Question

Let $k\ge 6$ be an integer number. Prove that the system of equations $\{\begin{array}{cc}{x}_1+x_2+\cdot \cdot \cdot +x_{k-1}=x_k,& \\ x_1^3+x_2^3+\cdot \cdot \cdot +x_{k-1}^3=x_k,& \end{array}$has infinitely many integral solutions.

• A. $X_6=-n$ and $x_i=0$ for all i >6.
• B. $X_6=1$ and $x_i=-1$ for all i >6.
• C. $X_6=-n$ and $x_i=0$ for all i >6.
• D. None of the above

Answer 1

Answer: (A)

$X_6=-n$ and $x_i=0$ for all i >6.

Consider the identity $(m+1{)}^3+(m-1{)}^3+(-m{)}^3+(-m{)}^3=6m$
If n is arbitrary integer, then $\frac{n-n^3}{6}$ is also an integer since
$n-n^3=-(n-1)n\times (n+1)$ and the product of three consecutive integers is divisible by 6. Setting $m=\frac{n-n^3}{6}$ in the above gives
${(\frac{n-n^3}{6}+1)}^3+{(\frac{n-n^3}{6}-1)}^3+{\left(\frac{n^3-n}{6}\right)}^3+\left(\frac{n^3-n}{6}\right)+n^3=n.$
On the other hand, we have ,
$(\frac{n-n^3}{6}+1)+(\frac{n-n^3}{6}-1)+\frac{n^3-n}{6}+\frac{n^3-n}{6}+n=n,$ yielding the following solution for
$k=6:x_1=\frac{n-n^3}{6}+1,x_2=\frac{n-n^3}{6}-1,x_3=x_4=\frac{n^3-n}{6},x_5=x_6=n.$ For $k>6$ we can take $x_1$ to $x_5$ as before,$x_6=-n$ and $x_i=0$ for all $i>6$