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

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.

• Really people? Three and a half years and mine is the first close vote. – MathematicsStudent1122 Nov 20 '16 at 0:01
• @MathematicsStudent1122 It seems that contest questions are taken a bit differently, as discussed here and probably in several other places on meta. I have edited the question to add the origin of the problem to have at least some context. (Although the origin was already clearly pointed out in the title.) – Martin Sleziak Nov 20 '16 at 2:42

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.

• Very interesting solution. should say $f(f(n)) = n + 1987$ at the second last line. – user58512 Mar 9 '13 at 15:33
• @user58512 Thanks for the correction! – Dan Shved Mar 9 '13 at 15:35
• +1 Great answer, mush easier to follow than the other ones. Only I would have said $B=f(f(\Bbb N))=\{\,i+1987\mid i\in\Bbb N\,\}$ right away, which avoids making a case distinction (and also like your answer avoids any commitment to where the natural numbers start exactly ;-). – Marc van Leeuwen Mar 9 '13 at 15:40
• @MarcvanLeeuwen I've updated the answer. Is this what you had in mind? If not, feel free to improve it! – Dan Shved Mar 9 '13 at 15:53
• @DanShved Why is the injectivity obvious? – Kugelblitz Aug 22 '15 at 8:33

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 $S=\sum_{i=0}^{l-1}(f(i)-i)$; then $l$ divides $S$ since $\sum_{i=0}^{l-1}f(i)\equiv\binom l2=\sum_{i=0}^{l-1}i\pmod l$, so put $s=S/l$.
• From the functional equation $\sum_{i=0}^{l-1}\bigl(f^k(i)-i\bigr)=\sum_{i=0}^{l-1}l=l^2$.
• By a telescopic sum, $f^k(i)-i=\sum_{j=0}^{k-1}\bigl(f^{j+1}(i)-f^j(i)\bigr)$.
• Since each $(\bar f)^j$ is a bijection, this implies $$\sum_{i=0}^{l-1}\bigl(f^k(i)-i\bigr) =\sum_{j=0}^{k-1}\sum_{i=0}^{l-1}(f^{j+1}(i)-f^j(i)\bigr) =\sum_{j=0}^{k-1}S=kS=kls.$$
• Therefore $l^2=kls$, so $l=ks$ and $k$ divides $l$.

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).

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.

I answered a minor variation of this question here: Functions such that $f(f(n))=n+2015$, unaware it was an IMO question. My proof might be a little shorter than those given here:

Suppose such a function exists. Then define the function $$g\colon \{1,\cdots,1987\}\to \{1,\cdots,1987\},\qquad g(n)=f(n)\mod{1987}.$$ Then $g$ is an involution of ${1,\cdots,1987}$. Since there are an odd number of elements, it follows that $g$ has a fixed point, say $k$. Then $f(k)=k+1987j$ for some $j$, whereupon $f(f(k))=k+1987\cdot 2j\not=k+1987$, contradiction.

• Wow, nice answer! – Sandeep Silwal Dec 31 '14 at 21:48
• Thanks. The same argument gives a slick proof of Marc van Leeuwen generalization posted above. – pre-kidney Dec 31 '14 at 22:34

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.