Does the antidiagonal in this square matrix always contain a prime? Does the antidiagonal in the square matrix with entries $1,2,\ldots,n^2$ row by row in that order always contain a prime?
For example: 
For n=2: $\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ the antidiagonal contains two primes (2,3). 
For n=3: $\begin{bmatrix}1 & 2&3\\4&5&6\\ 7&8& 9\end{bmatrix}$ the antidiagonal contains three primes (3,5,7). 
etc...
Is there a matrix for n>1 that doesn't contain a prime in the antidiagonal?
 A: As suggested in the comments, this question is equivalent to asking for primes of the form 1 +kn, where (n+1) is given and k is an integer at most n+1.  The current observations and theory suggest the answer is yes, for every n there is a k less than n+2 so that 1+kn is prime, but we do not have a proof.  Linnik's theorem is mentioned in a question link provided in a comment of Lucia, and later results give conditional and unconditional bounds on k.  Currently (as far as I know) your question is an open problem.
Much more can be said, however.  The question is one of a class of problems of the location of primes in certain arrays.  Two such problems are: is there a prime in every row of (a standard square arrangement of) a matrix of numbers from  1 to n^2 for every n > 1?  Are there large connected components of primes in an Ulam spiral starting the spiral from any integer n?  Your question inspires the following: How many diagonals (columns) contain primes in a (standard n by n) array?  Is it always at least $2\phi(n) (\phi(n))$?  Here one needs to nail down the definition of diagonal, but that is part of the fun of exploring.  Other arrangements of numbers can be considered.  As far as I know, hexagonal arrangements have been considered only for Eisenstein primes (cf Guy's UPINT book for more).
Gerhard "Opens Minds With Open Problems" Paseman, 2016.01.16
A: While it may be hard to prove precise results, it is extremely likely that there are quite a lot of primes on this antidiagonal. If $a$ is an integer small relative to $n$, then the number of primes $p \equiv 1$ (mod $a$) with $p \leq n^{2}$ is of the order of $\frac{1}{\phi(a)}\frac{n^{2}}{\log n^{2}}$ for $n$ large enough. This is according to the quantitative version of Dirichlet's Theorem for primes in arithmetic progression, and is proved. Here we are looking at the case $a = n-1$ which is not small relative to $n$. Nevertheless, it would be interesting to compare the number of primes on the antidiagonal with 
$$\left( \prod_{ {\rm prime} p | n-1} \frac{p}{p-1} \right) \frac{n}{2 \log n}$$ for $n$ large, and to examine the limiting behaviour of the ratio of the two quantities.
