I guess C
1) simplify the equation a bit to sqrt(P-1) = n and plug in a couple of primes
n being an interger is true for more than one prime
n=4 => P=5, for n=6 => P=37, etc.
2) every prime greater than 17 when squared is more than 200
Now two together, solve for equation sqrt(p-1)=n.... what prime number (greater than 17) when subtracted by 1 and brought to its square root, results in a interger....... 37
You can run a couple more prime numbers to check, 41, 43, none result in n being an interger!!
Answer two together : C
The Answer should be E.
In your explanation st 1 and st 2 and insufficient which is correct. Now when solving together we have two cases and just not one case:
Sqrt(37-1)=6 where both conditions are fulfilled ( 37^2 >200 and 37 = 6^2+1)
Sqrt(17-1)=4 where both conditions are fulfilled ( 17^2 >200 and 17 = 4^2+1)
Hence not sufficient.
Consider giving Kudos if my post helps in some way