I am trying to factorize the expression: $$(x+y+z)^p-[(-x+y+z)^p+(x-y+z)^p+(x+y-z)^p]$$ where $p$ is an odd prime and $x,y,z$ are any non-zero integers.
I know that it is divisible by $pxyz$.
How do I find/write the remaining factor(s) in summation notation? I tried to use the trinomial expansion but I keep getting lost. Any hints?
NOTE: I would write this as $A = -x+y+z,$ $B = x-y+z,$ $C=x+y-z,$ thus demanding that $A,B,C$ are all odd or all even. Then the expression is $$ (A+B+C)^p -A^p - B^p - C^p. $$ Probably good for something
Appears that I should have written $p=5$ as $$ 40xyz(2x^2 + 2 y^2 + 2 z^2) $$ This way, we get a consistent first term, $$ 8pxyz \left( \; \frac{p-1}{2} \, x^{p-3} + \mbox{other} \right) $$
bit of a mess to typeset. I did primes $3,5,7,11$ in gp-pari, wrote out the factors, each in a recognizable pattern. Don't see anything overall except for the even exponents and considerable symmetry.
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parisize = 4000000, primelimit = 500509
? p = 3; h = (x + y + z)^p; s = (- x + y + z)^p + (x - y + z)^p + (x + y - z)^p; d = h - s
%1 = 24*z*y*x
? p = 5; h = (x + y + z)^p; s = (- x + y + z)^p + (x - y + z)^p + (x + y - z)^p; d = h - s
%2 = 80*z*y*x^3 + (80*z*y^3 + 80*z^3*y)*x
? e = 80 * x * y * z * (x^2 + y^2 + z^2)
%3 = 80*z*y*x^3 + (80*z*y^3 + 80*z^3*y)*x
? d - e
%4 = 0
? p = 7; h = (x + y + z)^p; s = (- x + y + z)^p + (x - y + z)^p + (x + y - z)^p; d = h - s
%6 = 168*z*y*x^5 + (560*z*y^3 + 560*z^3*y)*x^3 + (168*z*y^5 + 560*z^3*y^3 + 168*z^5*y)*x
?
?
? d
%7 = 168*z*y*x^5 + (560*z*y^3 + 560*z^3*y)*x^3 + (168*z*y^5 + 560*z^3*y^3 + 168*z^5*y)*x
?
? e = 56 * x * y * z * (3 * x^4 + 3 * y^4 + 3 * z^4 + 10 * y^2 * z^2 + 10 * z^2 * x^2 + 10 * x^2 * y^2 )
%8 = 168*z*y*x^5 + (560*z*y^3 + 560*z^3*y)*x^3 + (168*z*y^5 + 560*z^3*y^3 + 168*z^5*y)*x
? d - e
%9 = 0
?
?
? p = 11; h = (x + y + z)^p; s = (- x + y + z)^p + (x - y + z)^p + (x + y - z)^p; d = h - s
%10 = 440*z*y*x^9 + (5280*z*y^3 + 5280*z^3*y)*x^7 + (11088*z*y^5 + 36960*z^3*y^3 + 11088*z^5*y)*x^5 + (5280*z*y^7 + 36960*z^3*y^5 + 36960*z^5*y^3 + 5280*z^7*y)*x^3 + (440*z*y^9 + 5280*z^3*y^7 + 11088*z^5*y^5 + 5280*z^7*y^3 + 440*z^9*y)*x
?
?
? d
%11 = 440*z*y*x^9 + (5280*z*y^3 + 5280*z^3*y)*x^7 + (11088*z*y^5 + 36960*z^3*y^3 + 11088*z^5*y)*x^5 + (5280*z*y^7 + 36960*z^3*y^5 + 36960*z^5*y^3 + 5280*z^7*y)*x^3 + (440*z*y^9 + 5280*z^3*y^7 + 11088*z^5*y^5 + 5280*z^7*y^3 + 440*z^9*y)*x
?
?
? e = 88 * x * y * z * (5 * x^8 + 5 * y^8 + 5 * z^8 + 60 * y^6 * z^2 + 60 * y^2 * z^6 + 60 * z^6 * x^2 + 60 * z^2 * x^6 + 60 * x^6 * y^2 + 60 * x^2 * y^6 + 126 * y^4 * z^4 + 126 * z^4 * x^4 + 126 * x^4 * y^4 + 420 * x^4 * y^2 * z^2 + 420 * x^2 * y^4 * z^2 + 420 * x^2 * y^2 * z^4 )
%12 = 440*z*y*x^9 + (5280*z*y^3 + 5280*z^3*y)*x^7 + (11088*z*y^5 + 36960*z^3*y^3 + 11088*z^5*y)*x^5 + (5280*z*y^7 + 36960*z^3*y^5 + 36960*z^5*y^3 + 5280*z^7*y)*x^3 + (440*z*y^9 + 5280*z^3*y^7 + 11088*z^5*y^5 + 5280*z^7*y^3 + 440*z^9*y)*x
?
? d - e
%13 = 0
?
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? p = 13; h = (x + y + z)^p; s = (- x + y + z)^p + (x - y + z)^p + (x + y - z)^p; d = h - s
e = 104 * x * y * z * ( 6 * x^10 + 6 * y^10 + 6 * z^10 +
110 * y^2 * z^8 + 110 * y^8 * z^2 + 110 * z^2 * x^8 + 110 * z^8 * x^2 + 110 * x^2 * y^8 + 110 * x^8 * y^2 +
396 * y^4 * z^6 + 396 * y^6 * z^4 + 396 * z^4 * x^6 + 396 * z^6 * x^4 + 396 * x^4 * y^6 + 396 * x^6 * y^4 +
1320 * x^6 * y^2 * z^2 + 1320 * x^2 * y^6 * z^2 + 1320 * x^2 * y^2 * z^6 +
2772 * x^2 * y^4 * z^4 + 2772 * x^4 * y^2 * z^4 + 2772 * x^4 * y^4 * z^2 )
By using multinomial theorem, it's easy to see that after cancellation, all terms in your expression will have coefficient divisible by $p$ (since only terms which don't have such coefficient when expanding, say, $(x+y-z)^p$ are $x^p,y^p,-z^p$, and these cancel out from all the sums). Now we need to show all terms not containing all of $xyz$ get cancelled out (as then all leftover terms will be of the form $kpx^ay^bz^c,k\in\Bbb Z,a,b,c>0$ which will end the proof). Suppose the term doesn't contain $z$, so is of the form $x^ay^{p-a}$. We have from binomial theorem $(x+y+z)^p=\text{(stuff containing $x$)}+(y+z)^p=\binom{a}{p}y^az^{p-a}+\text{(other terms)}$. It's easy to see that from $(-x+y+z)^p,(x-y+z)^p,(x+y-z)^p$ you will get, in some order, $\binom{a}{p}y^az^{p-a}+\binom{a}{p}y^az^{p-a}-\binom{a}{p}y^az^{p-a}+\text{(other terms)}$. So, all in all, terms containing $y^az^{p-a}$ will all cancel out and we are home.
