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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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You need to look at [Freshman's dream][1]. [1]: en.wikipedia.org/wiki/Freshman%27s_dream –  user17762 Jul 12 '12 at 4:50
    
@BabakSorouh I don't know but why do ask the question? –  user17762 Jul 12 '12 at 8:17

3 Answers 3

up vote 2 down vote accepted

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

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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Please try to prove or disprove part (a) also! –  Host-website-on-iPage Jul 12 '12 at 6:36
    
How x^p - x is divisible by p , its is only when x and p are co prime –  Arpit Bajpai Jul 12 '12 at 6:37
    
That's just group theory. Go modulo $p$. You then are in the multiplicative group modulo $p$, which has order $p-1$ so $x$ should have order dividing $p-1$ mod $p$, so that $x^{p-1}$ is 1 mod $p$ whenever $x$ is non-zero mod $p$. This implies $x^p$ is $x$ mod $p$ which is also true when $x$ is $0$ mod $p$ –  Host-website-on-iPage Jul 12 '12 at 6:37
    
Part (a) can also be proved or disproved quite easily. –  Host-website-on-iPage Jul 12 '12 at 6:40
    
Put a = 1, b = 3 and c = 5 and p = 2 –  Arpit Bajpai Jul 12 '12 at 6:40

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.

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