Find integral solutions to the equation \[(m^{2}-n^{2})^{2}=16n+1.\]

Prove that the sequence $a_{n}=\lfloor n\sqrt 2 \rfloor+\lfloor n\sqrt 3 \rfloor$ contains infintely many even and infinitely many odd numbers.

Let $ABC$ be a triangle such that $|AC|>|AB|$. Let $X$ be on line $AB$ (closer to $A$) such that $|BX|=|AC|$ and let $Y$ be on the segment $AC$ such that $|CY|=|AB|$. Intersection of lines $XY$ and bisector of $BC$ is point $P$. Prove that $\angle BPC+\angle BAC = 180^\circ$.

Given a finite string $S$ of symbols $X$ and $O$, we write $@(S)$ for the number of $X$'s in $S$ minus the number of $O$'s. (For example, $@(XOOXOOX) =-1$.) We call a string $S$ balanced if every substring $T$ of (consecutive symbols) $S$ has the property $-2 \leq @(T) \leq 2$. (Thus $XOOXOOX$ is not balanced since it contains the sub-string $OOXOO$ whose $@$-value is $-3$.) Find, with proof, the number of balanced strings of length $n$.

Let there be two circles. Find all points $M$ such that there exist two points, one on each circle such that $M$ is their midpoint.

$\displaystyle 2n$ students $\displaystyle (n \geq 5)$ participated at table tennis contest, which took $\displaystyle 4$ days. In every day, every student played a match. (It is possible that the same pair meets twice or more times, in different days) Prove that it is possible that the contest ends like this: - there is only one winner; - there are $\displaystyle 3$ students on the second place; - no student lost all $\displaystyle 4$ matches. How many students won only a single match and how many won exactly $\displaystyle 2$ matches? (In the above conditions)

Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \]

Positive integers $x>1$ and $y$ satisfy an equation $2x^2-1=y^{15}$. Prove that 5 divides $x$.

Click for solution Are there any integer solutions to this equation? Anyway, suppose what they want is not true. Of course, 2 divides $y+1$ and if we write the equation as $ ((y^5+1)/2)( y^{10}-y^5+1)=x^2$ we observe that $ ((y^5+1)/2, y^{10}-y^5+1)$ is a divisor of 3. If it is 1, then $y^{10}-y^5+1$ is a perfect square, impossible for y at least 2 (^we can put it between two consecutive squares). So, suppose it is 3. Then we can write $ y^5+1=6A^2$ and $y^{10}-y^5+1=3B^2$ with $x=3AB$. Since $ y^5-y$ is a multiple of 3, it follows that $ y+1$ is a multiple of 6. Thus, if we write $ ((y+1)/6)(y^4-y^3+y^2-y+1)=A^2$ we see that since 5 does not divide $A$, we have $ ((y+1)/6, y^4-y^3+y^2-y+1)=1$ and thus $ y^4-y^3+y^2-y+1$ is a square. Again, this is impossible by the fact that $ (2y^2-y)^2<4(y^4-y^3+y^2-y+1)<(2y^2-y+2)^2$.