# What is the maximum number of real roots in the interval (0,1) of a monic polynomial with integer coefficients and has degree 2022.

What is the maximum number of real roots in the interval (0,1) of a monic polynomial with degree 2022 and integer coefficients: $$f(x) = x^{2022} + a_{2021}x^{2021} ... + a_{0}$$?

Vieta's formulas, $$\sum_{i_{1}, k = 1,2...n where the i's are form the set {1...n}.

Show the upper bound is 2021 since, $$r_{1}r_{2}...r_{2022} = a_{0}$$ must have one root greater than one, otherwise $$a_{0} \notin \mathbb{Z},$$ since the left would be less than one.

A Quadratics maximum is 1 by the above, $$( x- (\sqrt{2}-1))(x -(-\sqrt{2}-1))$$ shows there can be one root in the interval. I assume the lower bound is 1011 by multiplying quadratics with distant roots in the interval to attain an example.

The rational root theorem shows the roots must be irrational. I can't find any progress in differentiating, integrating or treating roots as a module over the Integer Ring.

• Please include your question in the body of the question. Commented Jul 24 at 19:17
• Can you rescale one of the classic orthogonal polynomials? For a quadratic with $2$ roots how about $(x-\frac 12)^2-\frac 1{16}?$ Were you assuming that all the coefficients are integers? You didn't say so. Commented Jul 24 at 19:18
• I meant integer coefficients, thank you, I made the edit Commented Jul 24 at 19:25
• $(x-(\sqrt2-1))(x-(\sqrt2+1))=x^2-2\sqrt2x+1$. However, $(x-(\sqrt2-1))(x+(\sqrt2+1))=x^2+2x-1$.
– robjohn
Commented Jul 25 at 9:03
• I edited it thank you Commented Jul 25 at 21:19

For any $$N$$, we claim there exists some monic $$f \in \mathbb Z[X]$$ of degree $$N$$ with $$N-1$$ roots in $$(0,1)$$.

Indeed, consider the polynomial $$g(x) = \prod_{i=1}^{N-1} (x-\frac{i}{N})$$, and define $$y_j = \frac{2j-1}{2N}$$ for $$j=1,2,\dots, N$$. One can check via the intermediate value theorem on each interval $$(y_j, y_{j+1})$$ that

$$f(x) = x^N + M\cdot g(x)$$

works for all sufficiently large integers $$M$$ for which $$M\cdot g \in \mathbb Z[X]$$. Explicitly, we could take $$M$$ to be any multiple of $$N^{N-1}$$ greater than $$(\min_{1 \leq j \leq N} |g(y_j)|)^{-1}$$.

As you note, this is the best possible bound.

• Well, I learned something today. This construction shows more generally that we can arrange for $n-1$ of the roots of a monic integer polynomial of degree $n$ to be basically wherever we want (namely they must be arbitrarily close to the roots of $g(x)$, by continuity, if $M$ is sufficiently large, and moreover we can arrange for them to be real if the roots of $g$ are real). Commented Jul 24 at 23:00
• I can't see how the intermediate value theorem shows this with these y's . All of g's roots have multiplicity 1 so the sign of g changes at all N-1 roots. Let M be N^{N-1} and define c_{j} to be the extreme value (alternating sign) in every N partition of (0,1) the roots define. If after multiplying g by N^{N-1} ,$$\min_{1≤i≤N} c_{i}$$ is less than 1 then multiply g by integers until all absolute values of c_{i} are >= 1. Then it is clear that the positive values of f at the roots of g yield a root of f in between, since the addition of x^N wont take the negative c_{i}'s positive in f. Commented Jul 24 at 23:46
• @MaskedJaguar The point is that $f$ and $g$ have the same sign, and that $g$ alternates in sign. Qiaochu: corrected, thanks.
– RDL
Commented Jul 25 at 7:28
• Elegant solution: simple and consize (+) Commented Jul 25 at 9:58
• Since r{i} <= y{i+1} <= r{i+1} and your requirement it ensures the addition of the monic term wont change the fact f has N-1 roots in (0,1) since the when y{i} between two roots is negative , it stays negative. Is this correct ? Commented Jul 25 at 21:32

Here's a proof with an analytical flavor that the bound in the OP is sharp. (Edited to reflect a simplification proposed by @QiaochuYuan.)

Lemma: For any integer $$k>0$$, there is a polynomial of degree k with integer coefficients, all roots distinct, and all roots lying in $$(0,1)$$.
Proof: $$(1-2x)(1-3x)\cdots(1-(k+1)x)$$ suffices.

Proof of main proposition: Pick a polynomial $$p(x)$$ of degree $$2021$$ as in the lemma. Let $$r_j$$, $$j=1,2,\dots,2021$$ denote the roots of $$p(x)$$. Set $$r_0=0$$ and $$r_{2022}=1$$. Note that the sign of $$p(x)$$ alternates in successive regions $$(r_j,r_{j+1})$$.

There exists an $$\epsilon>0$$ such that for all $$j\in \{0,\dots,2021\}$$ there exists some $$x_0 \in (r_j,r_{j+1})$$ with $$|p(x_0)|>\epsilon$$ (since $$p(x_0)$$ is non-zero on each interval $$(r_j,r_{j+1})$$, and there are finitely many such intervals). Note that for any integer $$a>0$$, the rescaled polynomial $$p(ax)$$ has integer coefficients, its roots $$r_j/a$$ are distinct and lie in $$(0,1/a)$$, and between any two such roots there is again some point $$x_0$$ with $$|p(ax_0)|>\epsilon$$.

For $$a$$ large enough that $$1/a^{2022} < \epsilon/2$$, the polynomial $$p(ax) + x^{2022}$$ attains a value $$\pm \epsilon/2$$ somewhere in each interval $$(r_j/a,r_{j+1}/a)$$ (alternating in sign on successive intervals). So by the intermediate value theorem, $$p(ax) + x^{2022}$$ has 2021 roots in $$(0,1/a)\subset (0,1)$$.

• Sorry, it seems like the main argument is happening in the proof of the lemma but I don't think the statement of the lemma captures what you actually need from the proof. As written the lemma can be proven just by exhibiting e.g. $(1 - 2x)(1 - 3x) \dots (1 - (k+1) x)$. Commented Jul 25 at 20:27
• @QiaochuYuan Heh, good observation. I stand by the fact that the argument is correct as given, but you're observation allows for a significant simplification. I'll update shortly.
– Yly
Commented Jul 25 at 20:39
• @QiaochuYuan Updated. Thanks!
– Yly
Commented Jul 25 at 20:58

Not a complete answer, but here is a suggestion for a strategy. There's nothing special about $$2022$$, so it's natural to generalize the question to an arbitrary degree $$n$$. Let $$r_n$$ be the maximum number of real roots a monic integer polynomial of degree $$n$$ can have in $$(0, 1)$$. As you've observed, $$r_2 = 1$$, since we can construct monic quadratics one of whose roots lies in $$(0, 1)$$. In the cubic case, I messed around a bit and found the polynomial $$f(x) = x^3 - 6x^2 + 5x - 1$$, which has roots

$$0.308 \dots, 0.634 \dots, 5.05 \dots$$

which gives $$r_3 = 2$$. We can prove this by computing that $$f(0) = f(1) = -1$$ and $$f \left( \frac{1}{2} \right) = \frac{1 - 12 + 20 - 8}{8} = \frac{1}{8}$$, then appealing to the intermediate value theorem. Actually the polynomial $$f(x) = x^3 - (k+1) x^2 + kx - 1$$ works for $$k \ge 5$$, because we have $$f(0) = f(1) = -1$$ and

$$f \left( \frac{1}{2} \right) = \frac{1 - 2(k+1) + 4k - 8}{8} = \frac{2k - 9}{8}.$$

At this point we might guess that $$r_n = n-1$$ by generalizing this construction; so, we can try to write down a monic polynomial $$f$$ of degree $$n$$ such that $$f \left( \frac{k}{n} \right), 0 \le k \le n$$ is alternating in sign, so $$f$$ has a root between $$\frac{k}{n}$$ and $$\frac{k+1}{n}$$ for $$0 \le k \le n-1$$; if we could do this it would follow that $$r_n = n-1$$ and in particular $$r_{2022} = 2021$$. The sequence $$\frac{k}{n}$$ could potentially be replaced by some other sequence of reals between $$0$$ and $$1$$, I don't know what the right strategy will end up being. Also, if $$f(x)$$ is a polynomial with the maximum number of roots in $$(0, 1)$$ then so is $$f'(x)$$, which suggests the possibility of an inductive construction. It might be helpful to try the quartic or even the quintic cases before the general case to see if any useful patterns emerge.

• Great Idea, now I think see how Lagrange's Interpolation formula can be used to construct such a polynomial, will take me a few hours to iron out the details. Commented Jul 24 at 22:39