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$(a + b + c)^p - (a^p + b^p + c^p)$ is always divisible by

(a) $p - 1\quad$ (b) $a + b + c\quad$ ( c ) $p\quad$ ( d ) $p^2 - 1$

$p$ is prime

I am able to solve this by substituting values and by euler theorem by assuming $( a + b + c )$ are co prime with $p$.

But I am unable to solve it by expansion nothing is working

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    @BabakSorouh I don't know but why do ask the question?2012-07-12

3 Answers 3

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$(a+b+c)^p-(a^p+b^p+c^p)=(a+b+c)^p-(a+b+c)-(a^p-a)-(b^p-b)-(c^p-c)$

Since $x^p-x$ is 0 mod $p$, the above is always divisible by $p$ whether or not $a+b+c$ and $p$ are co-prime!

So option (c) is true

(b) and (d) are ruled out by setting $a=1,b=c=2,p=3$

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    (a) is ruled out by $a=b=1$, $c=0$, $p=5$.2012-07-12
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Expansion will do it, along with information about divisibility of certain multinomial or binomial coefficients by $p$.

But it is much cleaner to use Fermat's Theorem, in the version that says $x^p\equiv x\pmod{p}$. This holds with no restrictions on $x$. So in particular you need not separate out the case where $a+b+c$ is divisible by $p$.

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$(a+b+c)^p=a^p+b^p+c^p+$ terms containing some ${p\choose k}$ where $1\leq k\leq p-1$. As we know, ${p\choose k}=\frac{p}{k} {p-1\choose k-1}$ and as $p$ is a prime so $k\nmid p$ (as $k\lt p$), therefore, $\frac{{p-1\choose k-1}}{k}$ is an integer and thus $p\mid {p\choose k}$ and hence $(a+b+c)^p-a^p+b^p+c^p=$ terms containing some ${p\choose k}$ which is divisible by $p$ for sure.