Problem E:16386 from GMB 10/2022

          The problem proposed for class 6, on page 480.

In translation (thanks to Miss Google):

          "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