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A complex number $z$ is an integer if and only if $\sin(\pi z)=0$.

It follows that a complex number $z$ is an integer if and only $\sin^2(\pi z) = 0$. So for a real analytic function $f$ and any real number $x$, $x$ and $f(x)$ are both integers if and only if $\sin^2(\pi x) + \sin^2(\pi f(x)) = 0$.

Are there analytic functions that do the same over the complex numbers? Given a (complex) analytic function $f$, is there a way to assemble an analytic function $F$ so that $z$ and $f(z)$ are both integers if and only if $F(z) = 0$?

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While this can be done for any given f (as indicated by Angelo's answer), it can't be done in a way which is stable with respect to perturbations of f (which rules out most "analytic" answers, such as the one in the real case). This is because zeroes of a complex analytic function are stable wrt perturbations (Rouche's theorem), whereas the set of z for which z and f(z) are both integers is unstable (generically this set is empty). (In the real case, of course, zeroes are unstable wrt perturbation, especially for functions that are sums of squares.) –  Terry Tao Sep 17 '12 at 20:19
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2 Answers

up vote 4 down vote accepted

[Reposted as an answer, as requested. -T.]

While this can be done for any given $f$ (as indicated by Angelo's answer), it can't be done in a way which is stable with respect to perturbations of $f$ (which rules out most "analytic" answers, such as the one in the real case). This is because zeroes of a complex analytic function are stable wrt perturbations (Rouche's theorem), whereas the set of $z$ for which $z$ and $f(z)$ are both integers is unstable (generically this set is empty). (In the real case, of course, zeroes are unstable wrt perturbation, especially for functions that are sums of squares.)

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That's nice! Stated in a simple way you're saying there is no "master formula" like the sum of sin-squared's for the real case. –  Stephan Wehner Sep 18 '12 at 16:23
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Given any set $S$ of complex numbers without accumulation points, you can find an entire function with $S$ as its set of zeroes (this is known as Weierstrass's theorem). So the answer to your question is positive.

[Edit]: it seems to me that Terry's comment above answers the question much better than what I wrote. If Terry posts it an answer, I think that the OP should accept it.

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I'm very grateful for both answers! –  Stephan Wehner Sep 18 '12 at 16:17
    
Wondering if there are non-trivial examples with $F$ in closed form. (It's trivial if $f$ maps non-integers only to non-integers) –  Stephan Wehner Sep 18 '12 at 16:32
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