joi, 3 iulie 2025

Problem #1 from JBMO TEAM SELECTION TEST 2025 - GREECE

 Obtained from here.

         "Problem 1.

           (a) Let the positive integers  $p,\;q$  be prime numbers and let  $a$  be a positive

           integer. If  $a$  divides the product  $p\cdot q$ , and it holds that  $a>p$  and  $a>q$ , 

           prove that  $a=pq$.

          (b) Determine all pair  $(p,q)$  of prime numbers such that  $p^2+3pq+q^2$ 

           eqals a perfect square."


ANSWER CiP

(b)  $(3,\;7)$  and  $(7,\;3)$


Solution CiP

(a)  $a\mid p\cdot q\;\Rightarrow\;\;p\cdot q=a\cdot b$  for a certain  $b\in \mathbb{N}$. Noting that  $p$  divides the product $a\cdot b$ then, since it is prime, it follows

$$p\mid a\;\;\;or\;\;\;p\mid b.$$

      If  $p\mid a$, then  $a=p\cdot c$  for a certain  $c\in \mathbb{N}$. Then, from

$p\cdot q= (p\cdot c)\cdot b$  we obtain  $q=c\cdot b$. But  $q$  is also prime so we can have  $c=1$ (when  $q=b,\;p=a$  which contradicts  $a>p$)  or $b=1$ , in which case  $q=c$  and $a=pq$.

      If  $p\mid b$,  then $b=p\cdot d$  for a certain  $d\in\mathbb{N}$. Then, from  $p\cdot q=a(p\cdot d)$  we obtain  $q=a\cdot d$, but $q$ being prime and  $a>1$  it follow  $d=1,\;q=a$  wich contradicts  $a>q$.

With these,  (a)  is proven.


(b) If  $k\in \mathbb{N}$ is such that  $p^2+3pq+q^2=k^2$  then we get

$$pq=k^2-p^2-2pq-q^2=k^2-(p+q)^2=(k-p-q)\cdot (k+p+q)$$

From the above it can be seen that the number  $a:=k+p+q>p,\;q$  divides the product  $pq$. According to (a) we must have $a=pq$  that is, equivalent to

$k+p+q=pq\;\Rightarrow\;k=pq-p-q\;\;\Rightarrow\;\;p^2+3pq+q^2=(pq-p-q)^2\;\:\Leftrightarrow$

$\Leftrightarrow\;pq=p^2q^2-2p^2q-2pq^2\;\;\Leftrightarrow\;\;1=pq-2p-2q\;\Leftrightarrow\;5=pq-2p-2q+4\;\Leftrightarrow$

$$\Leftrightarrow\;\;\;5=(p-2)(q-2).$$ 

Then it follows that  $p-2=1$  or  $p-2=5$. We get the answer.

          Verification:  $3^2+3\cdot 3\cdot 7+7^2=9+63+49=121=11^2$.

$\blacksquare$

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