A polynomial equation on the integers

Question

A polynomial $p(x)$ is called self-centered if it has integer coefficients and $p(100) = 100.$ If $p(x)$ is a self-centered polynomial, what is the maximum number of integer solutions $k$ to the equation $p(k) = k^3$?

I am stuck, mainly because $p$ has to be an integer polynomial and the desired solutions must also be integers...

Answer 1

Suppose $S$ is the set of all $k$ that satisfy $p(k)=k^3$. Then we can write $$p(x)=x^3+\prod_{i\in S}(x-i)q(x)$$ But we also know that $p(100)=100$ so $$100-100^3=\prod_{i\in S}(100-i)q(100)$$ and since $100-100^3$ has $8$ prime factors (counting multiplicity), there can be at most $10$ integers in the product; thus the answer is $\fbox{10 }$ (which can be achieved by taking any ten such integers that work, and then making $q$ any appropriate polynomial.

Answer 2

Let $Q$ be the polynomial $Q(X) = p(X)-X^3$. We know that $$ Q(100)=p(100)-100^3=100-100^3=-100\cdot9999=-2^2\cdot 5^2\cdot 3^2\cdot 11\cdot 101\ . $$ So instead of working with $p$ (with the only restriction of being "centered") we can work equivalently with $Q$ (with the only above restriction), and want to count a maximal number $N$ of solutions of the equation $Q(x)=0$. We may split over $\Bbb Z$ in irreductible factors, then collect all non-linear factors in $R$ below, $$ Q(X)= (X-a_1)^{k_1}\cdots (X-a_N)^{k_N}\cdot R(X) \ , $$ so the condition on $Q(100)$ shows that there are maximally so many linear factors (i.e. $N$ is maximally equal to...) as different integer numbers that may "fit" in $Q(100)$.

A maximal case occurs only for $R$ constant, and moreover a(n integer) unit, so either $R(X)=+1$, or $R(X)=-1$ .

The case $R=+1$: We may write only $$ \begin{aligned} Q(100) &= (+2)(-2) (+5)(-5) (+3)(-3) \color{blue}{(+11)(-101)} (+1)(-1) \\ &= (+2)(-2) (+5)(-5) (+3)(-3) \color{red}{(-11)(+101)} (+1)(-1) \end{aligned} $$ as a product of different integers with a maximal number of factors.

The corresponding two choices of $Q$ are (assuming the leading coefficient one, and the match of the factors) $$ \begin{aligned} Q_1(X) &= \prod_{a\in\{\ \pm1,\pm2,\pm3,\pm5,\color{blue}{+11,-101}\ \}}(X-100+a)\ ,\qquad\text{ and} \\ Q_2(X) &= \prod_{a\in\{\ \pm1,\pm2,\pm3,\pm5,\color{red}{-11,+101}\ \}}(X-100+a) \ . \end{aligned} $$

The case $R=-1$: Similar situation, the "unpaired" values above, $\pm 11$, and $\pm 101$, have to be taken with the same sign.

So the maximal number of (distinct) integer solutions of $p(k)=k^3$ is $10$, and there are exactly four polynomials that match this maximal value.

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Computer check, one case only:

sage: var('X');
sage: p(X) = X^3 + prod([ X-100+a for a in [-1,1,-2,2,-3,3,-5,5,-11,101]]) 
sage: p(100)
100
sage: (p(X)-X^3).roots(ring=ZZ, multiplicities=False)
[111, 105, 103, 102, 101, 99, 98, 97, 95, -1]
sage: len(_)
10