1974 IMO Problems/Problem 6
Let be a non-constant polynomial with integer coefficients. If is the number of distinct integers such that prove that where denotes the degree of the polynomial
Lemma: Let be a polynomial with integer coefficients which is not constant. Then if obtains (or ) as its values for at least four times then ( or ) for all . Proof. Assume that for distince. Then if there's which then so where is a polynomial with the integer coefficients! So which is impossible cause can not presents as product of more than three distince numbers! This proved the lemma!
Back to our problem: For convinet put and . Firstly if then . Assume . If equation with more than three integer points (ie.. at least ) then equation implies so , ie... . The same case for equation . So . If then . Now assume that . In this case if then .
So let us show that . In fact if then has three integers distince roots, and the same for . So and where distince and distince and all with are integers! Then for all . So . Finally, we have for and because that can not presents as products of three distince numbers so , we may assume . Because so This means . So we must have which follows , which contracts!. So and we're done.
The above solution was posted and copyrighted by pluricomplex. The original thread for this problem can be found here: 
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