If $z_1,..z_n$ are roots of polynom $x^n+...+a_1x+a_0$, then $\prod_i |z_i|= |a_0|$. If $|a_0|\ge 1$ at least one root $|z_j|\ge 1$. If $a_0=0$, then only one root is 0 and $\prod_{z_i \neq 0}|z_i|=|a_1| \ge 1 \to \ exist \ j : \ |z_j| \ge 1.$

@above : The center of the disk isn't necessarily $0$.

A beautiful problem indeed. The key is to consider the discriminant (which must be an integer) and estimate its size.

angiland wrote: A beautiful problem indeed. The key is to consider the discriminant (which must be an integer) and estimate its size. Can you say a little more further ?

Consider the discriminant, which is the product of $(z_i-z_j)$ over distinct $i,j$. This is a non-zero integer, but using the condition we can upper bound it by $n^n \cdot 0.99^{n(n-1)}$, which is smaller than $1$ for large degree $n$.

How do you prove that if all the $z_i$ are contained in some disc of radius $0.99$ that \[\prod_{i\neq j}\left|z_i - z_j\right| \le n^n \cdot (0.99)^{n(n-1)} ?\]I can show that the bound on the right hand side is attained when the $z_i$ are all equally spaced around the boundary of a closed disc of radius $0.99$ (that is, the $z_i$ form a regular $n$-gon with as large a circumradius as possible), but I don't see an easy way to show that this has to be the maximum value of the discriminant. If we were just looking at the product of $\left|z_i - z_{i+1}\right|$ (indices taken modulo $n$) as $i$ goes from $1$ to $n,$ I don't think it's too difficult to show that the maximum value of the product is achieved for a regular $n$-gon (either using Jensen's inequality, or some smoothing argument, for example). However, it's not clear to me how to prove the upper bound for the discriminant, since in that case we need to look at the product of all pairwise differences of the $z_i.$ Edit: Solutions to this bound are presented by other AoPS users here.

any link for the official solution of the problem please

angiland wrote: Consider the discriminant, which is the product of $(z_i-z_j)$ over distinct $i,j$. This is a non-zero integer, but using the condition we can upper bound it by $n^n \cdot 0.99^{n(n-1)}$, which is smaller than $1$ for large degree $n$. With WA an upper bound on the "sufficiently large degree" is 645. Just FYI.

Whats wrong with the solution by RUST? Even if the center of the circle is not Origin but we can always have Meaningful transformations to get a new polynomial which features a circle with origin as center. Hope this helps I will try to obtain

No full solutions?

Ok, let us write a full solution, finally: 1. As mentioned previously, if the roots of such a polynomial are $z_1,...,z_n,$ then its discriminant $$ \prod_{i<j} (z_i-z_j) $$ is a non-zero integer. This can be proven, for instance, by induction on the degree. 2. We identify such a discriminant as a determinant of a Vandermonde matrix. Explicitly, if $M$ is the matrix whose lines are $(1,z_i,\dots,z_i^{n-1}),$ then it has determinant equal to the discriminant of the polynomial under scrutiny. 3. We use, in fact, that the discriminant of a polynomial is invariant under translations. That is, we might assume that the center of our disk is, in fact, zero, and that, for all $i$, $|z_i| \le 0.99.$ 4. Finally, we use Hadamard's inequality for determinants (see, e.g., https://en.wikipedia.org/wiki/Hadamard%27s_inequality): the determinant of a matrix might be bounded by the product of the norms of its column vectors. But we have that our matrix $M$ has column vectors $(1,...,1),\,(z_1,...,z_n),\,\dots,(z_1^{n-1},...,z_n^{n-1}).$ The norms of each of these vectors are bounded by $n,\,n\times 0.99,\,n \times (0.99)^2,\,\dots,n \times (0.99)^{n-1}.$ That means: $$|\det(M)| \le n^n (0.99)^{\frac{n(n-1)}{2}}.$$ It is then easy to see that the right hand side goes to $0$ as $n$ goes to infinity. $\square$

jowramos wrote: $$|\det(M)| \le n^n (0.99)^{\frac{n(n-1)}{2}}.$$ The right hand side goes to $0$ as $n$ goes to infinity But the RHS goes to infinity

govind7701 wrote: jowramos wrote: $$|\det(M)| \le n^n (0.99)^{\frac{n(n-1)}{2}}.$$ The right hand side goes to $0$ as $n$ goes to infinity But the RHS goes to infinity No, it does not: $n^n = e^{n \ln(n)},$ whereas $ (0.99)^{n^2/2} = e^{\ln(0.99)\cdot \frac{n^2}{2}}.$ Clearly, any quadratic function wins over any function of growth $n\ln(n)$ at infinity.

Very nice application of Hadamard's Inequality! A short fix is that the discriminant which is always an integer actually is $\prod_{i \neq j} (x_i-x_j)$ - but everything else in your solution works fine anyways since even though $\det(M)$ is not necessarily an integer, its square is an integer, so $|\det(M)| \ge 1$. Another way to go about proving the inequality $|\prod_{i<j} (x_i-x_j)| \le n^n R^{n(n-1)/2}$ for $x_i$ in a disk of radius $R$ is to use that the thing inside is a holomorphic function of each $x_i$, so its maximum is achieved when each $x_i$ is in the boundary of the circle (by something like the maximum modulus principle), then by connecting each point to the center of the disk and writing this product as a function of the angles between consecutive segments, the result follows from something like Jensen's inequality (with equality when all angles are the same). Also, responding to some earlier questions, the problem with just translating the disk to center zero is that the new polynomial won't necessarily have integer roots - indeed there are plenty of good polynomials of degree $>1$ like $x(x-1)(x^2-x+1)$.

Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ with distinct roots $r_1,\cdots,r_n$. The key idea is to consider $$\prod\limits_{i\ne j} (r_i-r_j)$$ It is a function with integer coefficients of $a_0,\cdots, a_{n-1}$ so it is an integer. Its modulus is at least 1. We bound this value analytically. By the maximal modulus principle, since this is a polynomial in one function, the max occurs when this $r_i$ lie on the circle of radius 0.99. Thus all $r_i$ lie on the same circle of radius $0.99$. By convexity the value of the expression is maximized when they form a maximal n-agon. We can see the modulus is $$\left((0.99)^{n-1}\prod\limits_{j=1}^{n-1} (1-\omega_n^j)\right)^n= \left((0.99)^{n-1}\lim_{x\to 1} \frac{x^n-1}{x-1}\right)^n=(n(0.99)^{n-1})^n<1$$