Find the nth Fibonnaci Prime, in the shortest code

C, 66

f(n,a,b){int i=2;while(a%i&&i++<a);return(n-=i==a)?f(n,b,a+b):a;}

C, 85, 81, 76

f(n){int i=1,j=0,k;for(;n;n-=k==i)for(j=i-j,i+=j,k=2;i%k&&k++<i;);return i;}

• borrowed code style of simplified prime number check from @Gautam

• self contained C function (no globals)

Testing:

main(int n,char**v){printf("%d\n",f(atoi(v[1])));}

./a.out 10
433494437

./a.out 4
13

Mathematica, 59 or 63 bytes

(a=b=1;Do[While[{a,b}={b,a+b};Length@Divisors@b>2],{#}];b)&
(a=b=1;Do[While[{a,b}={b,a+b};b~Mod~Range@b~Count~0>2],{#}];b)&

These are unnamed functions which take n as their input and return the correct Fibonacci prime. The shorter version uses Divisors. I'm not entirely sure whether this is allowed, but the other Mathematica answer even uses FactorInteger.

The second one doesn't use any factorisation related functions at all, but instead counts the number of integers smaller than n which produce 0 in a modulo operation. Even this version beats all valid submissions, but I'm sure just posting this answer will cause some people to provide competitive answers in GolfScript, APL or J. ;)