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Assume we have such a partition $ \mathbb{N}=A \cup B\cup C$ in nonempty subsets, such that $ 1\in A$ and the minimal $ n\in \mathbb{N}\setminus A$ lies in $ B$. Then $ n\ge 2$. 1) $ x,y,x+y$ can't lie in all three different subsets for any $ x,y\in \mathbb{N}$: Assume $ x\in A, y\in B, x+y\in C$, then by assumption $ z: =x^2-x(x+y)+(x+y)^2=y^2-y(x+y)+(x+y)^2$ lies in $ B$ and in $ A$. Contradiction! 2) $ n\ge 3$: If $ n=2$, then $ 1\in A,2\in B$ gives $ 3=1^2-1\cdot 2+2^2\in C$. But then $ 3=1+2$ contradicts 1). 3) We find $ k,l\in C$ with $ 0<l-k<n$: By 2) we have $ 1,2\in A, n\in B$, so for $ n>3$ we can take $ k: =n^2-2n+2^2,l: =n^2-n+1^2\in C$ with $ l-k=n-3$, so $ 0<l-k<n$. For $ n=3$ we have $ 1,2\in A$, $ 3\in B$, so $ 7=3^2-3\cdot 1+1^2\in C$. Then by 1) we have $ 4=1+3\not\in C$ and $ 4=7-3\not \in A$. So $ 4\in B$ and we can take $ k: = 4^2-4\cdot 2+2^2=12\in C$ and $ l: = 4^2-4\cdot 1+1^2=13\in C$. Finally choose $ (k,l)\in C^2$ with $ 0<l-k<n$ and $ k$ minimal. By minimal choice of $ n$ we have $ l>n$, so $ m: =l-n\in \mathbb{N}$. Also $ k-m=n-(l-k)$ and $ 0<l-k<n$ gives $ 0<k-m<n$, and again by minimal choice of $ n$ we have $ k-m\in A$. Together with $ n\in B$ and $ k,l\in C$ this gives $ m=l-n\not\in A$ and $ m=k-(k-m)\not \in B$ using 1). So we find $ (m,k)\in C^2$ with $ 0<k-m<n$, which contradicts the minimal choice of $ k$.

We can finish it in another way: let $a$ be smallest which is in set which not consists 1. Assume $1 \in A, a\in B$. We choose partition with smallest possible value of $a$. Now we see that $a^2-a+1$ is in $C$. So $a^2-a\not\in B$ (because $1, a^2-a,a^2-a+1$ aren't in three different subsets, like olorin showed). Now we make bijection $ka \rightarrow k$ and because $a \in B, a^2-a\not\in B$ we get contradiction.

@RaleD, what do you mean by "make bijection $ka \rightarrow k$"? @olorin: nice proof, but you the (WLOG) assumption that $1\in A$ and that the minimal $n\in\mathbb N\setminus A$ lies in $B$ should be done after step 1.

If $x$, $y$, and $x+y$ appear on different sets, then \[A=x^2+(x+y)^2-x(x+y)=y^2+(x+y)^2-y(x+y)\]means that $A$ is in the same set as both $x$ and $y$, contradiction. Thus, $x+y$ must be on the same set as either $x$ or $y$. Let our sets be $S_1$, $S_2$ and $S_3$. WLOG, let $1\in S_1$. Let $m$ be the smallest integer not in $S_1$, and let it be in $S_2$. Suppose $m=2$ then $1^2-1\cdot 2+2^2=3\in S_3$, contradiction. Suppose $m=3$ then $7\in S_3$, and following that, from the sum rule we found previously, $4\in S_2$, $5,6\in S_1$ and then $9,12,13\in S_3$ and so there is no place to put $8$. Thus, $m\ge 4$. Note that $m^2-m+1$ and $m^2-2m+4$ are in $S_3$. Let $k$ be the maximum positive integer such that $m^2-km+1$ and $m^2-(k+1)m+4$ are both in $S_3$. Note that $(m^2-km+1)-(m)$ must not be in $S_1$ and $(m^2-(k+1)m+4)-(3)$ must not be in $S_2$. They are the same number, so $m^2-(k+1)m+1$ must be in $S_3$. Similarly, now, $(m^2-(k+1)m+1)-(m-3)$ must not be in $S_2$ and $(m^2-(k+1)m+4)-(m)$ must not be in $S_1$, and since they are the same number, $m^2-(k+2)m+4$ must be in $S_3$. Contradiction. We are done.

What RaleD means, I think, is that if $a$ is the smallest which CAN be in a set without $1$. (Meaning that any smaller number cannot result in a valid partitioning) I guess the idea is that $x,y,x+y$ not being in three different sets is a homogeneous condition so we can multiply everything by $a$ so theoretically $a$ and $a^2-a$ must still be in the same set. I don't think it works and the reason is that while $x,y,x+y$ is a homogeneous condition, the proof also uses the fact that $a^2-a+1$ is in the other set, which is not homogeneous and will probably change the proof after multiply by another $a$.