P vs. NP and Pseudorandom Bit Generators According to an article on pseudorandom number generators (PRNG) by Jeff Lagarias, he states that trying to prove that a PRNG is unpredictable (secure) is just "as hard" as trying to prove that P!=NP. If it is considered impossible to prove mathematically that some bit sequence is random then it should be mathematically impossible to prove P!=NP, right? Are these two problems really equivalent? If they are then why all the hoopla over P vs NP, you can never prove it either way according to these definitions.
There is a bit of an issue with your question. You are using the words unpredictable and secure interchangeably, which would seem to imply that you mean "unpredictable to a polynomial-time bounded  adversary". This is not at all the same thing as being truly unpredictable, since clearly there is a function (in this case the PRNG) which generates them. So this is not related to any issue with proving that the sequence is 'truly' random in some sense. Rather, what you really need is that given the prior output of the PRNG it is computationally hard to predict the next number.

This formulation is clearly related to P vs NP since clearly calculating a sequence of $m$ pseudo-random numbers using the generator is in FP (the functional version of P) when the key $k$ to the PRNG is known. However given a set of $m$ sequential pseudo-random numbers outputted by the device, the problem of finding $k$ is clearly in FNP (the functional version of NP).

Thus proving that it is computationally easy to generate the pseudo random numbers but computationally hard to infer the key from a sample of the output of the PRNG would imply that $FP \neq FNP$ (and hence $P\neq NP$).