Hint

Maxim Bogdan wrote: Let $ \sqrt {23} > \frac {m}{n}$, with $ m,n$ positive integers. i) Prove that $ \sqrt {23} > \frac {m}{n} + \frac {3}{mn}.$ ii) Prove that $ \sqrt {23} < \frac {m}{n} + \frac {4}{mn}$ occurs infinitely often, and do exhibit(at least) three such examples. Dan Schwartz a) Use quadric congruence , it is easy to check that : $ (\frac{-k}{23})=-1,\forall 1\leq k \leq 6$ Thus the equation $ 23n^2-m^2=k$ has no solution for each $ 1\leq k\leq 6$ . Therefore $ 23n^2-m^2\geq 7$ ,then inequalities (a) follows from this . b ) Use the Pell's equation : $ 23n^2-m^2=7$ , it has solution $ (x,y)=(1,4)$ , therefore it has infinite solution . Pairs root $ (n_k,m_k)$ of this equation satisfy (b)

(-5/23)=(64/23)=1.

Sorry for my mistake .But i think the equation $ 23n^2-m^2=5$ has no integer solution . Proof of this statement use some results of Pell's equation . A lemma : We known that the equation $ x^2-dy^2=1$ always has solution in set of natural number (we only consider case $ d$ is not a perfect square . Call $ (a,b)$ is the smallest root of this equation . Consider the equation $ x^2-dy^2=n$ (2) .Let $ (x_0,y_0)$ is the smallest root of (2) . Then we have : $ y_0^2\leq \max\{nb^2,\frac{-na^2}{d}\}$ Now consider our problem . The equation $ x^2-23y^2=1$ has smallest positive solution is $ (x,y)=(24,5)$ Apply lemma , if $ (m_0,n_0)$ is the smallest solution of equation $ m^2-23n^2=-5$ then we must have : \[ n_0^2\leq \max\{-125,126\} \] Thus $ n_0\leq 10$ .Check $ n=1,..,10$ ,it is easy to see that equation (2) has no natural solution . My proof continues as above .

Sorry Maxim Bogdan,but can you tell me what is "Stars of Mathematics Bucharest 2008"?.If it is a contest could you please post all of its problem?

Problems posted. http://www.mathlinks.ro/viewtopic.php?t=1373872

there is no need for that lemma to prove that $ -5$ doesn't work, just use modulo $ 4$