Denote $P(m,n)$ the assertion that $f(m^2+mf(n))=mf(m+n)$. $P(0,n)$ shows that $f(0)=0$. $P(m,0)$ implies $f(m^2)=mf(m)$. Assume that $f$ is injective. Then $P(1,n)$ implies $f(f(n)+1)=f(n+1)$ i.e. $f(n)+1=n+1$ i.e. $f(n)=n$ for all $n$ which is indeed a solution. Assume now that $f$ is not injective i.e. there are $a \ne b$ such that $f(a)=f(b)$. Comparing $P(m,a)$ and $P(m,b)$ for $m \ge 1$ yields $f(m+a)=f(m+b)$ i.e. $f$ is eventually periodic. But in particular this implies that $f$ is bounded i.e. $f(n)<C$ for some constant $C$. But then $P(C,n)$ implies $C>f(C^2+Cf(n))=Cf(C+n)$ i.e. $f(C+n)<1$ i.e. $f(C+n)=0$ for all $n \ge 0$ i.e. $f$ is eventually just zero. In particular we then have $f(C)=0$ and hence $P(m,C)$ implies $mf(m)=f(m^2)=f(m^2+mf(C))=mf(m+C)$ and hence for all $m \ge 1 $it holds $f(m)=f(m+C)$ i.e. $f$ is periodic and thus identically zero which is also a solution. Altogether, we have two solution functions: $f(x) \equiv x$ and $f(x) \equiv 0$.

Sorry for no latex but here is another solution. Note f(0) = 0 Now, assume there is some a for which f(a) = 0 Thus, f(m^2) = mf(m+a) note that for n=0 f(m^2) = mf(m) Thus, if such a exists f(x) = 0 for all x If such a does not exist, only f(0) = 0 set m = -n f(n^2 -nf(n)) = 0 implies f(n) = n Thus, only solutions are f(x)=0 and f(x)=x

TheOneYouWant wrote: ...assume there is some $a$ for which $f(a) = 0$ Thus, $f(m^2) = mf(m+a)$ note that for $n=0$ $f(m^2) = mf(m)$ Thus, if such $a$ exists $f(x) = 0$ for all $x$ This conclusion is not true (or at least it is not directly clear). You can only derive $f(m+a)=f(m)$ which does not imply $f$ constant but rather $f$ periodic.

EDIT: we can proceed as Tintarn has done, i.e. F(x) < C to prove f is identically zero.

Let $P(m,n)$ denote the assertion that $f(m^2+mf(n))=mf(m+n)$, for all $m,n\in \mathbb{N}\cup\{0\}$. $\blacksquare$ $P(0,0) \implies f(0)=0$ $\blacksquare$ $P(m,0) \implies f(m^2)=mf(m)$ $\blacksquare$ Assume $f $ is not identically zero. We will establish injectivity of $f$. Say , $\exists a,b \in \mathbb{N}\cup\{0\}$ with $a < b$ so that $f(a)=f(b) $.(different from zero). $P(1,a)$ and $P(1,b)$ give that $f(1+a)=f(1+b)$.So, a simple induction gives $f(k+a)=f(k+b) $ for all $k\in \mathbb{N}\cup\{0\}$. Using this , we may obtain ( again after a simple induction) that $f(ka)=f(kb)$ for all $k\in \mathbb{N}\cup\{0\}$. So , $f(a^2)=f(ab)=f(b^2) \implies f(a^2)=f(b^2) \implies af(a)=bf(b) \implies a=b$ ( If $ab$ is different from zero) . This contradicts assumption and establishes injectivity for this case . If $ab=0$ then repeat the argument with $a+y$ and $b+y$ where $y>max{( |a| , |b| )}$ . We get injectivity here too. To conclude, $P(1,n)$ and injectivity imply $f(a)=a$ for all $a\in \mathbb{N}\cup\{0\}$. This indeed satisfies condition . Now, include $f \equiv 0$ as the other solution .

Click to reveal hidden text

Denote $P(m,n)$ as the assertion that $f(m^2+mf(n))=mf(m+n)$ for $m,n \in \mathbb{N} \cup \{0\}$ $f(x) \equiv 0$ is a trivial solution. Assume that there exists an $n$ such that $f(n) \not= 0$. $P(0,0)$ gives $f(0)=0$ and $P(m,0)$ gives $f(m^2)=mf(m)$. I claim that $f$ is injective. Suppose otherwise and let $f(a)=f(b)$, $a<b$. Now $P(m,a)$ and $P(m,b)$ gives us $$mf(m+a)=f(m^2+mf(a))=f(m^2+mf(b)) = mf(m+b) \implies f(m+a)=f(m+b)\text{ } \forall m \in \mathbb{N}$$so combine with $f(a)=f(b)$ to get $f(k)=f(k+b-a)$ for all $k \ge a$. This implies that $f$ is periodic starting from $f(a)$. This implies that $f$ is bounded above by $\text{max}(f(0),f(1),f(2), \cdots f(b))$. Meanwhile, assume that for all $N \in \mathbb{N}$, there exists an $n>N$ such that $f(n) \not= 0$. This gives us that $f(n^2) = nf(n) \ge n > N$, which contradicts the bounding argument above. Therefore, there exists an $N$ such that $f(n)=0$ for all $n > N$. As $f$ is periodic starting from $f(a)$, we easily get that $f(n)=0$ for all $n \ge a$. Now $$mf(m)=f(m^2)=f(m^2+mf(a)) = mf(m+a) = 0 \implies f(m)=0 \text{ } \forall m \in \mathbb{N}$$so combine this with $f(0)=0$ to get that $f(n)=0$ for all $n \in \mathbb{N} \cup \{0\}$, a contradiction. Therefore, $f$ is injective. Take $P(1,n)$, which gives us $f(f(n)+1)=f(n+1)$, so $f(n)=n$. This is another solution. Therefore, the answer is $\boxed{f(n) = 0 \text{ }\forall n \in \mathbb{N} \cup \{0\}}$ and $\boxed{f(n) = n \text{ }\forall n \in \mathbb{N} \cup \{0\}}$. $\blacksquare$

f(m)=0 is a solution lets check for others. First notice that f is strictly increasing function as f(m^2+mf(n))>f(m +n) and m^2 +mf(n)>m+ n. Now, Let P(m ,n) be the assertion. P(0,0) gives that f(0)=0. Claim f is injective : Let a ,b be two numbers such that f(a)=f(b) Subtracting P(1,a) from P(1,b) gives: f(1+f(a))=f(1+b) now if f(a) is not =b then it will contradict the increasing property of f as one them will get bigger than other. Therefore f is injective. P(1,m) gives f(1+f(n))=f(1+n) and then using injectivity we get: f(n)=n . So in all the functions satisfying this equation is f(n)=0 or f(n)=0

Solution: Only working functions are $f \equiv 0$ and $f \equiv \mathrm{id}$. They work without a doubt, we will now show their uniqueness. Let $P(m,n)$ be the assertion to the functional equation. If $f$ is an injective function, then $P(1,n)$ would show that $f \equiv \mathrm{id}$. For the rest of the solution, assume that $f$ is a non-injective function. Let $f(a) = f(b)$ for $a>b$ since $f$ is not injective. $P(m,a)$ and $P(m,b)$ reveals that $f$ is periodic with period $T = a-b$. Finally $P(T,n)$ gives us that \[f(T^2 + Tf(n)) = Tf(n+T) \iff f(n) = \frac{f(T^2)}{T}.\]So for non-injective $f$, it must be constant. This concludes the solution as the only possible constant solution is $f \equiv 0$. $\blacksquare$

The answer is $f(x)=x$ and $f \equiv 0$ only, which both work. Let $P(m,n)$ denote the assertion. $P(0,n)$ implies that $f(0)=0$. Suppose that $f(a)=f(a+d)$ for some $d>0$. Then by comparing $P(m,a)$ with $P(m,a+d)$ for $m \neq 0$ implies that $P(m+a)=P(m+a+d)$, hence $P$ is periodic with period $d$ from $a$ onwards, hence bounded. On the other hand, by plugging in a large $m$ and letting $n$ vary we get that $f$ must be eventually $0$ for size reasons. It follows that $f$ is periodic with period $N$ for all large enough $N$, hence it's identically $0$, which works. Otherwise, $P(1,n)$ implies $f(f(n)+1)=f(n+1) \implies f(n)=n$, which is the other solution. $\blacksquare$

hajimbrak wrote: Find all functions from $\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ such that $f(m^2+mf(n))=mf(m+n)$, for all $m,n\in \mathbb{N}\cup\{0\}$. Let $P(m,n)$ be the assertion that $f(m^2+mf(n)) = mf(m+n)$ $P(0,n)\implies f(0) = 0$ $P(m,0)\implies f(m^2) = mf(m)$ $P(1,n)\implies f(f(n) + 1) = f(n+1)$ $P(m,f(n)+1)\implies f(m+f(n)+1) = f(m+n+1)\quad\forall m,n\in\mathbb N$ In other words, $f(m) = f(m + |f(n) - n|)$ for all $m > \text{min}(n,f(n))$ Let $n\in\mathbb N$, and $m>n$ $P(m + |f(n)-n|, 0)\implies |f(n)-n|f(m) = 0$ But if $f(m) = 0\quad\forall m>n$ then repeated application of $f(n) = f(n^2)/n$ yields that $f$ vanishes everywhere. So either $\boxed{f(n) = 0\quad\forall n}$ or $\boxed{f(n)=n\quad\forall n}$