Hmm, looks like it went through French? Or maybe Romanian and French share some of the same common noun-mathematical noun pairings that English doesn't have. Here's a rewording with some minor vocabulary and grammatical fixes: Given $n \geq 2$ a natural number, $(K, +, \cdot)$ a (commutative) field such that $ \underbrace{1+\ldots+1}_{m}\ne 0$ for $m=2,\ldots,n$, $G$ a proper subgroup of the additive group $(K, +)$, and $f\in K[X]$ a polynomial of degree $n$. Show that there is $a \in K$ such that $f(a) \not\in G$. (This is one of many results with the moral that the multiplicative structure of a field looks random with respect to its additive structure.)

Let on the contrary assume $\forall x\in K\,:\, f(x)\in G $. Denote $g(X)=f(X+1)-f(X)$. Due to the given condition (EDITED: $char(K)=0$ or $char(K)>n$ ) it follows that $g$ is a polynomial of degree $n-1$. We also have: $\forall x\in K\,:\, g(x)=f(x+1)-f(x) \in G$. Thus we can consecutively drop the degree of the polynomial till we get a polynomial $p(X)=aX+b,\, a\neq 0$ for which $\forall x\in K\,:\, ax+b \in G$. But take some $y\in K\setminus G$ , and we will get a contradiction by setting $x=a^{-1}(y-b)$.

Probably you should make explicit where you use the (very much necessary) condition that the characteristic of the field is not in the interval $[2, \deg(f)]$. (The counter-examples we're trying to avoid are things like $x^q - x$ when the field is a finite field with $q$ elements.)

JBL wrote: Probably you should make explicit where you use the (very much necessary) condition that the characteristic of the field is not in the interval $[2, \deg(f)]$. Don't you see JBL !! If $f(X)$ is of degree $m,\, 2\leq m \leq n$ we have $f(X)= a_0X^m +a_1 X^{m-1}+\ldots + a_m $ , where $a_0\neq 0$. Therefore: $f(X+1)-f(X)=m\cdot a_0X^{m-1}+\ldots $ and $m\cdot a_0 = a_0(\underbrace{1+1+\ldots+1}_{m})\neq 0$. This guarantees that dropping the degree ot the polynomial, at the end we will have a polynomial of degree $1$. btw, i wrote "$char(K)=0$" because i did'n see the last "$n$" in the expression: " $\underbrace{1+1+\ldots+1}_{m}$ for $m=2,\ldots,n $ " . But it is not so essential, the important thing is that either $char(K)=0$ or $char(K) >n$. Edited.

JBL wrote: Hmm, looks like it went through French? Or maybe Romanian and French share some of the same common noun-mathematical noun pairings that English doesn't have. The same second meaning exists in german, too (field = "Körper" = body). I would actually guess that this is not just an accident, but that there was some intention behind using this word. A similar example would be "ring" (which in german, as in english, also denotes a torus-shaped object [it's funny that mathematics has found another word for the geometric shape of a ring]). As far as I know, this goes back to Hilbert using "Zahlring" (number ring), but I never understood why he used that term.

No special intention; in Romanian the word is "corp", for either a field, which in English is supposed to be commutative, or not, which in English is named a skew field, or division ring. However, the statement of the problem, as asked, specified "corp comutativ", i.e. commutative division ring, so it should be translated into English by field, not "body", which is a barbarism, if not just a "false friend".

mavropnevma wrote: No special intention "This" in my post refers to the sentence before, i.e. why the inventor of the word corp/Körper/field/etc. used exactly that word and not something else.

Assume the contrary. Then there exists a nonconstant polynomial $P\in K[x]$ of minimal degree for which $P(a)\in G$ for all $a\in K$. Note that $\deg P\leq n$. Suppose FTSOC $\deg P\geq 2$. Then $Q(x):=P(x+1)-P(x)$ has degree $\deg P-1\geq 1$: if the leading coefficient of $P(x)$ is $c$ then the leading coefficient of $Q$ is $c\cdot\deg P\neq 0$. Since $G$ is a group, we also have $Q(a)\in G$ for all $a\in K$, contradicting the minimality of $P$. Thus $P$ is linear so it is a permutation of $K$. But now $P$'s range is $K$ so it contains an element not in $G$, a contradiction. $\square$