The problem proposed for class 6, on page 480.
"Let $p$ and $q$ be two distinct prime numbers. Prove that if $15(p+q) >8pq$, then
$$93pq \leqslant 174(p+q) \leqslant 145 pq."$$
ANSWER CiP
One of the numbers is $2$
and the other can be $3,\;5,\;7,\;11,\;13,\;17,\;19,\;23,\;29.$
SOLUTION CiP
Natural numbers that verify the condition
$$15(p+q)>8pq \tag{1}$$
must be in finite quantity. Indeed, the condition (1) can be written equivalent to
$$\frac{1}{p}+\frac{1}{q}>\frac{8}{15}. \tag{2}$$
If none of the numbers $p,q$, is equal to $2$, then $\frac{1}{p}+\frac{1}{q} \leqslant \frac{1}{3}+\frac{1}{5}=\frac{8}{15},$ contradicting (2).
Assuming $p=2$ the condition (1) is written $15(2+q)>16q\;\;\Leftrightarrow\;\;q<30$. So $3\leqslant q \leqslant 29$ and $q$ is a prime number. Then we have
$$\frac{1}{2}+\frac{1}{29}\leqslant \frac{1}{p}+\frac{1}{q}\leqslant \frac{1}{2}+\frac{1}{3}\;\Leftrightarrow\;\frac{31}{58}\leqslant \frac{p+q}{pq}\leqslant \frac{5}{6}$$
and multiplying the last inequalities by $6 \cdot 29=174$ we get the conclusion.
$\blacksquare$
REMARK CiP
The condition (1) can also be written, after multiplying by 8, $$64pq-120p-120q<0\;\Leftrightarrow$$
$\Leftrightarrow \;(8p-15)(8q-15)<225$. If $p\geqslant 3$, we have $8p-15 \geqslant 9$ then we deduce $8q-15<\frac{225}{8p-15}\leqslant \frac{225}{9}=25$, so $8q<40$, $q<5$. We only have the possibility $q=3$ or $q=2$. When $q=3$, from (1) we deduce $p<5$. We can only have $p=3$ but then $p=q$, impossible. When $q=2$, from (1) we deduce $p<30$ and we find the answer.
$\square$ end REM
