75th Romanian Mathematical Olympiad

Since $c+d+1=a+b$, the number $a+b+c+d$ is odd. Therefore either three of the numbers $a,b,c,d$ are even (and, being primes are equal to $2$), or exactly one of them $a,b,c,d$ is even (therefore equal to $2$).

If three of the numbers $a,b,c,d$ are equal to $2$, condition (2) implies that the square of the fourth number is $3531$ – impossible. So exactly one of the numbers $a,b,c,d$ is equal to $2$. Now $a\le b$ and $c\le d$ yields that either $a=2$, or $c=2$.

The three numbers different from $2$ have the sum of their squares $3543-4=3539=M_3+2$. Since a perfect square is $M_3$ or $M_3+1$, one of the squares must be $M_3$, so the corresponding number, being prime, must be $3$.

This leaves two primes $x,y$ whose sum of their squares equals $3539-9=3530$.

If $x\le 41$ and $y\le 41$, then $x^2+y^2\le 2\cdot 41^2=3362<3530$. So at least one of $x,y$ is $\ge 43$ – say $x\ge 43$. If $x\ge 61$, then $x^2+y^2>61^2>3530$, So, $x$ can be only $43$, $47$, $53$ or $59$. Checking these values, $x=47$ and $x=53$ give no solution, $x=43$ gives $y=41$, and $x=59$ gives $y=7$.

So $a,b,c,d$ are $2,3,41,43$ or $2,3,7,59$. Finally, condition (1) is fulfilled only by $a=2,b=43,c=3,d=41$.