Least But First!

$1^2 + 11 = 12$ not prime

$2^2 + 11 = 15$ not prime

$3^2 + 11 = 20$ not prime

$4^2 + 11 = 27$ not prime

$5^2 + 11 = 36$ not prime

$\boxed{\large\color{#D61F06}6^2 + 11 = 47}$ prime number

The answer is $\large\boxed{\color{#20A900}\large6}$ .

$1 ^ { 2 } + 11 = 12 = 2 ^ { 2 } \times 3$ , so not prime.

$2 ^ { 2 } + 11 = 15 = 3 \times 5$ , so not prime.

$3 ^ { 2 } + 11 = 20 = 2 ^ { 2 } \times 5$ , so not prime.

$4 ^ { 2 } + 11 = 27 = 3 ^ { 3 }$ , so not prime.

$5 ^ { 2 } + 11 = 36 = 6 ^ { 2 }$ , so not prime.

$6 ^ { 2 } + 11 = 47$ , so prime.

Proof: $\displaystyle 6 < \sqrt { 47 } < 7$ , so $47$ must be divisible by the prime numbers less than $6$ .

Then, $2, 3, 5$ can be.

So, $47 \equiv 1 ( \mod 2 )$ .

$47 \equiv 2 ( \mod 3 )$ .

$47 \equiv 2 ( \mod 5 )$ .

Hence, $47$ is prime.

And, this way is called prime theorem .

(6)^2 +11=47 which is a prime no.

The simple solution is 6²+11 = 47. As 47 is prime, the value of a is 6.