(USO 1974) Let \(a,b,c\) be three distinct integers, and let \(P(x)\) be a polynomial whose coefficients are all integers. Prove that it is not possible that the following three conditions hold at the same time: \(P(a)=b, P(b)=c,\) and \(P(c)=a\).
For a polynomial \(P(x)=ax^2+bx+c\), consider the following two kinds of transformations:
Swap coefficients \(a\) and \(c\). Hence the polynomial \(P(x)\) becomes \(cx^2+bx+a\) after this transformation.
For any number \(t\) of your choice, change the variable \(x\) into \(x+t\). For example, with the choice of \(t=1\), after this transformation, the polynomial \(x^2+x+1\) becomes \((x+1)^2+(x+1)+1=x^2+3x+3\).
Is it possible, using only a sequence of these two transformations, to change the polynomial \(x^2-x-2\) into the polynomial \(x^2-x-1\)?
In this example we will discuss division with remainder. For polynomials \(f(x)\) and \(g(x)\) non-zero with \(\deg(f)\geq \deg(g)\) there always exists polynomials \(q(x)\) and \(r(x)\) such that \[f(x)=q(x)g(x)+r(x)\] and \(\deg(r)<\deg(g)\) or \(r(x)=0\). This should look very much like usual division of numbers, and just like in that case, we call \(f(x)\) the dividend, \(g(x)\) the divisor, \(q(x)\) the quotient, and \(r(x)\) the remainder. If \(r(x)=0\), we say that \(g(x)\) divides \(f(x)\), and we may write \(g(x)\mid f(x)\). Let \(f(x)=x^7-1\) and \(g(x)=x^3+x+1\). Is \(f(x)\) divisible by \(g(x)\)?