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Given, $f(x)=\sum_{i=0}^na_{i}x^{n-i}$ and $a_0=1$ have integral coefficients.If there exist four distinct integers a,b,c and d such that $f(a)=f(b)=f(c)=f(d)=5$,show that there is no integer $k$ such that $f(k)=8$

Source: An Excursion In Mathematics: For RMO and INMO

gaufler
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    Have you tried anything yet? – Mark Bennet Aug 07 '15 at 12:49
  • I have tried plugging in the values and subtracting one from the other. But could not understand anything. I think induction needs to be applied. But don't know how – gaufler Aug 07 '15 at 12:50
  • If it were just three distinct integers $f(x)=x^3-3x^2-x+8$ then $f(-1)=f(1)=f(3)=5$ and $f(0)=8$ – Mark Bennet Aug 07 '15 at 12:52
  • This question is different from what you're asking, but it does at least give some insight as to how you might prove that integer-coefficient polynomials satisfying integer-value restrictions can be shown to have no integer-valued solutions at integer points. So actually the problems are remarkably similar and the tools developed there may be of some use to you. – Eric Stucky Aug 07 '15 at 12:53
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    What's the source of this problem, please? – Gerry Myerson Aug 07 '15 at 12:55
  • Clearly all of $a,b,c,d$ need to have the same parity. Otherwise, see this problem, $f(m)$ will be odd for all $m$. – Jyrki Lahtonen Aug 07 '15 at 13:23

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Hint: Write $g(x)=f(x)-5$ by hypotesis $g(x)=(x-a)(x-b)(x-c)(x-d)h(x)$. Thus if there exists such $k$ we have $3=(k-a)(k-b)(k-c)(k-d)h(k)$, and you can conclude.

Euler88 ...
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