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Consider a function

$$ f: \Bbb{Z} \rightarrow \Bbb{Z} $$

Over the integers. Furthermore consider a number E such that there doesn't exist an integer R such that $f(R) = E$ or formally stated

$$ E | \lnot \left( \exists R| f(R) = E \right)$$

Is it ever possible that for each natural number $i$ there exists $w_i$ such that

$$ f(w_i) \equiv E \mod i $$

In other words,

$$ f^{-1}(E) \not \in \Bbb{Z}$$

Yet

$$ f^{-1}(E) \mod i \in \{{x \mod i}\} \forall i $$

This is a generalization of the question:

Do there exist Artificial Squares?

To now arbitrary functions.

Clearly for squares this is not the case, and the answer is fairly easy to generalize for any function of the form $f(x) = x^{a}$ but making a statement such as this over all possible functions seems bold.

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Sidharth Ghoshal
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    That notation is very sloppy - $f^{-1}(E)$ is a set, and doesn't mean different things depending on what you are trying to include it in. – Thomas Andrews Mar 21 '15 at 04:40
  • i read it three times trying to understand why this is a question at all ... good @Thomas posted an answer confirming it was trivial – Mirko Mar 21 '15 at 04:49
  • I still have issues with the answer though, namely that $|x|$ isn't well behaved. I don't think this is the question I really mean to ask, but I'm trying to figure what question I should be asking. There needs to be restriction on f in this case – Sidharth Ghoshal Mar 21 '15 at 04:50
  • See my comment underneath @ThomasAndrews for the tricky part with |x| – Sidharth Ghoshal Mar 21 '15 at 04:50

1 Answers1

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$f(x)=1+|x|$ is never zero, but $f(|i|-1)\equiv 0\pmod i$ for all $i\neq 0$.

Thomas Andrews
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  • This gives us $1 + |k| \mod R$ for some integer $0 \le k \le R-1$ and another integer $R$. There is not integer for which it is 0. But for every R there exists a k (namely R-1) such that is 0. Ah! thats a good one :) – Sidharth Ghoshal Mar 21 '15 at 04:46
  • I guess the trick is defining $|x|$ for the integers modulo N. For example $|-1| = 1$, yet $-1 \equiv 7 \mod 8 \rightarrow |-1| \equiv |7| \rightarrow 1 \equiv 7 \mod 8 $ and that is clearly no true – Sidharth Ghoshal Mar 21 '15 at 04:48
  • How to reconcile these two? @Thomas Andrews – Sidharth Ghoshal Mar 21 '15 at 04:51
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    No, the function is a function on $\mathbb Z$. You didn't not require a function on $\mathbb Z/N$. Those functions are much more complicated. – Thomas Andrews Mar 21 '15 at 04:51
  • In the even the function is defined cleanly ${\Bbb{Z}}/{\Bbb{N}}$ does that change things significantly? – Sidharth Ghoshal Mar 21 '15 at 04:53
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    The set of such functions are all functions such that $m\equiv n\pmod d \implies f(m)\equiv f(n)\pmod d$. I've written up a lot about these functions, but they are fairly complicated. http://thomasoandrews.com/math/congruent/ – Thomas Andrews Mar 21 '15 at 04:54
  • @frogeyedpeas: Are you trying to say that you want $f$ to be a polynomial rather than just any function? – hmakholm left over Monica Mar 21 '15 at 04:55
  • They include integer polynomials, but also non-integer polynomials like $f(n)=\frac{n(n-1)(n-2)(n-3)}{2}$ and uncountably many non-polynomial functions. – Thomas Andrews Mar 21 '15 at 04:56
  • I wanted it to be as general as possible while still being consistent with the laws of modular arithmetic. In other words if $a \equiv b \mod N$ then $f(a) \equiv f(b) \mod N$ do there exist non polynomial functions that satisfy that? – Sidharth Ghoshal Mar 21 '15 at 04:56
  • Yes, there are non-polynomial functions of that sort. @frogeyedpeas – Thomas Andrews Mar 21 '15 at 04:57
  • @ThomasAndrews Yep! I think you had the exact same line of thought back then as I am beginning to explore onw – Sidharth Ghoshal Mar 21 '15 at 04:57
  • The freedom of choice in each of the choices in constructing such functions I believe gives you the freedom to ensure that the function is never $0$ but can be made divisible by arbitrary numbers. I'd have to work that out, however. – Thomas Andrews Mar 21 '15 at 05:06