Example 3. Given three statements:

1) the equation $x+\frac{1}{x}=a$ has no real solutions;
2) the equality $\sqrt{a^{2}-4 a+4}=2-a$ holds;
3) the system $\left\{\begin{array}{l}x+y^{2}=a, \\ x-\sin ^{2} y=-3\end{array}\right.$ has a unique solution.

For which values of $a$ will two of these statements be true, and the third one false?
Solution. Rewrite the second equality as follows: $\sqrt{(a-2)^{2}}=2-a$, or $|a-2|=2-a$.

Therefore, $2-a \geqslant 0$, i.e., $a \leqslant 2$.

Transform the equation $x+\frac{1}{x}=a$ into the form

$$
x^{2}-a x+1=0
$$

For this equation to have no real roots, it is necessary and sufficient that its discriminant $a^{2}-4<0$, i.e., $-2<a<2$.

Now let's move to the last point. Suppose the pair of numbers $\left(x_{0} ; y_{0}\right)$ is one of the solutions of the given system. Obviously, the pair of numbers $\left(x_{0} ;-y_{0}\right)$ also satisfies the system. Therefore, for the condition of the uniqueness of the solution of the system to be met, it is necessary that $y_{0}=-y_{0}$, i.e., $y_{0}=0$, then $a=x=-3$. Substituting the number -3 for $a$ in the first equation of the system, we verify that when $a=-3$ the system has a unique solution. Thus, $a=-3$.

Now we see that if $-2<a<2$, then equations 1) and 2) are satisfied, but statement 3) is not.

If, however, $a=-3$, then statements 2) and 3) are satisfied, but statement 1) is not.

Finally, statements 1) and 3) cannot be satisfied simultaneously. Therefore, two statements will be true, and one false when $a=-3$ or when $a \in]-2 ; 2[$.

## EXERCISES

183. Factorize:

a) $1+x+x^{5} \quad$ (into 2 factors);

b) $1+x^{4}+x^{8} \quad$ (into 3 factors).

184. Calculate $a^{1978}+\frac{1}{a^{1978}}$, if $a^{2}+a+1=0$.
185. Prove that if $x+\frac{1}{x}=2 \cos \alpha$, then

$$
x^{n}+\frac{1}{x^{n}}=2 \cos n \alpha \quad(n \in \boldsymbol{N})
$$

186. Given three statements:
1) the quadratic trinomial $x^{2}+x+a$ is non-negative for all $x$;
2) the function $y=\log _{2 a} x$ is decreasing;

$6 \mathrm{C}$

3) the system $\left\{\begin{array}{l}x^{2}+y^{2}=a, \\ y+\cos x=2 \text { has a unique solution. }\end{array}\right.$

For what values of $a$ are two statements true and one false?
187. $m$ and $n$ are positive integers. It is known that out of the following four statements:

1) $m+1$ is divisible by $n$
2) $m$ equals $2 n+5$
3) $m+n$ is divisible by 3
4) $m+7 n-$ is a prime number,

three are true, and one is false. Find all possible pairs $m, n$.

188. Find the number $x$, if it is known that out of the following three statements:

a) $x$ is an integer;

b) $x^{2}-3 x$ is a negative integer;

c) $x+\frac{1}{x}$ is a positive integer, one is false.

189. Regarding $\triangle A B C$, it is claimed that:
1) triangle $A B C$ is a right triangle;
2) $\widehat{B A C}=30^{\circ}$
3) $|A B|=2|B C| ; \quad$ 4) $|A C|=2|B C|$.

It is known that two of these statements are true, and two are false. Find the perimeter of the triangle if $|B C|=1$.

## § 5. "Game" and Logical Problems

Example 1. A game involving two players consists of the following. Two piles of matches are taken. Players alternately discard one of them and split the remaining one into two, and so on until there is one match left in each pile. The player who makes the last move wins. Who will win (the player starting the game or his partner) if at the beginning of the game there were 20 matches in one pile and 25 in the other?

Solution. Let's solve the problem in general terms. Suppose at the beginning of the game, each pile contains an odd number of matches. Then, with the correct play, the player who starts loses. Indeed, after the player who starts discards one of the piles, there will be a pile with an odd number of matches, which he splits into two piles: one with an odd number of matches and the other with an even number. Now the second player discards the pile with an odd number of matches and splits the remaining pile into two so that each contains an odd number of matches. Continuing the game, the second player will be the last to split a pile of two matches into two parts, each containing one match, and thus he wins.

Thus, a losing position is a situation where there are two piles, each containing an odd number of matches.

If, however, one pile contains an even number of matches (for example, 20), and the other an odd number (for example, 25), then, to win, the player who starts must discard the second pile and split the remaining pile of an even number of matches into two piles, each containing an odd number of matches. Thus, the second player will be put in a losing position.