What does $ k= \overline{n1}$ mean?

That would be a number, but you overline it because it has unknown digits in it (n in here) $ 10=\overline{10}$ You can't write $ n1$ you must write $ \overline{n1}$

Hmm, I wasn't sure about this, cause in the version orl sent to me, they write: Problem NT3. Find all pairs of positive integers $ (m,n)$ such that $ m^2=nk +2$, where $ k$ is the number obtained from $ n$ by writing 1 on the left side of its decimal representation. Wouldn't that mean, that $ k= \overline{1n}$?

Go to JBMO Shortlists in National Olympiads forum and see. I don't know

$ m^2 = n(10n + 1) + 2 \implies 40m^2 - 79 = (20n + 1)^2 \implies$ $ (40m + 20n + 1)(40m - 20n - 1) = 79$, possible factor pairs are (1,79) and (-1,-79), no solution.

Sorry, I've found out it's $ k=\overline{1n}$.

Your solution is wrong! test the numbers, 40 m^2 isn't a squre.

Having bust my brains out trying (and failing) to do the thing as originally stated in JBMO with the mistake in it, I feel justified in reviving this thread to post a hopefully correct answer to the corrected problem, with $ k= \overline{1n} $, (rather than $ k= \overline{n1}$) If $n$ is a single digit, trying all the possibilities gives $(m,n) = (11,7)$ as a solution. If $n$ is two digits, then $k=n+100$ so $nk=n^2+100n$ Since $100 = 0 mod(4)$, $nk$ must be the same as $ n^2 mod(4)$ which is either $0$ or $1 mod (4)$ since $n^2$ is a square. So $nk + 2 = 2$ or $3 mod(4)$ , which cannot be a perfect square. The same argument applies when n is more than 2 digits. Hence $(m,n) = (11,7)$ is the only solution. Merlin

(11,7) is apparently NOT a solution! from the equation we get $(20n+1)^2-40m^2=-79$ let us consider the (OC's) equation $x^2-40y^2=-79$ its solutions $x_1=9,x_2=411,x_{n+1}=38x_n-x_{n-1}$ hence it's 2-periodic modulo 20 with the period (9,11),so it cant be $20n+1$,yielding no solution to the original equation.

$(11,7)$ is a solution, since $11^2 = 121 = 119 + 2 = 7\cdot 17 + 2$. The problem statement has been amended, from $k = \overline{n1}$ to $k = \overline{1n}$.

mavropnevma wrote: $(11,7)$ is a solution, since $11^2 = 121 = 119 + 2 = 7\cdot 17 + 2$. The problem statement has been amended, from $k = \overline{n1}$ to $k = \overline{1n}$. oh gosh!the problem has been typed uncorrectly! but the original problem can also be a divine one~

İf it is 1n, then the only solution is (m;n)=(11;7)