Powerful, Power-packed, Power-driven..........Primes!

Let us solve this by making some cases:

Case-1: When $n=2$ .

Consider the odd primes first, the only primes that satisfy the given condition are congruent to $1$ mod $4$ . There are $11$ of them that are $5,13,17,29,37,41,53,61,73,89$ and $97$ .

Now, we look at our only even prime number $2$ which can be expressed as $1^{2}+1^{2}=2$ . So, it also satisfies the condition.

Hence, there are $12$ such primes.

Case-2: When $n=3$ .

If $n=3$ , our expression becomes $a^{3}+b^{3}$ which can be factorized further into $(a+b)(a^{2}-ab+b^{2})$ . Only, $1^{3}+1^{3}=2$ satisfies the condition but $2$ has already been counted in the above case. So, we will not count it again.

Case-3: When $n=4$ .

When $n$ reaches $4$ , there are very few cases to check as the sum $a^{4}+b^{4}$ exceeds $100$ except when $a$ and $b$ are $1,2$ and $3$ . Putting the values, we get: $1^{4}+2^{4}=17$

$1^{4}+3^{4}=82$

$2^{4}+3^{4}=97$

Out of these, $17$ and $97$ are the only primes but they have already been counted.

Case-4: When $n≥5$ .

In this case, the permissible values of $a$ and $b$ are only $1$ and $2$ . Putting the values, we always get a composite number except $2$ which has already been counted.

So, we can observe that there are no different primes in any case.

Hence, our final answer is $\huge{12}$ .

We can also do this the long way by writing down all 25 primes.

Next, we make 4 columns, for n =2,3,4,5 and write down the possible values. 1-2 for n=2, 1-4 for n=3, 1-3 for n=4 and 1-2 for n=5. For each number see if any two numbers of the same column can add up to that number.

Not so long..maybe about 5-6 minutes.