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

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

• What is the source of this problem? – wythagoras Aug 28 '16 at 8:39
• 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, $f(256,81)=2f(4,3)+2f(3,4)-3$... – Did Aug 28 '16 at 8:53

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

• perhaps $(2^{2n},1)$? – Takahiro Waki Aug 28 '16 at 12:16
• @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. – Sangchul Lee Aug 28 '16 at 15:01
• Is this all possible function? – Takahiro Waki Aug 28 '16 at 16:56
• @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. – Sangchul Lee Aug 28 '16 at 18:27