Rioplatense Mathematical Olympiad

The answer is 3. First, notice that the last digit of Magalí's initial number, name it $n_1$, determines the last digit of the following numbers. This is explained in the following table, where we name $n_2=n_1^2+6$, $n_3=n_2^2+6$, $n_4=n_3^2+6$ and $n_5=n_4^2+6$:

$n_1$	$n_2$	$n_3$	$n_4$	$n_5$
1	7	5	blocked ⋆	blocked ⋆
2	0	blocked ⋆	blocked ⋆	blocked ⋆
3	5	blocked ⋆	blocked ⋆	blocked ⋆
5	1	7	5	blocked ⋆
7	5	blocked ⋆	blocked ⋆	blocked ⋆
9	7	5	blocked ⋆	blocked ⋆

For example, if $n_1$ had 4 as its last digit, then $n_2=n_1^2+6$ would have the same last digit as $4\times 4+6=22$, which is 2. We omitted the numbers with their last digit being even but different from 2 because they cannot be prime. Now, observe on the table that no matter which prime number Magalí chooses initially, in at most three steps she reaches a number with last digit 0 or 5, and so a multiple of 5. Since this number is greater than 5, it is not a prime number. Therefore, she can never press the ⋆ button more than three times. It is possible that Magalí presses ⋆ three times. Indeed, if she starts with $n_1=5$, the following numbers are $n_2=5^2+6=31$, $n_3=31^2+6=967$ and $n_4=967^2+6$. Since 5, 31 and 967 are prime numbers and $967^2+6$ is a multiple of 5, this proves the desired result.