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Polynomial # 1 (Posted on 2006-05-26) Difficulty: 3 of 5
Let f(x) be a nonconstant polynomial in x with integer coefficients and suppose that for five distinct integers a1, a2, a3, a4, a5, one has f(a1)= f(a2)= f(a3)= f(a4)= f(a5)= 2.

Find all integers b such that f(b)= 9.

No Solution Yet Submitted by Ravi Raja    
Rating: 5.0000 (1 votes)

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Some Thoughts re(2): Solution (spoiler) | Comment 6 of 7 |
(In reply to re: Solution (spoiler) by Richard)

Actually, that h(x) has integer coefficients, in this case, is simple, and I didn't need Gauss' Lema.

First, (x-a1)(x-a2)(x-a3)(x-a4)(x-a5) is a polynomial with integer coeefficients, since a1 through a5 are integers. Let's write this product as r(x)= x^5+r4.x^4+...+r1.x+r0.

Suppose we have f(x)=fn.x^n+...+f2.x^2+f1.x+f0 with f0 through fn all integers.

Finally write h(x)=hm.x^m+...+h2.x^2+h1.x+h0. (In this case, m=n-5, but that doesn't matter.)

When we multiply r(x) and h(x) to get f(x), the highest order  term (which must be fn) is hm... so hm is an integer. The next highest order term (f(n-1)) is r4.hm+h(m-1), so h(m-1) is also an integer. Keep working this way, and all h's turn out to be integers.




Edited on May 27, 2006, 8:11 am
  Posted by e.g. on 2006-05-27 08:09:37

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