How many Friday the 13ths can there be in a year (non-leap year)?
In the days of the year, the 13th of each month occupies the ranks: 13, 44, $72, 103, 133, 164, 194, 225, 256, 286, 317, 347$. To determine which day of the week each day corresponds to, it is sufficient to take these values modulo $7: 6,2,2,5,0,3,5,1,4,6,2,4$. If day 1 of the year (January 1st) is a Friday, then there will only be one Friday the 13th in August. If day 4 is a Friday, there will be 2 Friday the 13ths, in September and December. If day 2 is a Friday, there will be 3 Friday the 13ths. On the other hand, it is noted that it is impossible for there to be no Friday the 13th or more than 3.

Solution to the exercise \& Let's look at the equation modulo 4. If $x$ and $y$ are odd, then we would have $1+1=z^{2}(\bmod 4)$, which is impossible because a square cannot be congruent to 2 modulo 4.