- 127.
- Let

and

where $n>6$ is a natural number. Prove that the fraction $A/B$ is reducible.

When $(x,y)=(2,3)$, ${x}^{2}-\mathrm{xy}+{y}^{2}=7$, while, when $(x,y)=(4,5)$, ${x}^{2}-\mathrm{xy}+{y}^{2}=21=7\times 3$. It follows that, for each $n>6$, both $A$ and $B$ are divisible by 7 and the result follows.

and

so that both $A$ and $B$ are divisible by 7 and $A/B$ is reducible.

- 128.
- Let $n$ be a positive integer. On a circle, $n$ points are marked. The number 1 is assigned to one of them and 0 is assigned to the others. The following operation is allowed: Choose a point to which 1 is assigned and then assign $(1-a)$ and $(1-b)$ to the two adjacent points, where $a$ and $b$ are, respectively, the numbers assigned to these points before. Is it possible to assign 1 to all points by applying this operation several times if (a) $n=2001$ and (b) $n=2002$?

(b) Observe that the parity the number of ones does not change.
If we start with a single one, then there must be an odd number of ones,
so we cannot solve the problem when
$n=2002$.

- 129.
- For every integer $n$, a nonnegative integer $f(n)$ is assigned such that

- (a) $f(\mathrm{mn})=f(m)+f(n)$ for each pair $m,n$ of natural numbers;

- (b) $f(n)=0$ when the rightmost digit in the decimal representation of the number $n$ is 3; and

- (c) $f(10)=0$.

- Prove that $f(n)=0$ for any natural number $n$.

Let
$n=10k+1$, then
$0=f(30k+3)=f(3)+f(10k+1)=0+f(n)$. Let
$n=10k+7$. Then

Finally, let $n=10k+9$. Then

If follows that $f(n)=0$, whenever $n$ and 10 are coprime.

Since
$0=f(10)=f(2)+f(5)$, and
$f(2)\ge 0$,
$f(5)\ge 0$,
if follows that
$f(2)=f(5)=0$. The result now follows by
applying (a) to the prime factorization of a given number
$n$.

whence $f(n)=\mathrm{rf}(2)+\mathrm{sf}(5)+f(b)=0$.

- 130.
- Let $\mathrm{ABCD}$ be a rectangle for which the respective lengths of $\mathrm{AB}$ and $\mathrm{BC}$ are $a$ and $b$. Another rectangle is circumscribed around $\mathrm{ABCD}$ so that each of its sides passes through one of the vertices of $\mathrm{ABCD}$. Consider all such rectangles and, among them, find the one with a maximum area. Express this area in terms of $a$ and $b$.

We can circumscribe a rectangle each of whose right triangles is
isosceles, and whose heights are
$a/2$ or
$b/2$ and areas are
${a}^{2}/4$ or
${b}^{2}/4$. Thus, the maximum area of the circumscribed
rectangle is

- 131.
- At a recent winter meeting of the Canadian Mathematical Society, some of the attending mathematicians were friends. It appeared that every two mathematicians, that had the same number of friends among the participants, did not have a common friend. Prove that there was a mathematician who had only one friend.

- 132.
- Simplify the expression