k}) = x_k$ and $f(2^k)=2^k$ for $k=0,1,\dots$. Then, let \[ f\left( a \cdot 2^k + \frac bc \right) = a f\left( 2^k \right) + \frac bc \] for odd integers $a$, $b$, $c$. Check that this works. --- EDIT: Some motivation for this (since someone asked) This construction is not terribly unnatural. If you guess the answer is no, then a natural place to start looking is powers of two -- that is, define $f(2^{-k})$ for each integer $k$. You get that $2x_{k+1} - x_k \in \{0,1\}$ must hold for each $k$. If you make them all $0$ or all $1$ then you just get a linear function, but making them alternate $0$ / $1$ works fine. And in fact any combination of $0$'s and $1$'s works except for those which are eventually all $1$ or eventually all $0$. Once you've nailed down the powers of two you can just extend it to all rational numbers -- there's really only one reasonable way to do this. EDIT: Looks like I botched the extension from powers of two to all rationals... thanks Iurie for pointing that out. I've amended the post.

Does anybody have another construction?

This one is nice in that it's actually a formula: $f(\frac{p}{q})=\frac{p}{q}(1!+2!+\cdots+q!)$ This works because mod $1$, you're essentially saying $f(\frac{p}{q})=\frac{p}{q}(1+2!+3!+...)$, and so the function is "linear".

I originally had the same solution as gurev, but after seeing a different solution I thought a little harder and realized that there are many different-seeming constructions only because of https://www.expii.com/partial-fraction-decomposition/ (or more fancily written, the fact that $\mathbb{Q}/\mathbb{Z} = \bigoplus_{p} \mathbb{Z}[1/p]/\mathbb{Z}$---a direct sum decomposition as abelian groups, or equivalently, as $\mathbb{Z}$-modules). (The reason $\mathbb{Q}/\mathbb{Z}$ is relevant is that we can WLOG, by scaling (how, exactly?), restrict attention to $f$ such that $f(\mathbb{Z})\subseteq \mathbb{Z}$. Then we get a well-defined (why?) function $g\colon\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ such that $g(x\pmod{1}) = f(x)$ for any rational number $x$.) Since there's more than one prime, that means you can, using the partial fraction decomposition into $\mathbb{Z}[1/p]/\mathbb{Z}$ (fractions with denominators all powers of $p$) parts, "glue together" the "independently chosen" solutions for each prime together to get a construction. This is what some people did, essentially. (For example, you can define the prime-$p$ part $g_p\colon \mathbb{Z}[1/p]/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}$ as $g_p(x_p) = \alpha_p x_p$ for any integer $\alpha_p$, and then add over all primes together to define $g$---how, exactly, and why is the (a priori infinite) sum well-defined? Then, if the $\alpha_p$ are not all the same, then $g$ will work as a construction.) EDIT: Someone asked for a concrete example of a function over $\mathbb{Z}[1/p]/\mathbb{Z}$, the set of residue classes (mod $1$) of fractions of the form $\frac{a}{p^n}\pmod{1}$ (for integers $a$ and nonnegative integers $n\ge0$). For example, take $p=2$. Then one example is $g_2\colon \mathbb{Z}[1/2]/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}$ is given by mapping the residue class $0.a_1a_2a_3\ldots \pmod{1}$ (decimals in binary) to $0.a_2a_3a_4\ldots\pmod{1}$. However, if you replace "$f\colon \mathbb{Q}\to \mathbb{Q}$" with "$f\colon\mathbb{Z}[1/p]\to\mathbb{Q}$" (replacing the domain of $\mathbb{Q}$ with the smaller set of fractions with denominators all powers of $p$) for any fixed prime $p$, then there is essentially only one kind of construction: you can no longer glue together independent solutions. (Any construction for the $\mathbb{Z}[1/p]$ version of the problem will have to use an "infinite chain of divisibilities" like $p^1\mid p^2\mid p^5\mid p^7\mid\cdots$, which is in the same spirit as gurev's $1!\mid 2!\mid 3!\mid 4!\mid\cdots$.) If you're interested, you should work out the details yourself; they're not hard.

For those familiar with a bit of higher algebra, this is a manifestation of the fact that $\widehat {\mathbb Z}$ is not equal to $\mathbb Z$. And in fact describing all functions with the required property is pretty much equivalent to classifying the Pontryagin dual of $\mathbb Q/\mathbb Z$ - which is $\widehat {\mathbb Z}$ (Recall that $\widehat {\mathbb Z}$ is the set of all sequences $(x_n)$ with $x_n\in \mathbb Z/n\mathbb Z$ subject to compatibilities $x_n\equiv x_m\pmod m$ if $m\mid n$. Now $\widehat {\mathbb Z}$ acts on $\mathbb Q/\mathbb Z$ via $\hat{x} \cdot \frac mn=\frac{x_n m}n$. (The compatibility above implies choosing another denominator changes the result by an integer, that's why he have to quotient by $\mathbb Z$).) The function $f$ can be described as follows: choose an $\alpha\in \widehat {\mathbb Z}$ and let $f_0$ be any lift of multiplication by $\alpha$ $m_\alpha\colon \mathbb Q/\mathbb Z\to \mathbb Q/\mathbb Z$ to a function $f_0\colon \mathbb Q\to \mathbb Q$ (i.e. $f_0(x)\equiv \alpha x\pmod {\mathbb Z}$. Now $f(x)=f_0(x)+\beta x$ where $\beta$ is any rational number. This description is unique up to noting that for $n\in \mathbb Z$, $(\alpha+n, \beta-n)$ give the same result (so we can for example specify $0\leq \beta<1$ then the description is unique). gurev's construction uses $\alpha=1!+2!+\ldots$ - this is a convergent sum in $\widehat {\mathbb Z}$ because $1!+\ldots+n!$ stabilizes modulo every base. Evan's construction uses the isomorphism $\widehat {\mathbb Z}\cong \prod \mathbb Z_p$ (proved by the Chinese Remainder Theorem). He chooses $\alpha$ whose $2$-component is $2+8+32+\ldots$ (which equals $2/3$) and whose other components are all 1. EDIT: Actually, Evan's solution is wrong as stated in the post (though my interpretation of it isn't; it is the "correct" way his solution should have been). We see that $x_0=1, x_1=1, x_2=1/2$ therefore $f(1/3)+f(1/4)=1/3+1/2=5/6$ on the other hand $f(1/3+1/4)=f(7/12)=7/3f(1/4)=7/6$. His original construction yields a function for which $f(x+y)-f(x)-f(y)$ has no powers of 2 in the denominator. One corrects it by writing each rational number as $x=a2^{k}+b/c$ where $a,b,c$ are odd, then setting $f(x)=af(2^k)+b/c$; this is well-defined up to an integer. Victor's remark that $\displaystyle \mathbb Q/\mathbb Z=\bigoplus \mathbb Z\left[1/p\right]/ \mathbb Z=\bigoplus \mathbb Q_p/\mathbb Z_p$ is the dual of the above isomorphism $\widehat {\mathbb Z}\cong \prod \mathbb Z_p$. As I've mentioned, it is a well-know example of the Pontryagin Dual that all additive functions $\mathbb Q/\mathbb Z\to \mathbb Q/\mathbb Z$ are described by $\widehat{\mathbb Z}$ as above. From here is follows that all additive functions $\mathbb Q\to \mathbb Q/\mathbb Z$ are described by $\widehat {\mathbb Z}\oplus \mathbb Q/N$ where $N$ is the anti-diagonal copy of $\mathbb Z$ inside it (i.e. all numbers of form $(n,-n)$. The functions that satisfy the problem condition are arbitrary lifts of these to functions in $\mathbb Q\to\mathbb Q$.

A diiferent approach:

gurev wrote: This one is nice in that it's actually a formula: $f(\frac{p}{q})=\frac{p}{q}(1!+2!+\cdots+q!)$ This works because mod $1$, you're essentially saying $f(\frac{p}{q})=\frac{p}{q}(1+2!+3!+...)$, and so the function is "linear". Does $p$ and $q$ have to be relatively prime or something else? Thanks a lot!

Yes, $\gcd(p,q)=1$ and $q>0$.

How would you prove that construction works? For example, if we have $x=\frac{p}{q}$ and $y=\frac{r}{s}$, then $x+y=\frac{ps+qr}{qs}$. But there is a chance that $ps+qr$ is not relatively prime to $qs$, which just makes everything more complicated. e.g. $\frac{1}{6}+\frac{1}{3}=\frac{3}{6}=\frac{1}{2}$. Thanks.

there's no need to worry about that suppose that $q<s$, then we note that mod 1, $f(p/q)=(p/q)(1!+2!+...+qs!)$ and $f(r/s)=(r/s)(1!+2!+...+qs!)$ so $f(p/q)+f(r/s)=(p/q+r/s)(1!+2!+...+qs!)=f(p/q++r/s)$

gurev wrote: This one is nice in that it's actually a formula: $f(\frac{p}{q})=\frac{p}{q}(1!+2!+\cdots+q!)$ This works because mod $1$, you're essentially saying $f(\frac{p}{q})=\frac{p}{q}(1+2!+3!+...)$, and so the function is "linear". That's linear alright, but why is it a contradiction to the required conclusion?

Note: $f(x)=\lfloor x \rfloor$ is not a counter-example, because we can take $c=0$.

A slightly different approach. To construct a counterexample, we first define $f$ at $0$ as $f(0)=0$. Then, at every step we take some $x_i\in \mathbb{Q}$, where $f$ is not yet defined and extend $f$, but in a such way that $f(x_i)-r_ix_i$ is not integer, where $r_1,r_2,\dots$ is some enumeration of the rationals. In that way, at every step we keep $f$ being additive mod $1$, in the same time rule out the possibility $c$ in question to be $r_i$. Finally we get a function $f$, in agreement with the requirements, for which such constant $c$ doesn't exists. The details. Enumerate $\mathbb{Q}$ as $r_1,r_2,\dots$ and let $A_0 := \{0\}$. We define $f(0):=0$. Let $i=1$. The extension step. We take some $x_i\in \mathbb{Q}, x_i\not\in A_{i-1}$. There are two possibilities. Either there exists $n\in\mathbb{Z}$ with $nx_i\in A_{i-1}$ or for all $n\in \mathbb{Z}, nx_i\not\in A_{i-1}$. Let's first consider the latter case. Consider $A_i=\{nx_i+a' : n\in \mathbb{Z}, a'\in A_{i-1}\}$. It's easy to see any $a\in A_i$ can be represent uniquely as $a=nx_i + a', n\in \mathbb{Z}, a'\in A_{i-1}$. We choose $y_i\in \mathbb{Q}$, such that $y_i-r_ix_i\not\in \mathbb{Z}$ and then extend $f$ by setting $f(nx_i+a'):=ny_i+f(a')$ The former case. Let $n_0\in \mathbb{Z}$ be the smallest in absolute value $n\neq 0$, satisfying $nx_i\in A_{i-1}$, that's, $n_0 := \min\{|n|,n\in\mathbb{Z}, n\neq 0, nx_i\in A_{i-1}\}$. It's easy to see that if $nx_i\in A_{i-1}$ then $n_0\mid n$. Consider $A_i=\{nx_i+a' : n\in \mathbb{Z}, 0 \leq n<n_0; a'\in A_{i-1}\}$. It's again an unique representation of $A_i$. Let $f(x_i)=y_i$, where $y_i$ be specified later, and set $f(nx_i+a'):= ny_i+f(a')$. We wish to choose $y_i$ such that: 1) this extention of $f$ is still additive on $A_{i}$; and 2) $y_i-r_ix_i\not \in \mathbb{Z}$; Thus, in order $f$ to be additive mod $1$ on $A_i$, it should satisfy $n_0y_i = f(a')+m$, where $a'=n_0x_i\in A_{i-1}$ and $m\in \mathbb{Z}$. Hence $y_i=\frac{f(a')+m}{n_0},m\in\mathbb{Z}$ and $$y_i-r_ix_i= \frac{f(a')+m}{n_0}-r_ix_i $$Since $n_0>1$, we can choose $m\in \mathbb{Z}$ such that the above value is not integer and hereby determine the value of $y_i$. Note that $A_{i-1}\subset A_i$ and $A_i$ is closed under addition/substraction. $\blacksquare$ Now, we increase the value of $i$ and apply again the extension step. Since $\bigcup_{i=0}^{\infty}A_i=\mathbb{Q}$, $f$ is defined over $\mathbb{Q}$ and is additive mod $1$. On the other hand, by the mere construction, there doesn't exists $c\in\mathbb{Q}$ such that $\forall x\in \mathbb{Q}, f(x)-cx\in \mathbb{Z}$.

Lemma. $\mathbb{Q}/\mathbb{Z} \cong \bigoplus_{p} \mathbb{Z}[1/p]/\mathbb{Z}$. Proof: Let the map $\phi : \bigoplus_{p} \mathbb{Z}[1/p]/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ be defined as \[\phi : \sum_{p} (\frac{a_p}{p^k} + \mathbb{Z}) \mapsto (\sum_{p} \frac{a_p}{p^k}) + \mathbb{Z}\]Because the left side only has finitely many nonzero terms, the map is well-defined. It's not hard to show that it is both surjective and injective (details can be found e.g. here). For each prime $p$, choose some integer $a_p$ such that $a_2,a_3,a_5,\dots$ are not all the same. Then, we define $f_p: \mathbb{Z}[1/p] / \mathbb{Z} \to \mathbb{Q}$ as \[f_p(1/p^k) = \frac{a_p}{p^k}\]for any prime $p$ and $k\in\mathbb{N}$. Then because of the lemma, we can define $f$ by adding up the $f_p$s (this is ok because every rational number can be written as the sum of finitely many multilples of prime powers). The construction is then not equal to $cx$ for any $c\in\mathbb{Q}$, because the $a_i$ are not identical.

Let's extend this question to the reals. More precisely, define $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ to be the circle group (i.e. equivalence classes of reals modulo 1) and consider the functional equation \[f(x+y) = f(x)+f(y)\]where $f$ is a function from $\mathbb{T}$ to $\mathbb{T}$. There exist the trivial solutions $f_n(x) := nx$ for all integers $n$, but the TST problem shows that there exist nontrivial solutions for $f$. Simply take a construction for the TST problem and extend it to the reals (well, more precisely $\mathbb{R}/\mathbb{Z}$) using a Hamel basis. I came across this extended question while reading a physics book which claims ``the only 1-dimensional representations of $U(1)$ are the representations $\rho_n: x \rightarrow x^n$''. I think this is false. But in physics we implicitly assume that everything is continuous, so maybe the true statement is ``the only continuous solutions for $f$ are the trivial solutions''? Edit: In physics we assume the representation of a Lie group respects the differential structure; i.e. is a diffeomorphism.

No. Define \[g:(0,1) \to (0,1), g(t):=\begin{cases} t\ge \frac 12&\frac t2\\ t<\frac 12&\frac{1+t}2\end{cases}.\]Define $s_1,s_2,\cdots$ by $s_0=1,s_1=\frac 12$, $s_k = g(s_{k-1})$ for $k\ge 2$ so $2s_k - s_{k-1} \in \{0,1\}$. Then let $f(2^{-k} \cdot \frac pq) = p/q\cdot s_k$ where $2\nmid q$ and either $2\nmid p$ or $k=0$. This definition is rigged so that $f(cx) - \frac{f(x)}{x}\cdot cx\in\mathbb Z$ when $c$ is an integer. So this directly implies $f$ satisfies the condition $f(x+y)-f(x)-f(y)\in \mathbb Z$. But it is also clear that $s_k2^k, -2^k+s_k2^k$ are unbounded so we indeed have a contradiction.

We claim that there are no such functions by creating a counter example function. FTSOC, assume a constant exists. Then by the problem condition, for all integers $k$, we have that \[{f(kx)}={kf(x)}\] Fix $f(1)$. Now for every prime $q$, let $f(\frac{1}{q})=\frac{f(1)}{q}+\frac{2^q}{q}$. Then all other values of $f(x)$ are fixed by the above statement. Notice that if $c$ exists, we have that for all $q$, we have that \[\frac{f(1)}{q}+\frac{2^q}{q}-\frac{c}{q}\]is an integer. Since we know that $f(1)-c$ must be an integer (if we plug in $x=1$ to the original condition statement), we know that $\frac{2^q+k}{q}$ is always an integer, where $k$ is the fixed integer $f(1)-c$. However, we see that as we change $q$, we see that $k$ must be infinite, a contradiction, and we are done. $\blacksquare{}$

Wow this is so weird The answer is no. The idea is to expression $f$ as an infinite series $$f(p/q)=\frac{p}{q}(a_1+a_2+a_3\cdots)$$for fixed $a_i.$ Note that any function of this form is "linear" so it satisfies the condition. We will use the series $$f(p/q)=\frac{p}{q}(1!+2!+3!\cdots).$$This is not well-defined, since it involves a divergent sum. However, note that if we distribute the $p/q$, there will be a point after which all terms after that are integers (since eventually all factorials are divisible by $q$). Since we only care about the fractional part, we can "stop" the infinite sum at that point. Therefore, we define $f(p/q)$ as adding terms of the infinite series up to a point until we reach the part where it becomes all integers. This is actually well defined since it stops after a finite number of terms for any particular $p$ and $q$. However, since we have not changed the fractional part, it still satisfies the first condition. There is clearly no value of $c$ for this function, so we are done.