Prove that there is a function $f$ from the set of all natural numbers into itself such that $f(f(n))=n^2$ for all $n \in \mathbb{N}$.

Find all surjective functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: \[m \vert n \Longleftrightarrow f(m) \vert f(n).\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(n+1) > f(f(n)).\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(m)+f(n))=m+n.\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f^{(19)}(n)+97f(n)=98n+232.\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(n))+f(n)=2n+2001 \text{ or }2n+2002.\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+6f(n)=3f(f(n))+4n+2001.\]

Find all functions $f: \mathbb{N}_{0}\rightarrow \mathbb{N}_{0}$ such that for all $n\in \mathbb{N}_{0}$: \[f(f(n))+f(n)=2n+6.\]

Find all functions $f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $n\in \mathbb{N}_{0}$: \[f(m+f(n))=f(f(m))+f(n).\]

Find all functions $f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $m,n\in \mathbb{N}_{0}$: \[mf(n)+nf(m)=(m+n)f(m^{2}+n^{2}).\]

Find all functions $f:\mathbb{N} \to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: $f(2)=2$, $f(mn)=f(m)f(n)$, $f(n+1)>f(n)$.

Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m\in \mathbb{Z}$: \[f(f(m))=m+1.\]

Click for solution There are no such functions. Indeed, plugging $m=f(l)$ in our equation we get \[f(f(f(l)))=f(l+1)\] but on the other hand \[f(f(f(l)))=f(l)+1\] and thus \[f(l+1)=f(l)+1\] which means that our function has form \[f(l)=l+k\] where $k\in N$. Plugging this into the equation we obtain $2\cdot k=1$, which is contradiction.

Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that for all $m\in\mathbb{Z}$: $f(m+8) \le f(m)+8$, $f(m+11) \ge f(m)+11$.

Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m,n\in \mathbb{Z}$: \[f(m+f(n))=f(m)-n.\]

Find all functions $f: \mathbb{Z}\to \mathbb{Z}$ such that for all $m,n\in \mathbb{Z}$: \[f(m+f(n)) = f(m)+n.\]

Click for solution $ m = n = 0$ gives $ f(f(0)) = f(0)$. $ n = 0$ gives $ f(m+f(0)) = f(m)$, but $ n = f(0)$ also gives $ f(m+f(0)) = f(m)+f(0)$. From the previous two statements we conclude that $ f(0) = 0$. Then $ m = 0$ gives $ f(f(n)) = n$. Suppose $ f(k) = 0$ for some $ k$. Then $ n = k$ gives $ f(m) = f(m)+k$, so $ k = 0$. Now put $ m = f(1), n =-1$ in to get $ f(f(1)+f(-1)) = 0$, so $ f(1) =-f(-1)$. We now prove by induction that $ f(k) = kf(1)$ for all positive integers $ k$. Base case $ k = 1$ is trivial. Suppose it is true for $ k$. Then $ m = k, n = f(1)$ gives $ f(k+1) = (k+1)f(1)$, completing the induction. We also prove by induction that $ f(-k) =-kf(1)$ for all positive integers $ k$. Base case $ k = 1$ has already been established. Suppose it is true for $ k$. Then $ m =-k, n = f(-1)$ gives $ f(-k-1) =-kf(1)+f(-1) = (-k-1)f(1)$, completing the induction. So we have $ f(k) = kf(1)$ for all integers $ k$. Let $ k = f(1)$, and we get $ f(1)^{2}= f(f(1)) = 1$. So $ f(1) = 1$ or $ f(1) =-1$. We see that $ f(k) = k$ and $ f(k) =-k$ both work, and they are the only solutions.

Find all functions $h: \mathbb{Z}\to \mathbb{Z}$ such that for all $x,y\in \mathbb{Z}$: \[h(x+y)+h(xy)=h(x)h(y)+1.\]

Find all functions $f: \mathbb{Q}\to \mathbb{R}$ such that for all $x,y\in \mathbb{Q}$: \[f(xy)=f(x)f(y)-f(x+y)+1.\]

Find all functions $f: \mathbb{Q}^{+}\to \mathbb{Q}^{+}$ such that for all $x,y \in \mathbb{Q}$: \[f \left( x+\frac{y}{x}\right) =f(x)+\frac{f(y)}{f(x)}+2y, \; x,y \in \mathbb{Q}^{+}.\]

Find all functions $f: \mathbb{Q}\to \mathbb{Q}$ such that for all $x,y \in \mathbb{Q}$: \[f(x+y)+f(x-y)=2(f(x)+f(y)).\]

Find all functions $f:\mathbb{Q}^{+} \to \mathbb{Q}^{+}$ such that for all $x\in \mathbb{Q}^+$: $f(x+1)=f(x)+1$, $f(x^2)=f(x)^2$.

Let ${\mathbb Q}^{+}$ be the set of positive rational numbers. Construct a function $f:{\mathbb Q}^{+}\rightarrow{\mathbb Q}^{+}$ such that \[f(xf(y)) = \frac{f(x)}{y}\] for all $x, y \in{\mathbb Q}^{+}$.

A function $f$ is defined on the positive integers by \[\left\{\begin{array}{rcl}f(1) &=& 1, \\ f(3) &=& 3, \\ f(2n) &=& f(n), \\ f(4n+1) &=& 2f(2n+1)-f(n), \\ f(4n+3) &=& 3f(2n+1)-2f(n), \end{array}\right.\] for all positive integers $n$. Determine the number of positive integers $n$, less than or equal to 1988, for which $f(n) = n$.

Consider all functions $f:\mathbb{N}\to\mathbb{N}$ satisfying $f(t^2 f(s)) = s(f(t))^2$ for all $s$ and $t$ in $N$. Determine the least possible value of $f(1998)$.

The function $f: \mathbb{N}\to\mathbb{N}_{0}$ satisfies for all $m,n\in\mathbb{N}$: \[f(m+n)-f(m)-f(n)=0\text{ or }1, \; f(2)=0, \; f(3)>0, \; \text{ and }f(9999)=3333.\] Determine $f(1982)$.

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: \[f(f(m)+f(n))=m+n.\]

Find all surjective functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(n) \ge n+(-1)^{n}.\]

Find all functions $ f: \mathbb{Z}\setminus\{0\}\to \mathbb{Q}$ such that for all $ x,y \in \mathbb{Z}\setminus\{0\}$: \[ f \left( \frac{x+y}{3}\right) =\frac{f(x)+f(y)}{2}, \; \; x, y \in \mathbb{Z}\setminus\{0\}\]

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\]

Click for solution Peter wrote: Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\] We have that $f(f(f(1)))+f((1))+f(1)=3$,but every positive integer number is greater or equal to $1$,therefore $f(1)=1$. Let's suppose that we have proved that $f(k)=k$ for all $n\geq{k}$, and let's prove that $f(n+1)=n+1$. If $f(a)=f(b)$ then $3a=f(f(f(a)))+f(f(a))+f(a)=f(f(f(b)))+f(f(b))+f(b)=3b$ so $a=b$. We also have that $f(f(f(n+1)))+f(f(n+1))+f(n+1)=3(n+1)$,but if $n\geq{t}=f(n+1)$ then $f(n+1)=t=f(t)$ $\Rightarrow$ $n+1=t$,contradiction. Therefore, $f(n+1)\geq(n+1)$. Same way if $n\geq{t}=f(f(n+1))$ $\Rightarrow$ $f(f(n+1))=t=f(t)$ $\Rightarrow$ $f(n+1)=f(t)$ $\Rightarrow$ $n+1=t$ contradiction. And again $f(f(n+1))\geq{n+1}$. In the similar way we can prove that $f(f(f (n+1)))\geq{n+1}$. So we get that \[f(n+1)\geq{n+1}\] \[f(f(n+1))\geq{n+1}\] \[f(f(f(n+1)))\geq{n+1}\] and consequently $3(n+1)=f(f(f(n+1)))+f(f(n+1))+f(n+1)\geq3(n+1)$ $\Rightarrow$ $f(n+1)=n+1$. $\Rightarrow$ \[f(n)=n\] for all $n\in{N}$.

Find all strictly increasing functions $f: \mathbb{N}\to \mathbb{N}$ such that \[f(f(n))=3n.\]

Find all functions $f: \mathbb{Z}^{2}\to \mathbb{R}^{+}$ such that for all $i, j \in \mathbb{Z}$: \[f(i,j)=\frac{f(i+1, j)+f(i,j+1)+f(i-1,j)+f(i,j-1)}{4}.\]

Find all functions $f: \mathbb{Q}\to \mathbb{Q}$ such that for all $x,y,z \in \mathbb{Q}$: \[f(x+y+z)+f(x-y)+f(y-z)+f(z-x)=3f(x)+3f(y)+3f(z).\]

Show that there exists a bijective function $ f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $ m,n\in \mathbb{N}_{0}$: \[ f(3mn+m+n)=4f(m)f(n)+f(m)+f(n). \]