MHB Proving Divisibility with the Power of 5

[anemone]

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]

Spoiler

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 $5(a-b)(b-c)(c-a)$

 

[anemone]

Spoiler

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 $5(a-b)(b-c)(c-a)$

Sorry for the late reply, kaliprasad and thanks for participating!

I like your solution because of the way you introduced a polynomial function for the problem and well done!(Sun)

 

[caffeinemachine]

Given unequal integers $a, b, c$ prove that $(a-b)^5+(b-c)^5+(c-a)^5$ is divisible by $5(a-b)(b-c)(c-a)$.

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.

 

[anemone]

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.

Hey caffeinemachine,

Thanks for participating and that's another trick for me to learn today!(Sun)