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  1. If a Polynomial $f$ with complex coefficients takes integer values at $n+1$ consecutive integers ( where the degree of $f$ is $n$ ) , then we can conclude that $f$ takes integer values at all integers.

  2. "Polynomial $f$ has rational coefficients" $\iff$ "Polynomial $f$ sends at least $n+1$ integers to integers" ( where the degree of $f$ is $n$ )
    Here the $n+1$ integers don’t have to be consecutive.

PS: I came up with these two statements and I think they are correct , but I have no idea how to prove them and I don’t find any counter examples.
Examples: $x(x+1)/2$ takes all integers to integers.(statement 1)

$\sqrt{2}（x^2-x-2)$doesn’t have rational coefficients and it takes two integers to $0$,but it need to take at least 3 integers to integers.(statement 2)

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    You claim that you could not get Counter Examples , which is good. But you should list the Examples you got. You think these are Correct : Why ? – Prem Apr 19 '23 at 11:07
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    The properties of the Vandermonde matrix should help you prove or disprove these statements. – eyeballfrog Apr 19 '23 at 11:11
  • A related post which showcases how you can try to use eyeballfrog's hint (it does not answer your question directly though): https://math.stackexchange.com/a/516566/1104384 – Bruno B Apr 19 '23 at 11:18
  • The two statements are very different; the second one is basic number theory (choose an input that clears all denominators) while the first one follows from difference calculus (no matrices required): https://encyclopediaofmath.org/wiki/Finite-difference_calculus – JBL Apr 19 '23 at 11:39

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