Prove $f$ not continuous at SEEMOUS Contest

Question

Let $n$ be a nonzero natural number and $f:\mathbb{R}\to\mathbb{R}\setminus\{0\}$ be a function such that $f(2014) = 1 − f(2013)$. Let $x_1,x_2,x_3,...,x_n$ be real numbers not equal to each other. Prove $f$ not continuous if $$\begin{vmatrix} 1+f(x_1) & f(x_2) &f(x_3) &\ldots & f(x_n) \\ f(x_1) & 1+f(x_2) &f(x_3) &\ldots & f(x_n) \\ f(x_1) & f(x_2) &1+f(x_3) &\ldots & f(x_n)\\ \cdots & \cdots & \cdots & \ddots & \cdots \\ f(x_1) & f(x_2) &f(x_3) &\ldots & 1+f(x_n)\\ \end{vmatrix}=0.$$

Answer 1

First substract the last row from all rows, next replace the last column by the sum of all columns. The result is the matrix $$ \left(\begin{array}{ccccc} 1 & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ f(x_1) & f(x_2) & \cdots & f(x_{n-1}) & A \end{array}\right) $$ where $A = 1 + f(x_1) + f(x_2) + \cdots + f(x_n)$. Since the determinant is still $0$, we find $A = 0$ or $f(x_1) + f(x_2) + \cdots + f(x_n) = - 1$. Hence at least one of the values $f(x_i)$ is negative. However, $f(2013) + f(2014) = 1$ implies that at least one of the values $f(2013)$ and $f(2014)$ is positive. Since $f$ takes both positive and negative values whereas $0$ is not in the range of $f$, we conclude that $f$ cannot be continuous.