Polynomial always congruent to $0$ or $1$

Question

Problem: We are given a polynomial $P(x) \in \mathbb{Z}[x]$ and some prime $p$, such that $P(0)=0$, $P(1) = 1$ and for every natural number $n$, $P(n)$ is congruent to $0$ or $1$ modulo $p$. Prove $P(x)$ is of degree at least $p-1$.

My observations:

$a\equiv b \implies P(a) \equiv P(b)$ so we can only consider remainders of $P(0), P(1), \dots, P(p-1)$.
$P(n)$ must be divisible by $n$.
$P(n)^2 - P(n) \equiv 0$ for every $n$ so there must be a factor $x-n$ or some irreducible trinomial that is divisible by $p$ for $n$.
we could assume $\deg P(x) < p - 1$ and then show it has the same values in $p$ points as some other polynomial, then show that polynomial does not satisfy $P(0)=0$ and $P(1) = 1$

However, I don't know how to finish the solution.

Regarding the last point: You would want to assume that $\deg P(x) < p - 1$ since you want to prove that $\deg P(x) \geqslant p - 1$. — Aryaman Maithani, Jul 18 '21 at 08:47
The statement is unclear : you want to prove that for all natural number $n$ you have $P(n)$ congruent to $0$ or $1$ modulo $p$, or this is an assumption ? — Surb, Jul 18 '21 at 08:52
@Surb of course it does, so does $P(x) = x^{p-1}$ (it is either 0 or 1), but it is of degree exactly $p-1$ — Kangaroo976, Jul 18 '21 at 12:23
$P(0)=0$ implies the constant term in $P(x)$ is zero. Now consider $P(p)$. — CyclotomicField, Jul 18 '21 at 16:22
@CyclotomicField sure it is divisible by $p$. Anything else? — Kangaroo976, Jul 18 '21 at 18:11
@CyclotomicField we already know $P(1) = 1$ I don't know where you're going — Kangaroo976, Jul 18 '21 at 19:46
I would love to see an answer or a more elaborate hint (I assume you've solved the problem). — Kangaroo976, Jul 18 '21 at 19:53

Calvin Lin · Accepted Answer · 2023-02-26T14:03:16.607

We are working in $ \mathbb{Z}_p [x]$. Any rational number should be interpreted accordingly in $ \mathbb{Z}_p$.

Recall lagrange interpolation: If a polynomial satisfies $ P(i) = y_i$, then

$$P(x) = \left[\sum_i y_i \prod_{j\neq i } \frac{ x - j}{i - j}\right] +A(x)\prod_i(x-i). $$

We are given that $ y_i = 0, 1$ and specific values $ y_0 = 0, y_1 = 1$.

Claim: The polynomial in the parenthesis has degree $ p-1$.

Proof: The products $\prod_{j\neq i } \frac{ x - j}{i - j}$ each have degree $p-1$ and an (identical) leading coefficient of $\frac{1}{(-1)^{p-1} (p-1)!} \equiv (-1)^{p}$.
Hence, the coefficient of $ x^{p-1} $ in $\sum_i y_i \prod_{j\neq i } \frac{ x - j}{i - j}$ is $ - \sum y_i$.
This value is between $- (p-1)$ and $-1$, so in particular it is not 0 (in $\mathbb{Z}_p$).
Hence, the polynomial in the parenthesis has degree exactly $p-1$.

Corollary: Since the other term in $P(x)$ has degree at least $p$ (or 0), hence the problem statement follows.

Notes

When first approaching this problem, it's natural to consider the special case where $f(i) = 1$ and $f(j) = 0$ for $i\neq j$. We can determine that (EG through brute force or otherwise) this polynomial has degree $p-1$ and leading coefficient $p-1$. (This is in fact the lagrange basis polynomials $\prod_{j\neq i } \frac{ x - j}{i - j}$.)
The necessary condition in the problem is $ \sum y_i \neq 0 \pmod{p}$. Any solution will likely have to make use of this.
- As an example, we can show that if $P(x)$ has degree $\leq p-2$, then $ p \mid P(0) + P(1) + P(2) + \ldots + P(p-1)$. Since this condition is not satisfied, thus the degree is at least $p-1$.
Another approach could be to show that the coefficient of $x^{p-1}$ is $ - \sum y_i$.
- This follows from using the method of differences to calculate the coefficient of $x^{p-1}$ as

$$ \frac{1}{(p-1)!} \left[ {p-1 \choose 0} y_1 - {p-1 \choose 1} y_2 + {p-1 \choose 2 } y_3 + \ldots + (-1)^{p+1}{p-1 \choose 0 } y_p \right]. $$

It's not immediately clear to me that OP's observation of $[P(x)]^2 - P(x) \equiv 0$ can lead to a solution. The condition that $y_i \in \{0, 1 \}$ isn't necessary, and we just need $y_i \in \{a, b\}$.

Well, the form of the polynomial you've used is certainly correct. But I think this is incomplete. You never explicitly used the fact $P(x)$ has integer coefficients, which suggests you're missing something. You've stated yourself that we need to prove $\sum y_i$ is not zero. I've tried something similar before and I've given up so thanks for your answer anyway, I suppose I will be able to finish from here. — Kangaroo976, Jul 19 '21 at 07:23
@Kangaroo976 A) The fact that $P(x)$ has integer coefficients is exactly why I wrote "any rational number should be interpreted accordingly in $\mathbb{Z}_p$". IE $ 1/2 \equiv (p+1) / 2 \pmod{p}$. In particular, that's what allowed me to calculate the coefficient (using Wilson's theorem), which is how I'm explicitly using it. B) We have $ p-2$ values of $y_i$, each of which is either 0 or 1, 1 value of 1 and 1 value of 0. Hence the summation is between 1 and $p-1$ (inclusive), thus is not zero in $ \mathbb{Z}_p$. — Calvin Lin, Jul 19 '21 at 15:45
C) See my first note for the intuition behind this solution. In particular, find a polynomial with integer coefficients that satisfies the special case that I brought up, and you will start to see what is happening. D) Yes, I glossed over how the lagrange interpolation formula which gives rational coefficients can be adapted to give integer coefficients by simply taking the multiplicative inverse of the denominator. — Calvin Lin, Jul 19 '21 at 15:49
Yeah, now I realised. I checked everything, great answer. Thank you. — Kangaroo976, Jul 19 '21 at 18:02
@Kangaroo976 It would be greatly appreciated if you upvoted and accepted answers, especially if you think they are great. — Calvin Lin, Jul 19 '21 at 22:27

Polynomial always congruent to $0$ or $1$

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