Show that for any $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ different integers:

$\displaystyle \frac {a^{2008}}{(a-b)(a-c)}+\frac {b^{2008}}{(b-a)(b-c)}+\frac {c^{2008}}{(c-b)(c-a)}\in\mathbb{Z}$

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- Apr 12th 2011, 05:28 AMjames_bondShow that it is an integer
Show that for any $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ different integers:

$\displaystyle \frac {a^{2008}}{(a-b)(a-c)}+\frac {b^{2008}}{(b-a)(b-c)}+\frac {c^{2008}}{(c-b)(c-a)}\in\mathbb{Z}$ - Apr 13th 2011, 12:22 AMOpalg
I don't see a neat way to do this. The only method that I could find is to put it over a common denominator, to get

$\displaystyle -\dfrac{a^n(b-c)+b^n(c-a) + c^n(a-b)}{(b-c)(c-a)(a-b)} = -\dfrac1{(b-c)(c-a)(a-b)} \begin{vmatrix}a^n&b^n&c^n\\ a&b&c\\ 1&1&1\end{vmatrix}.$

Then doing row and column operations on the determinant I found that it is equal to $\displaystyle -(b-c)(c-a)(a-b)\sum a^ib^jc^k$, where the sum is taken over all triples (i,j,k) of nonnegative integers such that $\displaystyle i+j+k=n-2.$