We can take k such that it's even, it's equal to 2 (mod 3) and 9+k is composite, for example 5/9+k=3^2+k. This is because p^2=1 (mod 3) for any p different from 3 and we also consider the case p=3. Infinitely many such k exist from the Chinese remainder thm.

If we take $k=4a^2$,we can factorize $p^2+k$ with Sophie Germain.(a is an arbitrary number)

Unfortunately, you're wrong. Even if you take $k=4a^4$, the intended Sophie Germain factorization looks like $p^2 + k = p^2 + 4a^4 + 4a^2p - 4a^2p = (p+2a^2)^2 - (2a)^2 p$, but $p$ is no square. This idea would only work for $p^4+k$ ...

We choose k to be odd, then obviously for odd p (p^2)+k is composite. Let k=n-4 where n is an odd composite, this k satisfies the condition when p=2 also. Since there are infinitely many such n's there exists infinitely many such k too.

Have you not noticed it is clearly specified $k$ must be even? since for $k$ odd it is so trivial ...

But the problem statement requires $k$ to be even. It can easily be seen that $k=30a-4$ (for positive integers $a$) works.

Sorry I hadn't noticed that k has to be even.

For every natural $t>0$ $k=66t+2$ works . There is three cases: 1)If $p=2$ then $p^2+k=66t+4$ which is obviously an even integer bigger that two! 2)If $p=3$ then $p^2+k=66t+11$ again is divisible by eleven and bigger than it! 3)If $p>3$ then trivially $p^2\equiv 1\ (mod3)$ so $p^2+k$ divisible by three!

I think for $k=26+30*a$ for a natural $a$ we get $k+4=30*(1+a)$ and $k+9=5*(7+6*a)$ and for $p\ge5$ we have $p^2\equiv{1}(mod 3)$ so $p^2+k\equiv{2+1}\equiv{0(}mod 3)$. $\Box$