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Does there exist a bijection $f: \mathbb N \rightarrow \mathbb N$ such that if $f(a) = b$ then either $b = a^2$ or $a = b^2$?

Any help would be highly appreciated.

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  • $\begingroup$ What is $N$? Natural numbers? $\endgroup$ – Wuestenfux Jul 12 at 12:46
  • $\begingroup$ @Wuestenfux Yup $\endgroup$ – Ilan Aizelman WS Jul 12 at 12:47
  • $\begingroup$ Likely yes by induction $\endgroup$ – Quang Hoang Jul 12 at 12:48
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    $\begingroup$ No. Hint: Consider $35$. $\endgroup$ – Kumar Jul 12 at 12:59
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The answer is yes.

Let $S = \mathbb{N} \setminus \mathbb{N}^2$ be the set of numbers which are not perfect squares. Notice that $a\in S^m$ means that $a = b^m$ for some (unique) $b \in \mathbb{N}$ which is not a perfect square. Clearly $\mathbb{N} = \bigcup_{k \ge 0} S^{2^k}$ and this union is disjoint.

Define $f : \mathbb{N} \to \mathbb{N}$ as $$f(a) = \begin{cases} a^2 , &\text{ if $a \in S^{2^k}$ where $k$ is even } \\ \sqrt{a}, &\text{ if $a \in S^{2^k}$ where $k$ is odd } \\ \end{cases}$$

Then $f$ is bijective with inverse

$$f^{-1}(b) = \begin{cases} b^2 , &\text{ if $a \in S^{2^k}$ where $k$ is odd } \\ \sqrt{b}, &\text{ if $a \in S^{2^k}$ where $k$ is even } \\ \end{cases}$$


For some motivation, notice that if $a \in S$ then necessarily $f(a) = a^2$, and if $b \in S$ it must be $f(b^2) = b$. Therefore $f$ necessarily bijectively maps $S \mapsto S^2$ and $S^2 \mapsto S$. It remains to somehow bijectively map the union of $S^4, S^8, S^{16}, \ldots$ to itself. The most natural thing is $$S^4 \mapsto S^8, S^8\mapsto S^4$$ $$S^{16} \mapsto S^{32}, S^{32}\mapsto S^{16}$$ and so on.

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  • $\begingroup$ Thank you, this is perfect $\endgroup$ – Ilan Aizelman WS Jul 12 at 13:58

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