Polynomial Quest

Algebra Level 3

Find the number of polynomials $P(x)$ with integer coefficients such that $P(a)=b; P(b)=c; P(c)=a$ , where $a,b,c$ are distinct natural integers.

The answer is 0.

2 solutions

Surya Prakash
Jun 28, 2015

Given, $P(x)$ is a polynomial.
$P(a)-P(b)=b-c$ ;
$P(b)-P(c)=c-a$ ;
$P(c)-P(a)=a-b$ .
But $b-c|P(b)-P(c)$ .
So, $b-c|c-a$ which forces that $b-c\leq c-a$ .
Similarly, $c-a|a-b$ and $a-b|b-c$ .
So, $c-a\leq a-b; a-b\leq b-c$ .
So, $a-b=b-c=c-a$ . This implies that $a=b=c$ . But given that a, b, c are distinct.
So, there are no solutions of a, b, c.

Moderator note:

Is it possible to generalize this?

Same approach :-) +1

Kazem Sepehrinia - 5 years, 11 months ago

When we say that b - c divides p (b) - p (c) , aren't we assuming that p (x) is a polynomial with integer coefficients ?

Rohit Kumar - 5 years, 11 months ago

Thanks for remembering .... I forgot to add the main point that P(x) has integer coefficients.

Surya Prakash - 5 years, 11 months ago

thanks for editing it. nice problem

Rohit Kumar - 5 years, 11 months ago

Ravi Dwivedi
Jul 2, 2015

Suppose a,b,c are distinct integers such that p(a)=b,p(b)=c,and p(c)=a. Then p(a)-p(b)= b-c p(b)-p(c)= c-a p(c)-p(a)=a-b

But for any two distinct integers m and n, m-n divides p(m)-p(n).

Thus a-b|b-c, b-c|c-a, c-a|a-b

These force a=b=c, a contradiction. Hence there are no integers a,b and c such that p(a)=b,p(b)=c,p(c)=a.

So the answer is 0

Polynomial Quest

The answer is 0.

2 solutions

Moderator note:

0 pending reports