I noticed this trend in Wolfram Alpha for that $x-y$ divides $x^3-y^3, x^5-y^5, x^7-y^7$ so I was wondring if it is true for all $ k \in \mathbb{N}$?
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Does remainder theorem ring a bell? What happens if you substitute $x=y$ in any of the higher order expressions? This is one way to look at it... – imranfat Oct 26 '16 at 15:03
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3http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b/188710 – lab bhattacharjee Oct 26 '16 at 15:04
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Out of curiosity: why only the odd powers? It holds for even powers too. – sTertooy Oct 26 '16 at 15:04
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@SteamyRoot even was trivial for me – Postskjerm Oct 26 '16 at 15:05
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@labbhattacharjee thanks I understand now – Postskjerm Oct 26 '16 at 15:06
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@imranfat ah, I get it now, slightly embarrassed at the moment – Postskjerm Oct 26 '16 at 15:08
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In general, $x^{2k+1}-y^{2k+1} = (x-y)(x^{2k} + x^{2k-1} y + \dots + x y^{2k-1} + y^{2k})$.
$$ x(x^{2k} + x^{2k-1} y + \dots + x y^{2k-1} + y^{2k}) = x^{2k+1} + x^{2k} y + \dots + x^2 y^{2k-1} + x y^{2k} $$
$$ y(x^{2k} + x^{2k-1} y + \dots + x y^{2k-1} + y^{2k}) = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^{2k}y + x^{2k-1} y^2 + \dots + x y^{2k} + y^{2k+1}$$
so subtracting the two lines proves the desired result.
Nitin
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