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if a polynomial P with integer coefficients has three distinct integer zeroes, then show that P(n) not equal to 1 for any integer n.

10 years ago

Dear Aritra,

The question is incorrect,

There are so many examples to prove it incorrect such as :

let P = (x-1)(x-2)(x-3)* (1/6 )

its zeroes are 1,2,3 which are integers and its value at 4(an integer) is 1.

Best Of luck

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Thanks

Aman Bansal

9 years ago

Hi Aritra,

The above example P(x) = 1/6*(x-1)(x-2)(x-3) does not have integer co-efficients.

Let P(x) = (x-a)(x-b)(x-c)*Q(x). where A,a,b,c are integers with a,b,c being distinct. Q(x) is any polynomial with integer co-efficients.

Now for any integer n, consider P(n).

P(n) = (n-a)(n-b)(n-c)Q(n).

In the above expression Q(n) is any integer.

Assume P(n) = 1 for some n.

Since all are integers, only possibility is |n-a| = |n-b| = |n-c| = |Q(n)| = 1.

Consider only |n-a|=|n-b|=|n-c|=1 ------------------ (1)

Expression (1) can never be true for distinct integers a,b,c as explained below.

|n-a|=|n-b|

Only possibility n-a = b-n, ie a+b=2n.

similarly b+c=2n.

Subtraction a-c=0 or a=c --------- (contradiction).

Hence P(n) can never be equal to 1 for any integer n.

Regards,