p=2 is a trivial solution assume that p>=3 $m^3+n^3-4=(m+n)(p-mn)-4$ so $mn(m+n)+4=0(mod p)$ note $a=m+n , b=mn$ then p=a²-2b and ab+4=0(mod p) $a^3-2ab=0(mod p)$ $a^3+8=0(mod p)$ so p|(a+2)(a²-2a+4) note that $a=m+n<\sqrt(2p)$ so a+2<p if p>=6 ( we can see that 3 or 5 does not satisfy the condition) so p|(a²-2a+4)=(2b-2a+4) (mod p) p| (b-a)+2 so we have 0<a<p and 0<b<p so 0<(b-a)<p => 2<(b-a)+2<p+2 =>(b-a)+2=p =>a²-2b=b-a+2 => a²+a-2=3b => a²-4b=2-(a+b) we know that a²>4b so a+b<2 .... absurde so 2 is the Only Solution of the problem is it right the demo!?

I wonder this problem is from Silk Road Mathematical olympiad 2004

But p=5 and set m=2,n=1] OK] so p=2 and 5

p=13 is also a solution !! note that m, n are integers not necessarily positive