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Let $m, n, k$ be integers such that $(m-n)^2 + (n-k)^2 + (k-m)^2 = mnk$. Prove that, $m^3 + n^3 + k^3$ is divisible by $m + n + k + 6$.

We know,
$a^3 + b^3+c^3-3abc=\frac{1}{2}(a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\}$
$\therefore m^3+n^3+k^3-3mnk = \frac{1}{2}(m+n+k)\{(m-n)^2+(n-k)^2+(k-m)^2\}$
Or,$m^3+n^3+k^3-3mnk = \frac{1}{2}(m+n+k)\{mnk\}$
Or, $2m^3+2n^3+2k^3-6mnk= (m+n+k)(mnk)$
Or,$2(m^3+n^3+k^3)=(m+n+k)(mnk)+6mnk$
$\therefore 2m^3+2n^3+2k^3=(mnk)(m+n+k+6)$
Now at least one of the integers m,n,k must be even or else the equality doesn't hold because LHS even and RHS odd.So at least one of m,n or k must be even. So mnk must be even too. Let, $\frac{mnk}{2}=p$ where p is an integer then
$m^3+n^3+k^3=p(m+n+k+6)$
Thus $(m+n+k+6)$ divides $m^3+n^3+k^3$ [Proved]

This was too easy for a camp test problem....I also found problem 2 in exam to be really easy as well

We use the following well-known lemma:
For integers $a, b, c$, we have $$a + b + c | a^3 + b^3 + c^3 - 3abc$$

Notice that $(m+2)^3+(n+2)^3+(k+2)^3 - 3(m+2)(n+2)(k+2) = m^3+n^3+k^3$
Set $a = m + 2, b = n + 2, c = k + 2$ to get $m + n + k + 6 | (m + 2)^3 + (n + 2)^3 + (k + 2)^3 - 3(m + 2)(n + 2)(k + 2)$ and the result follows.

FuadAlAlam wrote: ↑

Mon Dec 07, 2020 1:23 pm

Let $m, n, k$ be integers such that $(m-n)^2 + (n-k)^2 + (k-m)^2 = mnk$. Prove that, $m^3 + n^3 + k^3$ is divisible by $m + n + k + 6$.

mnk=$(m-n)^2+(n-k)^2+(k-m)^2$
=> mnk=$2(m^2+n^2+k^2-mn-nk-mk)$
=> 3mnk=$6(m^2+n^2+k^2-mn-nk-mk)$

Now,
$m^3+n^3+k^3-3mnk$=(m+n+k)($m^2+n^2+k^2-mn-nk-mk$)
=>$m^3+n^3+k^3$=(m+n+k)($m^2+n^2+k^2-mn-nk-mk$)+3mnk
=>$m^3+n^3+k^3$=(m+n+k)($m^2+n^2+k^2-mn-nk-mk$)+$6(m^2+n^2+k^2-mn-nk-mk)$
=>$m^3+n^3+k^3$=(m+n+k+6)($m^2+n^2+k^2-mn-nk-mk$)
And we are done