Let $a, b, c$ be distinct integers, and let $P$ be a polynomial with integer coefficients. Show that it is impossible that $P(a)=b$, $P(b)=c$, and $P(c)=a$ at the same time.
Asked
Active
Viewed 67 times
0
-
What would the constant term of $P$ be under these conditions? – Sean Nemetz Jul 27 '17 at 02:36
-
I took some time posting my solution, I just wanted to share this problem with you guys. :) – problembuster Jul 27 '17 at 02:48
-
I request you to accept an answer and close the question. – Sarvesh Ravichandran Iyer Jan 08 '19 at 16:34
2 Answers
2
.Hint : By the remainder theorem, $P(x) - P(y)$ is divisible by $x-y$, for all $x,y$.
Assume that $a < b < c$, since they are distinct, and see that putting $x=c,y=a$ gives that $c-a$ divides $a-b$. Now, can you see the problem with this statement?
Sarvesh Ravichandran Iyer
- 76,105
1
Assume otherwise.
By the remainder theorem, $a-b$ divides $b-c$, $b-c$ divides $c-a$, and $c-a$ divides $a-b$.
Then, $b-c$ also divides $a-b$, therefore, $2a = b+c$.
$c-a$ also divides $b-c$, therefore $2c = a+b$
Substracting both equations:
$2(a-c) = -(a-c)$
$a-c=0$
$a=c$
Contradiction! Therefore, we are done.
problembuster
- 391
-
Your proof and mine are almost the same, but yours looks nicer, so $+1$. – Sarvesh Ravichandran Iyer Jul 27 '17 at 03:00