Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are positive integers.
As $a^{3} b-1=b\left(a^{3}+1\right)-(b+1)$ and $a+1 \mid a^{3}+1$, we have $a+1 \mid b+1$.

As $b^{3} a+1=a\left(b^{3}-1\right)+(a+1)$ and $b-1 \mid b^{3}-1$, we have $b-1 \mid a+1$.

So $b-1 \mid b+1$ and hence $b-1 \mid 2$.

- If $b=2$, then $a+1 \mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only solution in this case.
- If $b=3$, then $a+1 \mid b+1=4$ gives $a=1$ or $a=3$. Hence $(a, b)=(1,3)$ and $(3,3)$ are the only solutions in this case.

To summarize, $(a, b)=(1,3),(2,2)$ and $(3,3)$ are the only solutions.
$\mathbf{N 4}$. A rectangle in the $x y$-plane is called latticed if all its vertices have integer coordinates.

a. Find a latticed rectangle with area 2013 whose sides are not parallel to the axes.

b. Show that if a latticed rectangle has area 2011, then its sides are parallel to the axes.

Solution. a. The rectangle $P Q R S$ with $P(0,0), Q(165,198), R(159,203), S(-6,5)$ has area 2013.

b. Suppose that the latticed rectangle $P Q R S$ has area 2011 and its sides are not parallel to the axes. Without loss of generality we may assume that its vertices are $P(0,0), Q(a, b)$, $S(c, d), R(a+c, b+d)$ where $a, b, c, d$ are integers and $a b c d \neq 0$.

Then $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=2011^{2}$. As 2011 is a prime, $2011 \mid a^{2}+b^{2}$ or $2011 \mid c^{2}+d^{2}$. Assume that the first one is the case. Since $2011 \equiv 3(\bmod 4)$, this can happen only if $a=2011 a_{0}$ and $b=2011 b_{0}$ for some integers $a_{0}$ and $b_{0}$. Now we have $\left(a_{0}^{2}+b_{0}^{2}\right)\left(c^{2}+d^{2}\right)=1$. This means $c^{2}+d^{2}=1$ and hence $c=0$ or $d=0$, contradicting $c d \neq 0$.

Comment. The original problem proposal has the numbers 13 and 11 instead.