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Let P(x) be a polynomial with integer coeeficients of the form

$P(x)=x^5+ax^4+bx^3+cx^2+dx+e$

for some integers a,b,c,d and e.

If $$\left\{ \begin{array}{l} P(1)=1\\P(2)=2\\P(3)=3\\P(4)=4\\P(5)=5 \end{array} \right.$$

what is P(6)?

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  • $\begingroup$ You're basically given a system of linear equations that you need to solve in order to get $a,b,c,d$ and $e$. Then you'll be able to explicitly compute $P(6)$. $\endgroup$ Commented Dec 4, 2016 at 20:13
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    $\begingroup$ $P(x)-x$ is a polynomial of degree $5$ with leading coefficient $1$ and vanishes on $5$ points $1,2,3,4,5$, this means.... $\endgroup$ Commented Dec 4, 2016 at 20:19
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    $\begingroup$ @john_jerome Hmmm, it's possible to do it by solve a,b,c,d,e firstly then get P(6), but I guess we should not do it that way. so I guess some trick needed to solve P(6). $\endgroup$
    – lucky1928
    Commented Dec 4, 2016 at 20:19
  • $\begingroup$ See this answer. $\endgroup$ Commented Dec 4, 2016 at 21:15

1 Answer 1

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The polynomial $$P(x)-x$$ has $5$ roots, namely $1,2,3,4,5$. Since the degree of $P(x)-x$ is $5$ and the leading coefficient is $1$ , we have $$P(x)-x=(x-1)(x-2)(x-3)(x-4)(x-5)$$ This implies $$P(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+x$$Just insert $6$ to get the result.

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  • $\begingroup$ Following achille's hint. I did not notice this hint and had the same idea independently. In fact, you have a nice shortcut here. $\endgroup$
    – Peter
    Commented Dec 4, 2016 at 20:28
  • $\begingroup$ Awesome, so from 1,2,3,4,5, we can guess it's x, but if it's 1,10,3,4,5, how can we guess the P(x)? $\endgroup$
    – lucky1928
    Commented Dec 4, 2016 at 20:47
  • $\begingroup$ This method only works in special cases, of course $\endgroup$
    – Peter
    Commented Dec 4, 2016 at 20:49

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