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Does there exist a non-constant function $f:\mathbb N^2 \rightarrow \mathbb N$ such that $$f(x,y)+f(y,x)=f(x^2,y^2)+1$$ for all positive integers $x,y$?

I think that such a function does not exist. But I do not know how to prove

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  • $\begingroup$ What is the source of this problem? $\endgroup$ – wythagoras Aug 28 '16 at 8:39
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    $\begingroup$ Aren't we free to choose $f(x,y)$ when $x$ or $y$ is not a square, and then to complete the definition of $f$ using $f(x^2,y^2)=f(x,y)+f(y,x)-1$ recursively? For example, $f(256,81)=2f(4,3)+2f(3,4)-3$... $\endgroup$ – Did Aug 28 '16 at 8:53
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How about this function?

$$ f(x, y) = \begin{cases} 2^n + 1 & \text{if } (x, y) = (2^{2^n}, 1) \text{ or } (1, 2^{2^n}) \\ 1 & \text{otherwise} \end{cases} $$

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  • $\begingroup$ perhaps $(2^{2n},1)$? $\endgroup$ – Takahiro Waki Aug 28 '16 at 12:16
  • $\begingroup$ @TakahiroWaki, At each 'recursive step', the exponent is doubled: $ f(2^{2a}, 1) = 2f(2^a, 1) - 1$. So the exponent of the argument should be geometric itself. $\endgroup$ – Sangchul Lee Aug 28 '16 at 15:01
  • $\begingroup$ Is this all possible function? $\endgroup$ – Takahiro Waki Aug 28 '16 at 16:56
  • $\begingroup$ @TakahiroWaki As in Did's comment, you are free to choose any value for $(x, y)$ with at least one of $x$ and $y$ being not square. Then you can construct a solution from the functional equation. $\endgroup$ – Sangchul Lee Aug 28 '16 at 18:27

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