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anemone
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Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
kaliprasad said:Let us put $f(x) = (x-b)^5 + (b-c)^5 + (c-x)^5$
Putting x = b we get f(b) = 0
so (x-b) is a factor
so a-b is a factor of $(a-b)^5 + (b-c)^5 + (c-a)^5$
similarly we have (b-c) and (c-a) are factors
now $(a-b)^5 = a^5 – 5a b^4 + 10a^2b^3 – 10 a^3 b^2 +5 a^4 b – b^5 = a^5 – b^5 + 5m$ where $m = - a b^4 + 2a^2b^3 – 2 a^3 b^2 + a^4 b b^4$
similarly $(b-c)^5 = b^5 – c^5 + 5n$
$(c-a)^5 = c^5 – a^5 + 5k$
Adding we get $(a-b)^5 + (b-c)^5 + (c-a)^5 = 5 (m+n+k)$
So $(a-b)^5 + (b-c)^5 + (c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$
Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.anemone said:Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.
caffeinemachine said:Let $S_5=(a-b)^5+(b-c)^5+(c-a)^5$, $S^3=(a-b)^3+(b-c)^3+(c-a)^3$ and $S_2=(a-b)^2+(b-c)^2+(c-a)^2$.
From this thread http://mathhelpboards.com/challenge-questions-puzzles-28/prove-%5E5-b%5E5-c%5E5-5%3D-%5E3-b%5E3-c%5E3-3-%2A-%5E2-b%5E2-c%5E2-2-a-8276.html we know that
$$\frac{S_5}{5}=\frac{S_3}{3}\frac{S_2}{2}$$.
Note that since $(a-b)+(b-c)+(c-a)=0$, we have $S_3=3(a-b)(b-c)(c-a)$.
Thus $$S_5=5(a-b)(b-c)(c-a)\frac{S_2}{2}$$.
Clearly $2$ divides $S_2$ and thus we are done.
The Divisibility Challenge is a mathematical game where players are given a starting number and a set of rules to determine if the number is divisible by certain integers.
Players are given a starting number and a set of rules, such as "divisible by 2" or "not divisible by 5". They then must use these rules to determine if the starting number is divisible by the specified integers. The goal is to correctly identify all the divisibility rules and determine if the starting number is divisible by them.
The purpose of the Divisibility Challenge is to help players practice their mathematical skills and improve their understanding of divisibility rules. It can also be a fun way to challenge yourself or others and see how quickly you can solve the challenge.
Yes, the Divisibility Challenge can be played with any starting number. The rules may vary depending on the starting number, but the concept remains the same.
No, there is no time limit for completing the Divisibility Challenge. Players can take as much time as they need to solve the challenge, making it a great game for all levels of mathematical abilities.