3

Let $y_1,\dots,y_n$ be integers.

Conjecture:$\;$There exists $f\in\mathbb{Z}[x]$ such that $f(i)=y_i$ for $i=1,\dots,n\;$if and only if $(i-j){\,\mid\,}(y_i-y_j)$ for all $i,j$ with $1\le j < i\le n$.

I can prove the forward direction . . .

Suppose $f\in\mathbb{Z}[x]$ is such that $f(i)=y_i$ for $i=1,...,n$.

For some nonnegative integer $m$ we can write $$ f=\sum_{k=0}^m a_kx^k $$ where $a_0,...,a_m\in\mathbb{Z}$.

Let $i,j$ be integers with $1\le j < i\le n$. \begin{align*} \text{Then}\;\;& i\equiv j\;\bigl(\text{mod}\;(i-j)\bigr) \\[4pt] \implies\;& i^k\equiv j^k\;\bigl(\text{mod}\;(i-j)\bigr)\;\text{for}\;0\le k\le m \\[4pt] \implies\;& a_ki^k\equiv a_kj^k\;\bigl(\text{mod}\;(i-j)\bigr)\;\text{for}\;0\le k\le m \\[4pt] \implies\;& \sum_{k=0}^m a_ki^k\equiv \sum_{k=0}^m a_kj^k\;\bigl(\text{mod}\;(i-j)\bigr) \\[4pt] \implies\;& f(i)\equiv f(j)\;\bigl(\text{mod}\;(i-j)\bigr) \\[4pt] \implies\;& y_i\equiv y_j\;\bigl(\text{mod}\;(i-j)\bigr) \\[4pt] \implies\;& (i-j){\,\mid\,}(y_i-y_j) \\[4pt] \end{align*} as was to be shown, which completes the proof of the forward direction.

For the reverse direction, I considered induction on $n$, but while the verification for the case $n=1$ is instant, I don't see how the divisibility conditions of the conjecture can be used to achieve the inductive step.

Looking at some special cases, it's trivial to show that the conjecture holds for the case $n=2$, and with a little more work, it can be shown that it also holds for the cases $n=3$ and $n=4$.

Potentially relevant is the answer by @Chrystomath to

$\qquad$ Interpolating polynomial with integer coefficients

which gives a divisibility criterion that is necessary and sufficient for the existence of a qualifying $f$.

However I don't see how that divisibility criterion is implied by the divisibility conditions of my conjecture.

So the question remains . . .

Is my conjecture valid?

quasi
  • 61,115
  • 2
    "the conjecture holds for the case $n=2$" $,$ Unless I misread something, the interpolating quadratic for $,{ (0,0), (2,2), (4,8) },$ is $,\frac{1}{2} x^2,$, and the MO post linked in the question you quote implies that there is no higher degree interpolating polynomial with integer coefficients. – dxiv May 19 '22 at 06:15
  • 1
    @dxiv: Your counterexample is very simple -- thanks (although it's for the case $n=3$ so I must have made a mistake analyzing that case). If I can't find a simple fix, I'll delete the question. – quasi May 19 '22 at 06:29
  • 1
    @dxiv: In an attempt to repair the conjecture while retaining its essence, I've restricted the values of the inputs to $f$. – quasi May 19 '22 at 07:53
  • Typing in HTML is fun, but has its problems. – Rodrigo de Azevedo May 19 '22 at 08:12
  • 1
    @quasi I believe the answer is still negative e.g. interpolate[ (1,1), (2,16), (3,81), (4, 256), (5, 637)] $,= \frac{3}{2} x^4 - 5 x^3 + \frac{35}{2} x^2 - 25 x + 12,$. The points should satisfy the divisibility conditions since they are based off $,y=x^4,$ with the last value changed to $,y_5 = 5^4 + 12,$ where $,12 = \text{lcm}{i-j}=\text{lcm}(2,3,4)$. – dxiv May 19 '22 at 21:09
  • @quasi A more general counterexample for $,n \ge 5,$ is $,y_1=y_2=\dots=y_{n-1}=0,$ and $,y_n,$ a multiple of $\text{lcm}(1,2,\dots,n-1)$ but not of $,(n-1)!,$, then $p(x) = \frac{y_n}{(n-1)!}(x-1)(x-2)\dots(x-n+1) \not\in \mathbb Z[x],$. – dxiv May 19 '22 at 22:37
  • @dxiv: Thank you very much for the counterexamples and the insights! If you would transfer those comments to an answer, I would be happy to accept it. – quasi May 19 '22 at 23:01
  • @quasi Done, thanks. Question remains open whether there is a set of $,x_i,$ such that the conjecture holds true. One necessary condition would be for $,x_n - x_i,$ to be coprime, otherwise a counterexample similar to the above would work. Don't know that's enough, though, and my suspicion is that it's not. – dxiv May 19 '22 at 23:42

1 Answers1

3

Conjecture:$\;$There exists $f\in\mathbb{Z}[x]$ such that $f(i)=y_i$ for $i=1,\dots,n\;$if and only if $(i-j){\,\mid\,}(y_i-y_j)$ for all $i,j$ with $1\le j < i\le n$.

The reverse implication does not hold true in general for $\,n \ge 5\,$.

For a counterexample, consider the $\,(x_i,y_i)\,$ points $\,\{(1,0),(2,0),\dots,(n-1,0),(n,y_n)\}\,$. The Lagrange interpolating polynomial for these points is $\,p(x) = \dfrac{y_n}{(n-1)!}(x-1)(x-2)\dots(x-n+1)\,$.

For $\,n \ge 5\,$ it is possible to choose an $\,y_n\,$ so that $\,\text{lcm}(1,2,\dots,n-1) \mid y_n\,$ but $\,(n-1)! \not \mid y_n\,$. For such $\,y_n\,$ the Lagrange polynomial $\,p(x) \not\in \mathbb Z[x]\,$ since the leading coefficient $\,\dfrac{y_n}{(n-1)!} \not\in \mathbb Z\,$, even though the divisibility conditions are satisfied since $\,x_i-x_j \mid \text{lcm}(1,2,\dots,n-1) \mid y_n\;$ e.g. such a counterexample in the case $\,n=5\,$ is interpolate[ (1,0), (2,0), (3,0), (4,0), (5,12) ] $\,= \dfrac{1}{2} x^4 - 5 x^3 + \dfrac{35}{2} x^2 - 25 x + 12\,$.

Using the result proved on MO under Lagrange Interpolation and integer polynomials it follows that no higher degree interpolating polynomial with integer coefficients exists, either, since that would imply that the Lagrange interpolating polynomial must have integer coefficients as well.

dxiv
  • 77,867