Saudi Arabia booklet 2024

Let $f(a,b)=\text{gcd}(2024ab,a^2+254ab+b^2)$ then for 2 numbers $a,b$ on the board, the new number generated will be $f(a,b)$. We consider some cases: If $a,b$ are same parity then $f(a,b)$ is even. If $a,b$ are different from parity then $f(a,b)$ is odd. This means that the number of odd numbers either remains the same or decreases by 2 units; but initially, there are odd numbers, so the final number $x$ must be odd. Suppose that $x>1$, let $p$ be some prime divisor of $x$. Note that $$254=11\cdot 23+1\text{ and }2024=2^3\cdot 11\cdot 23.$$ If $p=11$, since $11|\text{gcd}(2024ab,a^2+254ab+b^2)$ then $11|a^2+ab+b^2$. Similarly, if $p=23$ then there is also $23|a^2+ab+b^2$. We have the following familiar lemma derived from Fermat's little theorem or quadratic residue: Lemma. If $p$ is a prime of form $3k+2$ and $p|a^2+ab+b^2$ then $p|a,b$. Since 11 has the form $3k+2$, according to the above lemma, one can get $11|a,b$. Continuing like that, the previous numbers on the board that generated $a,b$ also be divisible by 11, making all the original numbers divisible by 11, contradiction. Similarly for the case $p=23$. Finally, if $p\ne 11,23$ then $p|ab,p|a^2+506ab+b^2$ entails $p|ab,p|a^2+b^2$. Obviously from $p|ab$, we get that $p$ is a divisor of one of $a$ or $b$, but $p|a^2+b^2$ then $p|a,b$. We get the similar contradiction as the above argument. Therefore $x$ is odd and has no odd prime divisors, proving that $x=1$. ☐