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Without using the identity: $$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)(x^2+xy+x z+y^2+y z+z^2)$$ I am trying to find an argument to prove that for all $x,y,z$ there exists a nonzero integer $k$ such that:

$$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(x+z)(y+z)k$$ Proving $5$ divides the lefthand(LH) side expression is easy using FlT. How do I prove that $(x+y)(x+z)(y+z)$ divides the LH? Any hints?

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Using the factor theorem: the polynomial $P(x)$ is divisible by $x-a$ if an only if $P(a) = 0$.

Then $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+y$ given that $$P(-y) = (-y+y+z)^5-(-y)^5-y^5-z^5 =z^5+y^5-y^5-z^5 = 0$$

Similarly, $P(x)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $x+z$ given that $P(-z)=0$.

Finally, $P(y)=(x+y+z)^5-x^5-y^5-z^5$ is divisible by $y+z$ given that $P(-z)=0$.

Then we conclude that $(x+y+z)^5-x^5-y^5-z^5$ is divisible by $(x+y)(x+z)(y+z)$.

Do you agree?

Juan Ospina
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    It looks like a fair argument. Let's see what the mathstack community thinks. –  Sep 19 '15 at 23:21