Subject (1).

a) Solve the equation $x^{4}-y^{4}=16$ in the set of integers.

b) Find the largest natural number $n$ for which $\frac{n^{2}-2012}{n+1} \in \mathbb{Z}$.

c) Find the real numbers $a$ and $b$ if $\max \left\{a^{2}-b+1, b^{2}+a+1\right\} \leq \frac{3}{4}$.

Aurelia Caţaros \& Gabriela Ruse, Călăraşi
Solution.

a) We have $x^{4}-y^{4}=(x-y)(x+y)\left(x^{2}+y^{2}\right)=2^{4}=16$. From the very few remaining cases, only $(x, y)=( \pm 2,0)$ remains.

b) We have $\frac{n^{2}-2012}{n+1}=n-1-\frac{2011}{n+1}$. From $n+1 \mid 2011$ it follows that the largest natural number $n$ sought is $n+1=2011$, hence $n=2010$.

c) We have $\frac{3}{2} \geq\left(a^{2}-b+1\right)+\left(b^{2}+a+1\right)=\left(a+\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}+\frac{3}{2}$, so $\left(a+\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2} \leq 0$, from which $a=-\frac{1}{2}$ and $b=\frac{1}{2}$.