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The following result seems to be fairly well-known.

Suppose $p$ is a polynomial in one variable with positive integer coefficients, and suppose that $p$ is to be determined by giving several inputs in succession, which are allowed to depend on previous outputs. Then only two inputs are needed to determine $p$. [1, 2, 3, 4, 5, 6]

Is it also well-known that it still holds when $p$ is allowed to depend on any finite number of variables?


Suppose $p$ depends on $v$ different variables. Take $B > p(2, \dotsc, 2)$ and $D > \log_2 p (2, \dotsc, 2)$, which are bounds on the coefficients and on the degree, respectively. Now each coefficient of $p$ can be encoded as a digit of the base-$B$ representation of $$ p_2 = p(B^{D^0},B^{D^1}, \dotsc, B^{D^{v-1}}). $$ More specifically, to find the coefficient of $x_0^{d_0} \dotsm x_{v-1}^{d_{v-1}}$, consider its contribution to $p_2$, $$ (B^{D^0})^{d_0} \cdot (B^{D^1})^{d_1} \dotsm (B^{D^{v-1}})^{d_{v-1}} = B^{d_0 D^0 + d_1 D^1 + \dotsb + d_{v-1} D^{v-1}}. $$ This gives a power of $B$ (and therefore a digit in $p_2$) uniquely associated to the desired coefficient precisely because $D$ exceeds the degree of $p$.

  • Are you asking whether your solution is correct? It seems to be, upon a quick glance. – Greg Martin Feb 04 '14 at 22:23
  • @GregMartin This was part of an undergraduate research project. I'm wondering things like: Is it original? Is it considered trivial? Is it published somewhere such as a blog? (My advisor had not seen it before.) – A l'Maeaux Feb 04 '14 at 23:20
  • I had heard the one-variable version before, as a "chestnut" problem, though I don't know where to find it published. My opinion is that most mathematicians who know the solution to the one-variable version would be able to extend it to the several-variable version, although I have no idea if that extension ever been pointed out before. – Greg Martin Feb 05 '14 at 01:01
  • @GregMartin What's a chestnut problem? – A l'Maeaux Feb 05 '14 at 02:34
  • "An old chestnut" is a common idiom: http://idioms.thefreedictionary.com/an+old+chestnut (but I didn't mean to imply that it was stale or now boring - just that it had been passed around often) – Greg Martin Feb 05 '14 at 03:05
  • Actually, only $1$ input is needed. We can just look at $p(\pi)$ and since $\pi$ is transcendental in principle you can figure out the polynomial. – Peter Woolfitt Jan 23 '15 at 17:31

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