Prove that there exist infinitely many positive integers $k$ such that the equation $$\frac{n^2+m^2}{m^4+n}=k$$don't have any positive integer solution.

Let $ABC$ be a triangle and $P$ be an interior point. Let $\omega_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ internally and tangent to the circumcircle of $ABC$ at $A_1$ internally and let $\Gamma_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ externally and tangent to the circumcircle of $ABC$ at $A_2$ internally. Define $B_1$, $B_2$, $C_1$, $C_2$ analogously. Let $O$ be the circumcentre of $ABC$. Prove that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ and $OP$ are concurrent.

Let a $9$-digit number be balanced if it has all numerals $1$ to $9$. Let $S$ be the sequence of the numerals which is constructed by writing all balanced numbers in increasing order consecutively. Find the least possible value of $k$ such that any two subsequences of $S$ which has consecutive $k$ numerals are different from each other.

Initially given $31$ tuplets $$(1,0,0,\dots,0),(0,1,0,\dots,0),\dots, (0,0,0,\dots,1)$$were written on the blackboard. At every move we choose two written $31$ tuplets as $(a_1,a_2,a_3,\dots, a_{31})$ and $(b_1,b_2,b_3,\dots,b_{31})$, then write the $31$ tuplet $(a_1+b_1,a_2+b_2,a_3+b_3,\dots, a_{31}+b_{31})$ to the blackboard too. Find the least possible value of the moves such that one can write the $31$ tuplets $$(0,1,1,\dots,1),(1,0,1,\dots,1),\dots, (1,1,1,\dots,0)$$to the blackboard by using those moves.

Is it possible that a set consisting of $23$ real numbers has a property that the number of the nonempty subsets whose product of the elements is rational number is exactly $2422$?

On a triangle $ABC$, points $D$, $E$, $F$ are given on the segments $BC$, $AC$, $AB$ respectively such that $DE \parallel AB$, $DF \parallel AC$ and $\frac{BD}{DC}=\frac{AB^2}{AC^2}$ holds. Let the circumcircle of $AEF$ meet $AD$ at $R$ and the line that is tangent to the circumcircle of $ABC$ at $A$ at $S$ again. Let the line $EF$ intersect $BC$ at $L$ and $SR$ at $T$. Prove that $SR$ bisects $AB$ if and only if $BS$ bisects $TL$.