IMO 1987 - function such that $f(f(n))=n+1987$

Question

Show that there is no function $f: \mathbb{N} \to \mathbb{N}$ such that $$f(f(n))=n+1987, \ \forall n \in \mathbb{N}$$

This is problem 4 from IMO 1987 - see, for example, AoPS.

Answer 1

Here is an alternative approach. It is obvious that such an $f$ must be an injection. Now, look at sets $\mathbb{N}$, $A = f(\mathbb{N})$ and $B = f(f(\mathbb{N})) = \{n + 1987 \ | \ n \in \mathbb{N}\}$.

It is easy to see that $B \subset A \subset \mathbb{N}$, and also from the injectivity of $f$ that $f$ induces a bijection between the disjoint sets $\mathbb{N} \setminus A$ and $A \setminus B$. Therefore, $\mathbb{N} \setminus B = (\mathbb{N} \setminus A) \cup (A \setminus B)$ must contain an even number of elements. But $|\mathbb{N} \setminus B| = 1987$, which is a contradiction, and we are done.

Answer 2

Hint $\ $mod $\rm\,p\!:\ f(f(n)) = n+p\equiv n\:$ is an involution, hence $\rm\:p\:$ odd $\Rightarrow$ that it has a fixed point $\rm\:f(n)\equiv n,\:$ so $\rm\:f(n) = n\!+\!k p,\ k\in\Bbb Z.\:$ Now the orbit of $\rm\:f\:$ on $\rm\,n\,$ yields a contradiction

$$\rm\begin{array}{rlcl} &\rm n &\rm \to &\rm \color{#C00}{n+kp} \\ \to &\rm \color{#0A0}{n\!+\!p} &\rm \to &\rm n\!+\!(k\!+\!1)p \\ \to &\rm n\!+\!2p &\rm \to &\rm n\!+\!(k\!+\!2)p\\ \to &\rm n\!+\!3p &\rm \to &\rm n\!+\!(k\!+\!3)p\\ &\rm\ \cdots &\rm &\rm \quad\cdots \\ \to &\rm \color{#C00}{n\!+\!kp} &\rm \to &\rm n\!+\!(k\!+\!k)p = \color{#0A0}{n\!+\!p}\ \Rightarrow\ 2k=1\ \Rightarrow\Leftarrow\: k\in\Bbb Z\\ \end{array}$$

Remark $\ $ This was a hint I posted many years ago in another math forum. You can find many further posts here exploiting parity and involutions by searching on involution and Wilson (group theoretical form of Wilson's theorem).

Answer 3

Just a minor variation of the answer by user58512, for fun; like that answer it works even if we replace $\Bbb N$ by $\Bbb Z$. The equation $f(n+1987)=f(n)+1987$ in that answer means that $$f(n+1987)-(n+1987)=f(n)-n,$$ so it shows that $f(n)-n$ only depends on the class of $n$ modulo $1987$. Also the map $\bar f$ that $f$ induces on $\Bbb Z/1987\Bbb Z$ is surjective (the class of $n$ is in the image of $\bar f^2$) hence bijective: $f(n)$ runs over all congruence classes as $n$ runs over all congruence classes modulo $1987$.

Now let $S$ be the sum of $f(n)-n$ over all those congruence classes, obviously an integer. But computing the sum of $1987=f(f(n))-n=\bigl(f(f(n))-f(n)\bigl)+\bigr(f(n)-n\bigr)$ over all those congruence classes, one gets on one hand the odd number $1987^2$, and on the other hand the even number $2S$; contradiction.

Answer 4

Here is a generalisation.

Proposition. For positive integers $k,l$, the functional equation $$ f^k(n)=n+l\qquad\text{for all $n$} $$ has no solutions $f$ in the functions $\Bbb Z\to\Bbb Z$, or in the functions $\Bbb N\to\Bbb N$, unless $k$ divides $l$.

In the case $k\mid l$, one does of course have solutions, for instance $n\mapsto n+l/k$.

Proof. Assume that $f$ satisfies the functional equation.

• For all $n$ one has $f(n+l)=f^{k+1}(n)=f(n)+l$.
• Therefore $f(n+l)-(n+l)=f(n)-n$, and $f(n)-n$ is constant when $n$ varies over any congruence class modulo $l$.
• The image under $f$ of such a congruence class is therefore another congruence class, and $f$ gives rise to a function $\bar f:\Bbb Z/l\Bbb Z\to\Bbb Z/l\Bbb Z$.
• Since $(\bar f)^k(\bar n)=\bar n$ for all $\bar n\in\Bbb Z/l\Bbb Z$, this function $\bar f$ is a bijection.
• Put $S=\sum_{i=0}^{l-1}(f(i)-i)$; note that since $f(i)-i$ is constant on congruence classes modulo $l$, the same sum is obtained if one sums over a different set of representatives of the congruence classes modulo$~l$.
• In particular, since for any $j\in\Bbb N$ the set $\{\,f^j(i)\mid0\leq i<l\,\}$ is a set of representatives of the congruence classes modulo$~l$ (since $(\bar f)^j$ is a bijection), one has $\sum_{i=0}^{l-1}(f^{j+1}(i)-f^j(i))=S$.
• Also $\{\,f(i)\mid0\leq i<l\,\}$ is set of representatives of the congruence classes modulo$~l$, so $\sum_{i=0}^{l-1}f(i)\equiv\sum_{i=0}^{l-1}i\pmod l$, and hence $S\equiv0\pmod l$. So $S=ls$ for some integer$~s$.
• From the functional equation $\sum_{i=0}^{l-1}\bigl(f^k(i)-i\bigr)=\sum_{i=0}^{l-1}l=l^2$.
• One can decompose each term in that sum as a telescopic sum: $f^k(i)-i=\sum_{j=0}^{k-1}\bigl(f^{j+1}(i)-f^j(i)\bigr)$.
• Therefore $$l^2=\sum_{i=0}^{l-1}\bigl(f^k(i)-i\bigr) =\sum_{j=0}^{k-1}\sum_{i=0}^{l-1}(f^{j+1}(i)-f^j(i)\bigr) =\sum_{j=0}^{k-1}S=kS=kls.$$
• Simplifying by a factor $l$ one gets $l=ks$, proving that $k$ divides $l$.

Answer 5

We prove that if $f(f(n)) = n + k$ for all $n$, where $k$ is a fixed positive integer, then $k$ must be even. If $k = 2h$, then we may take $f(n) = n + h$.

Suppose,

$f(m) = n$ with $m= n (mod $ k $)$.

Then by an easy induction on $r$ we find $f(m + kr) = n + kr, f(n + kr) = m + k(r+1)$.

We show this leads to a contradiction.

Suppose $m < n$,

$n = m + ks$ for some $s > 0$ $\implies f(n) = f(m + ks) = n + ks$.

But $f(n) = m + k, so m = n + k(s - 1) >= n$. Contradiction.

So we must have $m\ge n$, so $m = n + ks$ for some $s \ge 0$.

But now,

$f(m + k) = f(n + k(s+1)) = m + k(s + 2)$.

But $f(m + k) = n + k$, so $n = m + k(s + 1) > n$. Contradiction again.

So if $f(m) = n$, then $m$ and $n$ have different residues $(mod $ k $)$.

Suppose they have $r_1$ and $r_2$ respectively.

Then the same induction shows that:

All sufficiently large $s= r_1 (mod k)$ have $f(s)= r_2(mod k)$,

and that all sufficiently large $s = r_2 (mod k)$ have $f(s) = r_1 (mod k)$.

Hence if $m$ has a different residue $r(mod k)$, then $f(m) cannot have residue $r_1$ or $r_2$.

For if f(m) had residue $r_1$, then the same argument would show that all sufficiently

large numbers with residue $r_1$ had $f(m) = r (mod k)$.

Thus the residues form pairs, so that if a number is congruent to a particular residue, then f of the number is congruent to the pair of the residue. But this is impossible for $k$ odd.