Part $a$ $A= \{3k | k \in \mathbb Z \},B= \{3k+1 | k \in \mathbb Z \},C= \{3k+2 | k \in \mathbb Z \}$ works and it is the only way to do this except from permutations.

SCP wrote: Part $a$ $A= \{3k | k \in \mathbb Z \},B= \{3k+1 | k \in \mathbb Z \},C= \{3k+2 | k \in \mathbb Z \}$ works and it is the only way to do this except from permutations. and if you proved a is the only way, b is easy ^^

Assume the three sums are disjunct again , also for $\mathbb Q$. If $A,B,C$ satisfy the question, then $kA,kB,kC$ do. As $A,B,C$ there are some numbers $a,b,c$ , taking the lcm of the denomenators and calling it $k$. We now have $kA,kB,kC$ contains all natural numbers and by part $a$, we know the form of this numbers. Proof there were only one way: assume $n,n+1$ are elements of the same set, wlog $A$. If here exist some sequence of the form $Yba_1a_2aa\cdots a_{n-1} a_n c$ $Y=a$ means $ca$ and $a_nb$ gives contradiction. $Y=c$ , then $Ya_2$ and $ba_1$ have same sum, contradiction. $Y=b$, then $Yc$ and $ba_n$ gives contradiction. Hence no two consecutive elements are of the same set. In the other case we have somewhere sth of the form $abaX$ (wlog) and then $X$ can't be $c$ and we get $babX$ and so on. Hence we wouldn't have three elements, contradiction. Now we can concluse we have $A,B,C$ are of the form as given in part $a$. Now look to the numbers in the set $S= \{z+0.5|z \in \mathbb Z\}$. If $S$ $aa$ we get together with $bc$ ... If $S$ contains elements of $A,B,C$ it is also of the form $abc \cdots$ or $acb \cdots$. In both cases, we will have sums which aren't disjunct. If $S$ contains only elements of wlog $A,B$ it has to be of the form $abab\cdots$ and then $b_1a$ together with $b_2c$ gives $b_1+c=b_2+a$ . Hence again contradiction. Conclusion: It is impossible for $\mathbb Q$.

or just use the LEMMA "partition of $\mathbb{Z}$ on such condition" is unique.(not too hard to prove) and think of the set $\{ \cdots,-\frac{2}{k} -\frac{1}{k}, 0 , \frac{1}{k},\frac{2}{k} \cdots \}$, and there is only one way to partite this set. Now, you can see $0,1$ are different integers, so they are contained in different sets, also $\frac{0}{3}, \frac{3}{3}$ are contained in the same set, contradiction.

Other solution for part b): Assume that such $A$, $B$, $C$ exist and that $a\in A$, $b\in B$, $c\in C$ and set $d:=a+b-c$. Then $a+b=c+d$ is in $A+B$ and can hence be neither in $A+C$ nor in $B+C$. Thus $d$ is neither in $A$ nor in $B$, hence in $C$. The same argument applied to $e:=d+b-a=2b-c$ shows that $e\in A$. We have thus shown: In every arithmetic progression $x$, $y$, $2y-x$ the three members are either all in the same set $A$, $B$ or $C$ or there is one member in each of the three sets. Now set $f:=(2a+b)/3$ and $g:=(a+2b)/3$. If $a$, $f$, $g$ are all in $A$, we find a contradiction considering $f$, $g$, $b$. If on the other hand $a$, $f$, $g$ are one in each of the sets $A$, $B$, $C$, then $f\in B$, $g\in C$ or vice versa. Anyway $f$, $g$, $b$ can neither be all in the same set nor in three different sets. This contradiction shows that such sets $A$, $B$, $C$ cannot exist.

Solution:

Proposed Generalization: what if we partitioned $\mathbb{Z}$ into $n$ sets, not just 3? What about $\mathbb{Q}$?

for part (a) , take $A={x : x=0(mod 3)} ,B={y : y=1(mod3) } ,C={z : z=-1(mod 3)}$

Apparently bad solution: (a) The answer is clearly yes. Let \[A=\{x \mid x \equiv 0 \pmod 3\}, B=\{x \mid x \equiv 1 \pmod 3\}, C=\{x \mid x \equiv 2 \pmod 3\}\] Then \[A+B=\{x \mid x \equiv 1 \pmod 3\}, A+C=\{x \mid 2 \pmod 3\}, B+C=\{x \mid 0 \pmod 3\}\] which are disjoint as desired. Furthermore, I claim that (up to permutation) this is the only solution. Since $A, B, C$ are nonempty, there exist $a,b,c$ such that $a \in A, b \in B, c \in C$. Assume for the sake of contradiction that $a+1 \in A$. Then $(a+1)+b \equiv A+B$. If $b+1 \in C$, then $a+(b+1) \in C$, contradiction. Thus $b+1 \notin C$. By the same logic, $c+1 \notin B$. If $b+1 \in A$ and $c+1 \in A$, then $b+(c+1) \in A+B$ and $c+(b+1) \in A+C$, contradiction. Thus at least one of $b+1 \in B$, $c+1 \in C$ must be true. WLOG assume that $b+1 \in B$. Then $c+(b+1) \in B+C$. If $c+1 \in A$, then $b+(c+1) \in A+B$, contradiction. Thus $c+1 \notin B, c+1 \notin A \implies c+1 \in C$. Following this logic, for any $k \in \mathbb{Z}$ we have $a+k \in A, b+k \in B, c+k \in C$. In particular $a+(b-a) \in A$, contradiction. By the same logic, if $a+2 \in A$, then for any $k \in \mathbb{Z}$ we have $a+2k \in A, b+2k \in B, c+2k \in C$. By Pigeonhole, at least two of $a,b,c$ have the same parity. WLOG assume $a,b$ are both even. Then $a=2x, b=2y \implies 2x+2(y-x) \in A$, contradiction. If $a+3 \notin A$, then assume WLOG $a+3 \in B$. Then $a+c+3 \in B+C \implies c+3 \in A$. Similarly, $b+3 \in C$. Then $a+c+6 \in A+B, a+b+6 \in B+C$. Thus $a+6 \in A$ or $a+6 \in B$, and $a+6 \in B$ or $a+6 \in C$. Thus $a+6 \in B$. Similarly, $c+6 \in A, b+6 \in C$. Thus for any $k \mathbb{Z}-\{0\}$, $a+3k \in B, b+3k \in C, c+3k \in A$. Therefore $a,b,c$ are pairwise distinct in modulo 3. WLOG $a \equiv 0 \pmod 3, b \equiv 1 \pmod 3, c \equiv 2 \pmod 3$. Then $a+b \equiv 1 \pmod 3$, but $B+C$ consists of all integers that are $1 \pmod 3$. Thus $a+b \in A+B, a+b \in B+C$, contradiction. Thus for any $a \in A$, we have $a+3 \in A, a+1,a+2 \notin A$. The result folows. b) I claim there are no such sets $A, B, C$. Assume for the sake of contradiction that there exist such sets $A, B, C$. Then the sets $kA, kB, kC$ also work, where $k$ is any rational constant and $kA=\{ka \mid a \in A\}$. Consider any two primes $p$ and $q$, neither of which is $3$. By the unicity of the solution in A, by considering $pA,pB,pC,qA,qB,qC$ we have \[\{\frac{1}{p},\frac{4}{p},\hdots\}, \{\frac{2}{p},\frac{5}{p},\hdots\}, \{\frac{3}{p},\frac{6}{p},\hdots\}\] are subsets of $A,B,C$ in some order. The same holds for $q$. Thus $(p+2q, 2p+q, p+3q, 3p+q, 2p+3q, 3p+2q)$ must be pairwise distinct mod 3. Since $p,q$ are primes that are not $3$, we have $p \equiv 1, 2 \pmod 3$ and $q \equiv 1,2 \pmod 3$. Thus either $p \equiv q \pmod 3, p \equiv 2q \pmod 3,$ or $2p \equiv q \pmod 3$. Thus $(p+2q, 2p+q, p+3q, 3p+q, 2p+3q, 3p+2q)=(p,q,2p,2q,p+2q,2p+q)$ are not pairwise distinct mod 3 (in fact the latter two entries are not even necessary), which is a contradiction. Therefore our original assumption is false, and there are no such sets $A,B,C$.

I don't understand why the solutions here are so complicated. I hope I am not doing something wrong: once you prove the partition for part 1 is unique up to permutations (wlog 0 in A, 1 in B, 2in C), it follows that if you have a rational number $r$ in set A, then $r+3k$ is in A, $r+3k+1$ is in B and $r+3k+2$ is in C for all integers $k$. Thus for every rational number $x$ there is some corresponding rational $y$ in [0,1] lying in set A. Thus A must contain all of [0,1] so A is given by $[0,1] +3k$ and the other sets are drtermined. But then it is obvious that this is not a partition.

Sorry to double post but can someone make sure I'm not misunderstanding the problem. Apparently even the official solution has a more complicated method to prove part b once you prove the partition for part a is unique.

viperstrike wrote: Sorry to double post but can someone make sure I'm not misunderstanding the problem. Apparently even the official solution has a more complicated method to prove part b once you prove the partition for part a is unique. I strongly believe your solution is right, as Iyukhson said, it should be really easy once you prove the partition for (a) is unique.

Does anybody know the country proposing?

Warning: the following proof contains a lot of WLOG, sometimes without mention. I think it's all correct, but I'm too tired to go and write it better. (a) Let $A=3\mathbb{Z}$, $B=3\mathbb{Z}+1$, and $C=3\mathbb{Z}+2$. It is an important lemma for (b) that this is essentially the only solution to (a). To do this, suppose $A,B,C$ satisfy the conditions of the problem. First, we show that the sets can't be bounded above. Suppose that both $B$ and $C$ were bounded above, so there is some $N$ such that $N+k\in A$ for all $k\ge 0$. Suppose $b\in B$ and $c\in C$. Then, $N+b+c=(N+c)+b\in A+B$ and $N+b+c=(N+b)+c\in A+C$< so we have a contradiction. The other case is that only $C$ is bounded above, so both $A$ and $B$ have arbitrarily large numbers in them. Thus, all numbers at least some $N$ are in either $A$ or $B$. Pick $N+\alpha\in A$ and $N+\beta\in B$, and let $c\in C$. We see that $2N+\alpha+\beta-c\in A$ or $2N+\alpha+\beta-c\in B$, and WLOG it's in $A$. Then, we have $2N+\alpha+\beta=(N+\alpha)+(N+\beta)\in A+B$ and $2N+\alpha+\beta=(2N+\alpha+\beta-c)+c\in A+C$, so we have the desired contradiction. Thus, no set is bounded above. We have the following lemma. Lemma: If $0\in A$ and $m\in C$ such that $m>0$ and $m'\not\in C$ for all $m'<m$ satisfying $m'\ge 0$, then $1,\ldots,m-1$ are either all in $A$, or are all in $B$. Proof of Lemma: Clearly, $1,\ldots,m-1$ are not in $C$ by the definition of $m$. Suppose $x\in A$ and $m-x\in B$. Then, $m=x+(m-x)\in A+B$ and $m=0+m\in A+C$, so this can't happen. We thus have that $x,m-x\in A$ or $x,m-x\in B$. Say we have $1,m-1\in A$. We now claim by induction that $x,m-x\in A$ for all $x\le m/2$. The base case of $x=1$ is true by definition. Suppose $x-1,m-x+1\in A$, and suppose $x,m-x\in B$. We see that $m+1=x+(m-x+1)\in A+B$, and $m+1=1+m\in A+C$, so we have a contradiction, so $x,m-x\in A$. Thus the induction is complete, and we can prove a similar statement when $1,m-1\in B$, so the lemma is proved. $\blacksquare$ We no claim that the situation in the lemma is impossible unless $m=1,2$. Suppose $m\ge 2$. By the lemma, the sets up to $m+1$ are one of the following 6. \[A=\{0,1,\ldots,m-1,m+1\}, B=\{\}, C=\{m\}\]\[A=\{0,1,\ldots,m-1\}, B=\{m+1\}, C=\{m\}\]\[A=\{0,1,\ldots,m-1\}, B=\{\}, C=\{m,m+1\}\]\[A=\{0,m+1\}, B=\{1,\ldots,m-1\}, C=\{m\}\]\[A=\{0\}, B=\{1,\ldots,m-1,m+1\}, C=\{m\}\]\[A=\{0\}, B=\{1,\ldots,m-1\}, C=\{m,m+1\}\]Since none of the sets are bounded above, the first case gives a problem as letting $x$ be the first number $\ge m+2$ that appears in $B$, we have a contradiction by the lemma with $B$ and $C$ swapped, since not all of $1,\ldots,x-1$ are together ($m$ and $m-1$ are separate for example). The same argument disposes of the third case. If $m\ge 3$, then in the second case we have $m+2=1+(m+1)\in A+B$ and $m+2=2+m\in A+C$, so it's a contradiction. Remember that this case still works for $m=2$. If $m\ge 2$, then in the fourth case we have $m+2=2+m\in B+C$ and $m+2=(m+1)+1\in A+B$, so the fourth case works only if $m=2$. The fifth case fails as $m+1=0+(m+1)\in A+B$ and $1+m\in B+C$. The final case fails as $1+m=m+1\in B+C$ and $0+(m+1)=m+1\in A+C$. Thus, if $m=2$, then we have only the second or fourth case, both of which imply that $0,1,2$ are in different sets. The only case remain to examine is if $m=1$. We would like to show that again $0,1,2$ are all in different sets, so suppose for sake of contradiction that $0,2\in A$ and $1\in C$. The sets aren't bounded above, so there is some $x\in B$ that is the smallest $x\ge 3$, and the same argument with the lemma shows that this case is impossible. Thus, moving to full generality, we've shown that $n,n+1,n+2$ have to all be in different sets, so easy induction shows that \[\{A,B,C\}=\{3\mathbb{Z},3\mathbb{Z}+1,3\mathbb{Z}+2\}.\] (b) Let $d$ be a number such that $dA$, $dB$, and $dC$ all contain an integer. We can do this by picking three random elements, one from each set, and letting $d$ be the lowest common denominator of all the elements. Let $A'=dA\cap\mathbb{Z}$, $B'=dB\cap\mathbb{Z}$, and $C'=dC\cap\mathbb{Z}$. We see that $A',B',C'$ satisfy the conditions of (a) as they clearly contain all integers, they are disjoint, they are nonempty, and have \[(A'+B')\cap(A'+C')\subseteq (A+B)\cap(A+C)=\emptyset,\]so $(A'+B')\cap(A'+C')=\emptyset$ and all cyclic variations. Thus, $0,1,2$ are in different sets, so $0$ and $1/d$ are in different sets of $A,B,C$. We can do the exact same argument replacing $d$ with $3d$, and get that $0,3$ are in the same sets of $3dA,3dB,3dC$. Thus, $0$ and $3/3d=1/d$ are in the same set of $A,B,C$, which is a direct contradiction. Thus, there are no sets $A,B,C$ for this part. $\blacksquare$

Maybe similar to above solutions.

Solution from Twitch solves ISL: For (a), just take $A = \{0 \bmod 3\}$, $B = \{1 \bmod 3\}$, $C = \{2 \bmod 3\}$. Actually, let's prove this construction for (a) is unique, up to relabeling the sets. We prove that: Claim: We have $A+B-C \subseteq C$. Proof. The hypothesis states that since if $a+b-c = a'$ for some $a,a' \in A$, $b \in B$, $c \in C$, we get a contradiction, and similarly. Similarly, we have $B+C-A \subseteq A$, $C+A-B \subseteq B$. $\blacksquare$ Let $a$, $a'$ be any two elements of $A$. Let $b_0$ be any fixed but arbitrary element of $B$. Consider any $c \in C$. Then \[ a + b_0 - c \in C \implies a' + b_0 - (a + b_0 - c) = \boxed{c + (a'-a) \in C}. \]This, together with symmetric variants, implies that $A$, $B$, $C$ must be arithmetic progressions with the same common difference (by considering the smallest gap between two consecutive elements in the same set). Hence that common difference must be $3$ and the uniqueness is proved. We show this implies the answer to (b) is negative. Assume for contradiction such a partition exists. Let $a_0$, $b_0$, $c_0$ be arbitrary elements of $A$, $B$, $C$. Choose a large integer $n$ such that $na_0$, $nb_0$, $nc_0$ are all integers divisible by $3$. Then consider $nA \cap {\mathbb Z}$, $nB \cap {\mathbb Z}$, $nC \cap {\mathbb Z}$. It is a partition satisfying (a) and with each set nonempty, but it is not the one we expected (because all three sets have multiples of $3$). This is a contradiction.

We prove that the only partition for (a) is labeling integers modulo $3$ (it obviously works). This easily implies the answer to (b) is no. Let $x \le y$ be integers and let $A' = A \cap [x, y]$, $B' = B \cap [x, y]$, $C' = C \cap [x, y]$. Choose $x$ to be small enough and $y$ to be large enough such that $A'$, $B'$, $C'$ are all nonempty. Then by applying IMO 2000/4, we get that $A'$, $B'$, $C'$ partition $[x, y]$ according to modulo $3$, or one is $\{x\}$, one is $\{y\}$, and the third is $[x+1, y-1]$. Let $x \to -\infty$ and $y \to \infty$. Clearly the latter situation can never happen, so we get that $A$, $B$, $C$ partition the integers modulo $3$.

Obviously for A) the construction $0$ mod $3,$ $1$ mod $3,$ and $2$ mod $3$ all work. Now we claim it is the only construction that works, up to stupid stuff. This is equivalent to the claims 1) No two consecutive integers (say $a,a+1$) are in a set (say $A$). 2) No two integers differing by $2$ (say $a,a+2$) are in a set (say $A$). Proofs: 1) Assume for contradiction's sake $a,a+1$ are in the same set. Now note that we cannot have $x$ in $B$ and $x\pm 1$ in $C;$ in other words, no two consecutive integers may be in different sets. Now assume $y,y+1$ are in $B,$ and note for $c$ in $C$ (this must exist as $C$ is nonempty) we may not have $c-1$ or $c+1$ in $A$ or $B$, so $c-1$ and $c+1$ are both in $C.$ But this implies $C$ is the set of all positive integers, which is a contradiction. Thus, if some element $k$ is in $B$ or $C$, then $k-1$ and $k+1$ are in $A$. Now we cannot have two elements $b$ and $c$, one in $B$ and the other in $C$, have a difference of $2.$ Now we claim that $k$ cannot belong to $B$ (and analogously not $C$ either). Note that if it did, this implies $k+1$ and $k-1$ must be in $A.$ Then note that for $c$ in $C,$ $c+1$ must be in $A,$ so $c+1+k-1\in C+A$ and $k+c\in B+C,$ contradiction. So all integers are in $A,$ contradiction. 2) Assume for contradiction's sake $a,a+2$ are in the same set. Thus if $b$ is in $B$ we cannot have $b-2$ or $b+2$ be in $C.$ Now assume $b$ and $b+2$ are both in $B$; then by our reasoning above, for $c$ in $C$ (which must exist), we have that $c-2$ and $c+2$ cannot be in $A$ or $B,$ so it must be in $C.$ Therefore all of the integers congruent to $c$ mod $2$ are in $C.$ This is bad because since $|A|,|B|>0,$ we must have values $a,b$ such that $|a-b|=2$ and there are values $c_1,c_2$ in $C$ such that $|c_1-c_2|=2$ for obvious reasons. Thus if some element $k$ is in $B$ or $C$, then $k-2$ and $k+2$ are both in $A.$ Now we claim that $k$ cannot belong to $B$ (and analogously not $C$ either). Note that if it did, this implies $k+2$ and $k-2$ must be in $A.$ Then note that for $c$ in $C,$ $c+2$ must be in $A,$ so $c+2+k-2\in C+A$ and $k+c\in B+C,$ contradiction. So all integers are in $A,$ contradiction. Now this nukes B. Note that the integers are distributed based on mod $3.$ But also note that the rationals $\frac{k}{3}$ (including $3\mid k$) must be distributed based on $k$ mod $3$ as well, so for instance $0$ and $\frac{3}{3}$ are in the same set but $0$ and $1$ are in different sets.

I hate this writeup. We use $X,Y,Z$ instead of $A,B,C$ because I wrote it this way and am too lazy to change. <3 Over $\mathbb Z,$ let $X := \{a \mid a\equiv 1\pmod 3\},$ $Y:=\{b \mid b\equiv 2 \pmod 3\}$ and $Z:= \{c\mid c\equiv 0 \pmod 3\}.$ Over $\mathbb Q,$ we claim the answer is no. Suppose otherwise and that $x\in X$ and $y\in Y$ with wlog $y>x$ and $y-x = d.$ Claim: The sequence of reals $x,y, y+d, y+2d, \dots$ must alternate as $x,y,z,x,y,z,\dots$ where $z\in Z.$ Proof. Consider some $z$ in the sequence. Define $x_1$ and $y_1$ as the first initial two $x,y.$ Then, for $z-d,$ it must must be in $Y,$ for otherwise $x_1 + z = y_1 + z-d$ is in both $X+Z$ and $Y+Z.$ Similarly, if a sequence has $y,$ it must be preceded by $x$ and $z$ must precede $x.$ Inductively, this implies the desired. $\blacksquare$ Now consider $x_1+2d/3.$ If $x_1+2d/3 \in Y\cup Z,$ then it must be true by the above claim that $x_1+3\cdot 2d/3 = x_1+ 2d \in X,$ but we have already established it must be in $Z,$ contradiction. Hence, $x_1+2d/3$ must be in $x.$ As a result, continuing the sequence with difference $d/3,$ we have $$x_1+2d/3 \in X, \quad x_1+d \in Y, \quad x_1+4d/3 \in Z, \quad x_1+5d/3 \in X, \quad x_1+2d \in Y,$$but we already established $x_1+2d \in Z,$ contradiction. $\blacksquare$

ISL marabot solve Split $\mathbb{Z}$ into $\{0\pmod 3\}, \{1\pmod 3\}, \{2\pmod 3\}$, which clearly works. So $(a)$ is $\boxed{\text{yes}}$. The key claim is: That is the only possible partition of $\mathbb{Z}$. Subclaim: $A+B-C\subseteq C$, and similarly $A+C-B\subseteq B$ and $B+C-A\subseteq A$. Proof: Suppose some $a+b-c=a'$, where $a'\in A$. Then $a+b=a'+c$, a contradiction to the problem statement. Similarly if $a+b-c=b'$. $\blacksquare$ Now let $a$ and $a'$ be in $A$, $b$ be some fixed element of $B$, and $c$ be any element in $C$. Then we get \[a+b-c\in C\implies a'+b-(a+b-c)=c+(a'-a)\in C\] Similarly, \[c+(b'-b)\in C, b+(c'-c)\in B, b+(a'-a)\in B, a+(b'-b)\in A, a+(c'-c)\in A.\] Let the smallest gap between consecutive elements in $C$ be $k$. So the smallest gap between consecutive elements in $A$ or $B$ is also $k$. So we have three arithmetic progressions with the same common difference, which implies that common difference must be $3$. The only way this is possible is if you split it up into $0\pmod3, 1\pmod3, 2\pmod3$. $\blacksquare$ The answer to $(b)$ is $\boxed{\text{no}}$. Suppose there existed such a partition of $\mathbb{Q}$. Then consider three values $a\in A, b\in B, c\in C$. Now choose $n$ so that $na, nb, nc$ are all divisible by $3$. Since we are now over the rationals, the partition $\{nA\cap \mathbb{Z}\}, \{nB\cap \mathbb{Z}\}, \{nC\cap \mathbb{Z}\}$ should work, but it's not the required mod $3$ partition as each set has a multiple of $3$.

Part A is trivial, just let $A,B,C$ consist of the numbers $0,1,2$ modulo $3$, respectively. Claim: This and permutations are actually the only solutions to Part A. Proof. First, we prove that two consecutive integers cannot be in the same set. WLOG let $0,1$ both be in $A$, then any number in set $B$ must be adjacent to a number in $A$ or $B$ and similar for $C$. If two consecutive numbers are both in $B$, the numbers adjacent to a number in $C$ must be in $C$, contradiction, and similar for $C$. Thus any number adjacent to a number in set $B$ or $C$ must be in set $A$. Let $b$ be in set $B$ and $c$ be in set $C$, then $A+B$ and $A+C$ are not disjoint because $b+(c-1) = c + (b-1)$, contradiction. Next, we prove two integers two apart cannot be in the same set. WLOG let $0,2$ be in $A$ and let $1$ be in $B$. Then $-1, 3$ cannot be in either $A$ or $C$ and thus also must be in $B$. Thus, if $c$ is in $C$, $c-2$ or $c+2$ cannot be in $A$ or $B$ and thus must be in $C$. This leads to a contradiction. $\blacksquare$ Now let's solve Part B. Let $a,b,c$ be an element in sets $A,B,C$, respectively. Consider the set of numbers that are a multiple of $D=\text{gcd}(c-a, b-a)$ away from $A$. This contains $a,b,c$ (a number in every set) and is $\mathbb{N}$ translated and dilated, so the claim applies. However, the claim also applies if we consider the set of numbers that are a multiple of $\frac{D}{3}$ away from $a$. These two results imply that $a+D$ and $a$ are both in the same set and not in the same set, contradiction.

Let $nX$ and $X-Y$ denote the sets $\{nx\mid x\in X\}$ and $\{x-y\mid x\in X,y\in Y\}$ respectively. $(\text a)$ Let $A=\{n\mid n\equiv0\pmod3\}$, $B=\{n\mid n\equiv1\pmod3\}$, $C=\{n\mid n\equiv2\pmod3\}$. In fact, we can show that these are the only solutions (up to switching around $A,B,C$). We claim that no set of $A,B,C$ contains two distinct integers with a positive difference less than $3$. First, we will show that $A+B-C\subseteq C$, $B+C-A\subseteq A$, and $C+A-B\subseteq B$. It suffices to show the first one. Let $a\in A$, $b\in B$, $c\in C$, then if $a+b-c\in A$ we have: $$A+B\ni a+b=(a+b-c)+c\in A+C,$$while if $a+b-c\in B$ then: $$A+B\ni a+b=(a+b-c)+c\in B+C,$$which are both impossible. Then we show that no set of $A,B,C$ contains two numbers with a positive difference of $1$. Suppose otherwise, and WLOG let $\{a,a+1\}\subseteq A$. Suppose $a+2\notin A$, then WLOG $a+2\in C$. Let $b\in B$. We have: $$c+1=(a+2)+c-(a+1)\in B+C-A\subseteq A,$$so $c+1\in A$. Also: $$a+3=(c+1)+(a+2)-c\in A+B-C\subseteq C.$$Then: $$a+1=a+(a+3)-(a+2)\in A+C-B\subseteq B,$$a contradiction. Then $a+2\in A$, so by induction $x\in A$ for all $x\ge a$. But by the same argument in the reverse direction, $x\in A$ for all $x\le a+1$, so $A=\mathbb Z$ and then $B$ must be empty, contradiction. Finally, we will show that no set of $A,B,C$ contains two numbers with a positive difference of $2$. Suppose otherwise, and WLOG let $\{a,a+2\}\subseteq A$. WLOG $a+1\in B$. If $a+3\in C$ then: $$a=(a+2)+(a+1)-(a+3)\in A+B-C\subseteq C,$$which is absurd. So $a+3\in B$. Suppose $\{x,x+2\}\subseteq A$ and $\{x+1,x+3\}\subseteq B$. We claim that $x+4\in A$ and $x+5\in B$. It suffices to show that $x+4\notin C$ and $x+5\notin C$. If $x+4\in C$, we have: $$x+3=(x+1)+(x+4)-(x+2)\in B+C-A\subseteq A,$$a contradiction. So $x+4\in A$. Similarly, if $x+5\in C$, we have: $$x+4=(x+5)+(x+2)-(x+3)\in C+A-B\subseteq B,$$which is also false, so $x+5\in B$. By induction, $y\in A\cup B$ if $y\ge a$, and the argument in the reverse direction shows that $y\in A\cup B$ if $y\le a+1$ too. Then $C$ must be empty, contradiction. Finally, let $a\equiv0\pmod3$ be arbitrary, WLOG $a\in A$. Then $a+1\in B\cup C$, so WLOG $a+1\in B$. We get $a+2\in C$, $a+3\in A$, $a+4\in B$, $a+5\in C$, etc. so reversing this argument gives that the only valid construction is the one given, ignoring rearrangements. $(\text b)$ Suppose such a partition into $A,B,C$ exists. For any $n$, $nA\cap\mathbb Z$, $nB\cap\mathbb Z$, and $nC\cap\mathbb Z$ satisfy part $(\text a)$. Let $m$ be three times the $\gcd$ of all denominators of elements in $A\cup B\cup C$. Then $mA\cap\mathbb Z$, $mB\cap\mathbb Z$, $mC\cap\mathbb Z$ are all subsets of (or equal to) $\{x\mid x\equiv0\pmod3\}$, a contradiction. So no such partition exists.

First part (a), I'll show the only way is to partition $\mathbb{Z} \pmod{3}$. Note that translating the partition by any number still works. Suppose there were two consecutive numbers in some set, say $A$, wlog assume $0$ and $1$ by translating. Suppose the numbers $0,1, \cdots, k$ are in $A$ and $k+1$ is in $B$. Then since $(k+1) + (1) = (k+2) + (0)$, the number $k+2$ must necessarily be in $A$ and by induction we get that every number at least $k+2$ is in $A$. Since $C$ is nonempty, let $x$ be some number in it, which must be negative. So choose an $n$ such that $n+x = k$ which gives a contradiction. Similarly, suppose both $0$ and $2$ were in $A$ and $1$ was in $B$. Then similar to above, we get that every number at least $3$ is in $A$ and so if $x$ (which is negative) is in $C$, picking $n$ such that $n+x = 1$ gives a contradiction. So we have that any three consecutive numbers are in different sets, which implies the only possible way to do (a) is mod $3$. For part (b), consider the numbers of the form $\frac{k}{3}$ with $k \in \mathbb{Z}$. By part (a), we must have that the partition contains things based on $k$ mod $3$, and specifically, that both $\frac{0}{3}$ and $\frac{3}{3}$ are in the same set, but this is impossible since $0,1$ need to be in different subsets, so there is no such way to partition $\mathbb{Q}$. $\blacksquare$

We claim that the only solution to $\mathbb Z$ is $A$, $B$, $C$ each being all integers of a single residue class. Suppose $a$, $a+r\in A$ where $r<3$. Note that if $a$, $a+r\in A$ for some positive integer $r$, then there do not exist $b\in B$ and $c\in C$ such that $|b-c|=r$. Otherwise, we can add $a+r$ to the smaller one and $a$ to the larger one to get that $A+B$ and $A+C$ are not disjoint. Call $A$ and $B$ $r$-disjoint if there do not exist $a\in A$ and $b\in B$ such that $|a-b|=r$. We claim that when $r=1$, there do not exist $b$, $b+kr\in B$ for all positive integers $k$. We proceed with induction. When $k=1$, we have $b$, $b+r\in B$ forces $A$ and $C$ to be $r$-disjoint. We already know that $B$ and $C$ are $r$-disjoint, so if $c\in C$ then $c-r$ and $c+r$ cannot belong to either of $A$ and $B$. They must both belong to $C$. Thus, $A$ and $B$ are $r$-disjoint. This means that $A$ has every $r$th integer, $B$ has every $r$th integer, and $C$ also does, absurd. Suppose it is true for $k=i-1$. Then, if $b\in B$ then $b-r\in A$ and $b+rj\in A$ for all $j\in \{1,2,\dots, i-1\}$. In particular, since $b-r$ and $b+ri-r\in A$, $B$ and $C$ are $ri$-disjoint. If $b+rk\in B$ then $A$ and $C$ are $ri$-disjoint so if $c\in C$ then $c+ri$ cannot belong to either $A$ or $B$, so it belongs to $C$. Thus, $A$ and $B$ are also $ri$-disjoint. We have that if $a\in A$ then $a+ri\in A$ and similarly for the other sets. (Until this point, $r=1$ is not used) Since $r=1$ and $b-1,b,b+1,b+2,\dots,b+i-1$ all don't belong to $C$, $C$ is empty. Contradiction. Now, we can replicate the same proof for $r=2$ up until the previous paragraph. In this case, no number that has the same parity as $b$ can be in $C$. If $C$ is every number of the different parity, then $A$ and $B$ are $2$-disjoint, absurd. Thus, of course there are values $A$ that are different parity from $B$. But since $b-2,b+2,\dots,b+2i-2$ are all in $A$, this different-parity number in $A$ must be adjacent to some other number in $A$, and we reduce to $r=1$. Thus, any two numbers in the same set differ by at least three, showing our claim. This is obviously a solution on the integers, but on the rationals, for each $k$, we are forced to use the same method on \[\frac{0}{k},\frac{1}{k},\frac2k,\dots\]so $0/6$ and $3/6$ are the same set but $0/2$ and $1/2$ are clearly not. Contradiction, no solution over $\mathbb Q$.

For (a) notice that splitting numbers by remainder modulo $3$ suffices. I claim that this is the only way to partition $\mathbb{Z}$. Let $x$ be the largest constant such that the probability a randomly chosen integer is in $A$ is at least $x$. Define $y$ and $z$ similarly. (Density?) Notice that $x+y+z=1$. We also get \[S=\max\{x,y\}+\max\{y,z\}+\max\{z,x\}\le 1\implies S=1\implies x=y=z=\frac{1}{3}.\] Take an element $a\in A$. Notice that \[a+B\subseteq A+B\]\[C+a\subseteq C+A\]thus if $s\in B+C$ then $s\not\in a+B$ and $s\not\in C+a$ which implies $s\in a+A$ or $s-a\in A$. Take $b_1$ and $b_2$ in $B$ and take $c\in C$. Then \[b_1+c-a\in A\]\[b_2+c-(b_1+c-a)=(b_2-b_1)+a\in A\]which implies that $a+k(b_2-b_1)\in A$ for any integer $k$. Now notice the difference $1$ never appears in any set; that is, elements $e$ and $e+1$ never appear in the same set. Otherwise by the above section every integer would appear in every set. If the difference $2$ never appears in any set, then each group of three consecutive integers has one integer in each set. This yields the modulo $3$ construction. If the difference $2$ appears in some set, then by the lemma we would find one of $x$, $y$, and $z$ to be at least $\frac{1}{2}$, a contradiction. Now if we could extend to $\mathbb{Q}$ then note $0$ and $1$ are in different sets and yet $\frac{0}{3}$ and $\frac{3}{3}$ should be in the same set, a contradiction. $\blacksquare$

WHAT!!!!!