Q) If \(P(x)\) is a polynomial with integer coefficients and \(a,b,c\) three distinct integers, then show that it is impossible to have \(P(a)=b, P(b)=c, P(c)=a\)
Solution:
The solution revolves around an important property of integer coefficient Polynomials,
For any 2 integers \(x,y\), \(P(x)-P(y)\) is always divisible by \(x-y\).
So from this we have \(a-b\) divides \(P(a)-P(b)\) which implies \(a-b\) divides \(b-c\), \(b-c\) divides \(c-a\), \(c-a\) divides \(a-b\).
Once we have this, the remaining solution is trivial. We see that the above scenario is possible when \(|a-b|=|b-c|=|c-a|\). Let’s assume \(a>b>c\), thus we have
\(a-b=b-c=a-c\), which violates the condition that \(a \ne b\ne c\).
How to come to the |a-b|=|b-c|=|c-a| conclusion? Is that a theorem by any chance?? Also do you have the inmo 1986 qsn paper?