"if P(x) takes value 0 for n+1 different values of x",  "then all its coefficients are actually zeros."

1) Then p(x) is of degree >=n+1

2) No reason , that all coefficients then are zero.

3) A polynomial can has zeros in some domain without be identically zero.  Please reformulate your additional information in the question.

"if P(x) takes value 0 for n+1 different values of x",  "then all its coefficients are actually zeros."

1) Then p(x) is of degree >=n+1

2) No reason , that all coefficients then are zero.

3) A polynomial can has zeros in some domain without be identically zero.  Please reformulate your additional information in the question.

Thanks Hanifa, but 1) n is the notation for the degree of P(x), so it cannot have any other degree, 2) all coefficients are zero because they can be calculated by the Lagrange formula, so 3) sometimes a polynomial is identical zero due to the fact that it takes enough values zero.

Dear Igor,

1) If it is of degree n, then it can not have n+1 different zeros. this assumption can not take place.

2) by definition of a polynomial: it is identically zero when all coefficients are zero. It means that the zero of the ring or the field which you define your polynomial over, is the only polynomial equal to zero (with degree =0 of course).

Thank you again Hanifa, of course you were completely right about the degree of polynomials. I have now inserted the beautiful sing `≤' in the text of my question.

My actual problem is: there is a big expression, depending on 10 complex variables and involving many square roots, and this expression is clearly identical zero — because it turns into zero for any random values of the variables. What remains is to propose a way for my computer to prove this within the life time of our Universe.

Huge thanks to all interested.

May be the idea of polynomials can not solve the problem, think in complex analysis of multivariables functions, unfortunately, I have no idea in that topic.  

An other idea, if you take the square of the expression, then try to write is as sums of complex quadratic forms , I don't know if it is easy, I hope it can help.

Здравствуйте, Игорь Германович! Рад Вас встретить здесь. Поскольку Вы надеетесь перебрать все "разумные" значения на компьюьере, то значит Вы надеетесь, что их будет конечное число. Что-то в это не верится. Для голоморфных функций нескольких переменных, наверное, надо обращаться к подготовительной теореме Вейерштрасса. В известной книге Шабата есть несколько упражнений на этй тему (стр 302, упр 10-12). В них "разумные" значения образуют плоскость или дугу.

Hi Victor Mikhailovich, and all people,

What follows is not exactly my problem, just to get the idea what it looks like.

Consider the following expression: 1/(1+sqrt(a)). Then take a computer algebra system and rationalize the denominator. You will get (sqrt(a)-1)/(a-1).

Now consider a more complicated expression: 1/(1+sqrt(a)+sqrt(b)). Do the same, and get (sqrt(b)*(b-a+2*sqrt(a)-1)-b+sqrt(a)*(-b+a-1)-a+1)/(b^2+(-2*a-2)*b+a^2-2*a+1).

Then you can proceed with 1/(1+sqrt(a)+sqrt(b)+sqrt(c)), 1/(1+sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)), and so on, and kindly inform me how many square roots your computer was able to handle.

Игорь Германович! Команда evala(simplify(1/(1+sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)))), в Maple находит это значение меньше чем за 2 сек. Однако, это похоже предел для этого примитивного инструмента. Наверное, надо пользоваться специализированными пакетами работы с алгебраическими числами (если они есть).

Victor Mikhailovich, thanks for your comment, but I didn't get it in full. Did you try also five roots? If yes, did it take more than 2 seconds? What is "primitive" with Maple? For the record: I work, chiefly, with Maxima and Singular. The latter is of course very specialized.

Игорь Германович! Для пяти корней дождаться окончания счета за 5 минут не удалось. Поскольку есть такое резкое различие во времени, я и решил, что 4 корня - это граница, когда задачу еще можно решить примитивными методами, т.е. с помощью команды evala.

При n=5 наверно надо искать минимальный многочлен для этой дроби, тогда ее можно будет представить через корень этого многочлена. Однако, этот корень может и не радикал. А вообще при   n=5 задача избавления от иррациональности в знаменателе и получения выражения через степени sqrt(a), sqrt(b),... имеет решение? У алгебраистов не спрашивали?

OK. Now you can assign numeric values — even very big if you like — to all letters, and feel the difference between the speed of symbolic and numeric calculations. 

Yes Peter, and it seems I will have to rely on such grids (in the case of algebraic functions, a grid will also work if chosen carefully).

Or maybe I will learn deeper properties somewhere...

Anyhow, Happy New coming Year to all!

There are different ways to construct so called "generic tuples" with the property that if F is polynomial and F(x1, ..., xn) = 0 then F is 0. A fist idea is to take x1, ... xn transcendental over the field generated by the coefficients and algebraically independent. But there are ways to define it in the basis field as well. There is always an integer M such that the tuple given by x1 = M, x(i+1) = 2^xi, is a generic tuple.

The issue is addressed in "Multinomial Points", Houston J Math, Vol 34, No 3 (2008), p 661, which is posted at https://www.scribd.com/document/85492144/Multinomial-Points. 

Look at Proposition 3.1. It is the analog of Professor Korepanov's initial statement. I have no doubt that there are other tests. Unless otherwise stated, we were interested in multivariable polynomials with integer coefficients.   

Dear Prof. Cornelius, your paper seems to be more related to the question about polynomials with rational coefficients taking only integral values. Is it so? Of course, it possibly touches the present issue as well.

Also, I am studying the very interesting work by Profs. Cornelius and Schultz. It will take time. But maybe somebody could answer the following question already now?

Let us return to the polynomial P(x) with integer coefficients, each ≤ 10 in absolute value. Then the two conditions P(3)=0 and P(4)=0 are enough to say that P(x) is identical zero (right?). The question is: is there any generalization of this fact to the case of, say, two indeterminates?

PS. It may be interesting also to consider conditions like P(⅓)=0, P(¼)=0, etc. :D

One more interesting question is: has anyone encountered a situation like mine? Namely, while developing a theory (in my case a sort of topological quantum field theory), you see that it will be beautiful if two huge expressions with many interdependent square roots coincide identically. And numeric calculations with precision of 100 or 300 digits agree with you — but the problem turns out too difficult for a straightforward computer algebra.