Quidditch wrote: Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfying $\text{(i)}$ $f(f(m)+n)+2m=f(n)+f(3m)$ for every $m,n\in\mathbb{Z}$. $\text{(ii)}$ there exists a $d\in\mathbb{Z}$ such that $f(d)-f(0)=2$, and $\text{(iii)}$ $f(1)-f(0)$ is even. Let $P(x,y)$ be the assertion $f(f(x)+y)+2x=f(y)+f(3x)$ Let $a=f(0)$ and $c=f(1)$ $P(x,0)$ $\implies$ $f(3x)=f(f(x))+2x-a$ and so $P(x,y)$ implies New assertion $Q(x,y)$ : $f(f(x)+y)=f(y)+f(f(x))-a$ $P(x,3x-f(x))$ $\implies$ $f(3x-f(x))=2x$ and so $2\mathbb Z\subseteq f(\mathbb Z)$ So $Q(x,y)$ implies new assertion $R(x,y)$ : $f(2x+y)=f(y)+f(2x)-a$ $R(x,2y)$ $\implies$ $f(2x+2y)=f(2y)+f(2x)-a$ and so $f(x)-a$ is additive over $2\mathbb Z$, and so linear and so $f(2x)=2kx+a$ for some integer $k$ $R(x,1)$ $\implies$ $f(2x+1)=f(2x)+c-a$ $=2kx+c$ Note trhat since $c-a$ is even, $f(2x)$ and $f(2x+1)$ both are even, or both are odd And then $P(x,y)$ implies that $f(x)$ is even $\forall x$ and so both $a,c$ are even. Since $a$ is even, $P(2x,0)$ $\implies$ $(2k^2-6k+4)x+(k-1)a=0$ and so $k=1$ And so $\boxed{f(2n)=2n+2u\text{ and }f(2n+1)=2n+2v\quad\forall n\in\mathbb Z}$, which indeed fits, whatever are $u,v\in\mathbb Z$

pco wrote: And so $\boxed{f(2n)=2n+2u\text{ and }f(2n+1)=2n+2v\quad\forall n\in\mathbb Z}$, which indeed fits, whatever are $u,v\in\mathbb Z$ Uhhh? What about $f(n)=2n$ ?

Direct solution but less efficient. For any integers $m$ and $n$, let $P(m, n)$ denote the statement $f(f(m) + n) + 2m = f(n) + f(3m).$ First, consider $P(0, n)$. It follows that $f(n + f(0)) = f(n) + f(0)$ for every integer $n$. From the second condition, suppose that $d$ is an integer such that $f(d) = 2 + f(0)$. Plugging this into $P(d, n)$, we have $$f(2 + f(0) + n) + 2d = f(n) + f(3d) \implies f(n + 2) - f(n) = f(3d) - 2d - f(0).$$Since $f(3d) - 2d - f(0)$ is constant, it directly implies that $$f(n + 2) - f(n) = f(2) - f(0).$$By using mathematical induction, we conclude that $$ f(n) = \begin{cases} f(1) + \frac{1}{2}(n - 1)\bigl(f(2) - f(0)\bigr), & \text{if } n \text{ is odd,} \\ f(2) + \frac{1}{2}(n - 2)\bigl(f(2) - f(0)\bigr), & \text{if } n \text{ is even.} \end{cases} $$Thus, we find that $f(n) = an + b$ for odd integers $n$, and $f(n) = an + f(0)$ for even integers $n$, where $a, b \in \mathbb{Q}$ are constant values: $$a \coloneqq \frac{f(2) - f(0)}{2}, \quad b \coloneqq f(1) - \frac{1}{2}\bigl(f(2) - f(0)\bigr).$$Suppose $x$ is any odd integer. Consider $P(x, n)$: $$f(ax + b + n) + 2x = f(n) + 3ax + b.$$Notice that there exist infinitely many integers $n$ such that $ax + b + n$ is odd. For such $n$, substituting yields $$a(ax + b + n) + b + 2x = f(n) + 3ax + b.$$If these $n$ are odd, then $$a^2x + ab + an + 2x = an + 3ax + b\implies (a^2 - 3a + 2)x = b - ab.$$Suppose, for contradiction, that $a^2 - 3a + 2 \neq 0$. We can choose a sufficiently large $x$ such that $$|x| > \left|\frac{b - ab}{a^2 - 3a + 2}\right|,$$leading to a contradiction. Hence, $a^2 - 3a + 2 = 0$. Similarly, if such $n$ are even, the same result holds, yielding $a^2 - 3a + 2 = 0$. Thus, $a \in \{1, 2\}$ and $b$ is an integer. Case 1: If $a = 1$, plugging in $P(1, 0)$ gives $f(b + 1) + 2 = f(0) + b + 3$. Observe that $f(1) - f(0) = 1 + b - f(0) \equiv b + 3 + f(0) \pmod{2}$. Thus, $2 \mid f(b + 1)$. If $b + 1$ is odd, $f(b + 1) = 2b + 1 \equiv 0 \pmod{2}$, which is impossible. Therefore, $b + 1$ is even, so $b$ is odd and $f(0)$ is even. In conclusion, $$f(x) = \begin{cases} x + 2k + 1, & \text{if } x \text{ is odd,} \\ x + 2l, & \text{if } x \text{ is even,} \end{cases}$$for some integer constants $k$ and $l$. Case 2: If $a = 2$, plugging in $P(1, 0)$ gives $f(b + 2) + 4 = f(0) + b + 6$. Observe that $f(1) - f(0) = b + 2 - f(0) \equiv b + 2 + f(0) \pmod{2}$. Thus, $2 \mid f(b + 2)$. If $b$ is odd, then $f(b + 2) = 2b + 4 + b$, so $2 \mid b$, which is impossible. Thus, $b$ is even, and $f(0)$ is also even. Plugging in $P(0, 0)$ gives $f(f(0)) = 2f(0) \implies 3f(0) = 2f(0)$, so $f(0) = 0$. Plugging in $P(1, 1)$ gives $f(b + 3) = 2b + 6 \implies 3b + 6 = 2b + 6$, so $b = 0$. In conclusion, $f(x) = 2x$ for all integers $x$. Verification Step: First Solution: $m$ is odd and $n$ is odd: $f(m + 2k + 1 + n) + 2m = 3m + n + 2(2k + 1) = f(n) + f(3m)$. $m$ is odd and $n$ is even: $f(m + 2k + 1 + n) + 2m = 3m + n + 2k + 1 + 2l = f(n) + f(3m)$. $m$ is even and $n$ is odd: $f(m + 2k + 1 + n) + 2m = 3m + n + 2k + 1 + 2l = f(n) + f(3m)$. $m$ is even and $n$ is even: $f(m + 2k + 1 + n) + 2m = 3m + n + 4l = f(n) + f(3m)$. Notice that $f(1+2l-2k)-f(0)=1+2l-2k+2k+1-2l=2$ and $f(1)-f(0)=2k-2l+2$, it is clear to be an even integer. Second Solution: $\mathrm{LHS}=f(2m+n)+2m=6m+2n=f(n)+f(3m)=\mathrm{RHS}$ for every integers $m$ and $n$. And obviously, $f(1)-f(0)=2$, in either way, it is also an even integer. In conclusion, the only functions that satisfy the conditions are: $$f(x) = \begin{cases} x + 2k + 1, & \text{if } x \text{ is odd,} \\ x + 2l, & \text{if } x \text{ is even,} \end{cases} \text{ for all }x\in\mathbb{Z}$$and $f(x)=2x$ for all $x\in\mathbb{Z}$.