A 7TH GRADE PROBLEM FROM NMO // O PROBLEMA DE CLASA A 7-A DE LA ONM

                     We will solve the following problem (It is Problem 3 from NMO 2026, proposed for grade 7 page 5):

               Determine the pairs  $(a,b)$ of nonzero natural numbers for which the

 numbers  $\frac{a^2+b}{a+b-1}$  and  $\frac{b^2+a}{a+b}$  are natural.

{author} Lucian DRAGOMIR, Oțelu Roșu

Vocabulary  :  NMO = National Mathematics Olympiad

ANSWER CiP

$$a=1,\;\;\;b=1$$

                         Solution CiP

               Let be       $\frac{a^2+b}{a+b-1}=m\in\mathbb{N},\;\;\;\frac{b^2+a}{a+b}=n\in\mathbb{N}\;\;\;\;\;\;\;\;\;(1)$

From (1)  we obtain :

$m(a+b-1)=a^2+b\;,\;\;\;n(a+b)=b^2+a \tag{2}$

          Subtracting the equations (2) we obtain :

$(a+b-1)\cdot m-(a+b)\cdot n=(a-b)(a+b-1) \tag{3}$

          We view  (3)  as a linear Diophantine equation with the unknowns  $m\; and\; n$. A particular, almost obvious solution of  (3)  is  $m_0=a-b\;,\;\;n_0=0.$

 Then, all integer solutions of the equation  (3)  will be :

$m_k=a-b+k\cdot (a+b)\;,\;\;\;n_k=k\cdot (a+b-1)\;\;,\;\;\;k\in\mathbb{Z} \tag{4}$

However, we are only interested in natural number solutions  $(m_k,n_k)\in\mathbb{N}^*\times \mathbb{N}^*$  so  $k>0.$

          The second equation in  (2)  is written

$k\cdot (a+b-1)\cdot (a+b)=b^2+a$

 from which we obtain :

$k=\frac{b^2+a}{(a+b)(a+b-1)} \tag{5}$

                    This is where I got stuck, although the intention was to see from  (5)  what the possible values ​​for  $k$  are.

     Examining the "First Solution" from the official Olympiad materials (see page 15, Problem 3) I found that I was on the right track. Moreover, the expression in  (5)  appears there with the notation  $"p"$  (which in Greek is written  $\pi$). “Diophantus would denote this number using the letter π, hence my choice to include it in the title of the post.”(Idea first shared with COPILOT.) That's how I chose, a little bit cool, sly, the title of the Post there.

"Returning to our sheep," as the proverb would say, we see that :

$k< 1\;\Leftrightarrow\;b^2+a< a^2+2ab+b^2-a-b\;\Leftrightarrow$

$\Leftrightarrow\;a(a-2)+b(2a-1)>0$

for  $a\geqslant 2$. So in the case  $a\geqslant 2$  there are no solutions.

          For  $a=1,\;\;m=\frac{1+b}{b}=1+\frac{1}{b}\;,\;\;n=\frac{b^2+1}{b+1}=b-1+\frac{2}{b+1}$. and  $m\;,n$  are integers if and only if  $b=1$. 

          We got the answer.

$\blacksquare$