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Difference between revisions of "1987 IMO Problems/Problem 4"

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Revision as of 07:43, 19 July 2016

Problem

Prove that there is no function $f$ from the set of non-negative integers into itself such that $f(f(n)) = n + 1987$ for every $n$.

Solution

Solution 1

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 $m \equiv n \mod k$. 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 $m < n$, so $n = m + ks$ for some $s > 0$. Then $f(n) = f(m + ks) = n + ks$. But $f(n) = m + k$, so $m = n + k(s - 1) \ge n$. 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.

So if $f(m) = n$, then $m$ and $n$ have different residues $\pmod k$. Suppose they have $r_1$ and $r_2$ respectively. Then the same induction shows that all sufficiently large $s \equiv r_1 \pmod k$ have $f(s) \equiv r_2 \pmod k$, and that all sufficiently large $s \equiv r_2 \pmod k$ have $f(s) \equiv r_1 \pmod k$. 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 $f(m) \equiv r \pmod k$. 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.

Solution 2

Solution by Sawa Pavlov:

Let $N$ be the set of non-negative integers. Put $A = N - f(N)$ (the set of all $n$ such that we cannot find $m$ with $f(m) = n$). Put $B = f(A)$.

Note that $f$ is injective because if $f(n) = f(m)$, then $f(f(n)) = f(f(m))$ so $m = n$. We claim that $B = f(N) - f(f(N))$. Obviously $B$ is a subset of $f(N)$ and if $k$ belongs to $B$, then it does not belong to $f(f(N))$ since $f$ is injective. Similarly, a member of $f(f(N))$ cannot belong to $B$.

Clearly $A$ and $B$ are disjoint. They have union $N - f(f(N))$ which is $\{0, 1, 2, \ldots , 1986\}$. But since $f$ is injective they have the same number of elements, which is impossible since $\{0, 1, \ldots , 1986\}$ has an odd number of elements.

Solution 3

Consider the function $g: \mathbb{Z}_{1987} \rightarrow \mathbb{Z}_{1987}$ defined by $g(x) = f(x\; {\rm mod }\; 1987) \;{\rm mod }\; 1987$. Notice that we have $f(k) + 1987 = f(f(f(k))) = f(k + 1987)$, so that $f(x) = f(y) \;{\rm mod }\; 1987$ whenever $x = y \;{\rm mod }\; 1987$, and hence $g$ is well defined.

Now, we observe that $g$ satisfies the identity $g(g(x)) = x$, for $x \in Z_{1987}$. Thus, $g$ is an invertible function on a finite set of odd size, and hence must have a fixed point, say $a \in \mathbb{Z}_{1987}$. Identifying $a$ with its canonical representative in $\mathbb{Z}$, we therefore get $f(a) = a + 1987k$ for some non-negative integer $k$.

However, we then have $f(f(a)) = a + 1987$, while $f(f(a)) = f(a + 1987k) = 1987k + f(a) = 2\cdot1987k + a$ (where we use the identity $f(x + 1987) = f(x) + 1987$ derived above, along with $f(a) = f(a) + 1987$. However, these two equations imply that $k = \dfrac{1}{2}$, which is a contradiction since $k$ is an integer. Thus, such an $f$ cannot exist.

Note: The main step in the proof above is that the function $g$ can be shown to have a fixed point. This step works even if 1987 is replaced with any other odd number larger than 1. However, for any even number $c$, $f(x) = x + \dfrac{c}{2}$ satisfies the condition $f(f(x)) = x + c$.

--Mahamaya 21:15, 21 May 2012 (EDT)

1987 IMO (Problems) • Resources
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Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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