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Let $x,y,z$ be integers and $p$ an odd integer.How can I prove that $$\sum_{i+j+k=p}{p\choose i+j+k}x^iy^jz^k\equiv x^p+y^p+z^p\pmod{(x+y)(y+z)(z+x)}$$

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    Hint: The sum is $(x+y+z)^p$. – 2017-02-11

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We have to prove that the polynomial $f=(x+y)(x+z)(y+z)$ divides the polynomial $g=(x+y+z)^p-(x^p+y^p+z^p).$ It is enouth to show that all roots of $f$ are roots and for $g.$ Take, for example, a root $x=-y$ and put into $g$. We get $z^p-( (-y)^p+y^p+z^p)=0$ for odd $p$. Thus $g=(x+y)g_1$ for some polynomial $g_1.$ In a similar way working with $x=-z$ and $y=-z$ we get that $g=0 \mod (x+y)(x+z)(y+z).$

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    Thanks for input. Would you happen to figure out what the summation expression of that polynomial of degree n-3 is? – 2017-02-11
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    what is $n?$... – 2017-02-11
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    Let say an odd prime – 2017-02-11
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    I can choose any n and wolfram will do the rest. But I wanna know what the general expression is like. – 2017-02-11
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    I mean - in your question there is no any $n.$ So, what is $n?$ – 2017-02-11
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    Sorry, I meant p. – 2017-02-11
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    No matter. Just $p-3$ must be odd. – 2017-02-11
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    Nope. $p-3$ is even since p is odd. – 2017-02-11
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    Well, in your question $p$ is odd and that formula is wrong for $p-3.$ – 2017-02-11
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    suppose that p=5, f(x,y,z)= 5(x^2+y^2+z^2+ xy+ yz+zx). What is the summation notation of that expression? – 2017-02-11
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    $$(x+y+z)^p-(x^p+y^p+z^p)=f(x,y,z)(x+y)(y+z)(z+x)$$ I would like to know the summation notation of $f(x,y,z)$. I hope this is much clearer. Let me give you your due credit and ask another question. Thanks. – 2017-02-11