No Function with $f(f(n)) = n + 2025$ — Answer

Let $c = 2025$ for convenience. Suppose for contradiction that $f: \mathbb{Z} \to \mathbb{Z}$ satisfies $f(f(n)) = n + c$ for all $n$.

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Step 1: Periodicity — $f(n + c) = f(n) + c$

Apply $f$ to both sides of $f(f(n)) = n + c$: $f(f(f(n))) = f(n + c)$

But also, applying the functional equation to $f(n)$ in place of $n$: $f(f(f(n))) = f(n) + c$

Combining: $f(n + c) = f(n) + c$ for all $n \in \mathbb{Z}$.

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Step 2: $f$ descends to a map $\bar{f}$ on $\mathbb{Z}/c\mathbb{Z}$

Since $f(n+c) = f(n) + c$, the residue $f(n) \bmod c$ depends only on $n \bmod c$. So define: $\bar{f}: \mathbb{Z}/c\mathbb{Z} \to \mathbb{Z}/c\mathbb{Z}, \qquad \bar{f}(n \bmod c) = f(n) \bmod c$

This is well-defined. From $f(f(n)) = n + c \equiv n \pmod{c}$, we get: $\bar{f}(\bar{f}(n)) = n \quad \text{in } \mathbb{Z}/c\mathbb{Z}$

So $\bar{f}$ is an involution ($\bar{f}^2 = \mathrm{id}$), hence a bijection.

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Step 3: $\bar{f}$ has no fixed points

Suppose $\bar{f}(k) = k$ for some $k \in \mathbb{Z}/c\mathbb{Z}$, i.e., $f(n) \equiv n \pmod{c}$ for some $n$ with $n \equiv k$.

Then $f(n) = n + mc$ for some integer $m$. Applying $f$ again using Step 1: $f(f(n)) = f(n + mc) = f(n) + mc = n + 2mc$

But $f(f(n)) = n + c$, so $2mc = c$, giving $m = \tfrac{1}{2}$ — not an integer. Contradiction.

So $\bar{f}$ is a fixed-point-free involution on $\mathbb{Z}/c\mathbb{Z}$.

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Step 4: Contradiction

A fixed-point-free involution partitions every element into a 2-element orbit $\{k,\, \bar{f}(k)\}$ (since $\bar{f}(k) \neq k$ and $\bar{f}(\bar{f}(k)) = k$). So the orbits partition $\mathbb{Z}/c\mathbb{Z}$ into pairs, requiring $|\mathbb{Z}/c\mathbb{Z}| = c$ to be even.

But $c = 2025 = 45^2$ is odd. Contradiction. $\blacksquare$

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Why 2025 specifically?

The proof works for any odd $c$. For even $c$, no contradiction arises — and indeed $f(f(n)) = n + 2$ has a solution: $f(n) = n + 1$.

The key is not any special property of 2025 beyond its being odd.

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Summary of the argument chain

$f(f(n)) = n + c \implies f(n+c) = f(n)+c \implies \bar{f} \text{ is a fixed-point-free involution on } \mathbb{Z}/c\mathbb{Z} \implies c \text{ is even}$