In fact, thanks to Bertrand's Postulate, we can be certain there are millions of primes between those two.

The Canadian Senate has very little power and influence. And in any case, the Conservative Party currently holds a majority in the senate, making it very unlikely that the Senate would attempt to block a bill passed by the House.

Coverage Mapper is doing some of what you describe.

We also know that all even perfect numbers must have the form 2^(p-1) * (2^p - 1), where 2^p - 1 is a Mersenne prime. It's not known whether an odd perfect number exists, but we at least know that there are none less than 10^1500.

It is far easier to test Mersenne numbers for primality than general integers, thanks to the Lucas-Lehmer primality test. Weeding out composite Mersenne numbers by trial factoring is also faster, thanks to a theorem that eliminates most of the candidate factors.

As for numbers of the form 2^n+1, it's easy to show that if 2^n+1 is prime, then n must be a power of two. Such numbers are known as Fermat numbers, and there is also a fast primality test (Pepin's test) for numbers of this form. But because of the power of two in the exponent, each Fermat number is double the size (in terms of the number of digits) of the previous one. It doesn't take long before you exceed current computing capabilities. And so far, all the Fermat numbers we've managed to test have turned out to be composite, apart from the first five. Furthermore, it is conjectured that there are only finitely many Fermat primes, so it's possible we've already found them all. On the other hand, it is conjectured that there are infinitely many Mersenne primes.

I think you meant Quantum Computing. Quantum Crypto already exists and in no way threatens the security of classical crypto.

Indeed, it's pretty much the whole point of jailbreaking. :-) And as far as I know, the some of the jailbreaking tools exploit arbitrary code execution vulnerabilities to do their job.

Actually a Diffie-Hellman key over a prime field can be broken in about the same time as a similarly sized RSA key using, for example, the index calculus algorithm. The amount of computation required to break a 2048-bit Diffie-Hellman key is about 2^112, not 2^1024.

There are still plenty of reasons to jailbreak: high-quality Youtube videos over 3G, video recording on pre-3GS devices, unix utilities, a flashlight app that sets the screen to maximum brightness, the ability to run any C64 game on Manomio's emulator, plenty of apps that aren't available in the app store...

Just jailbreak, unlock, and buy a prepaid SIM card in the country you're visiting. I used a couple hundred voice minutes and 2GB of data while in Bulgaria and it only cost me about $60. The jailbreak/unlock process is quite simple these days.

Will it be brighter than iridium flares, which can reach an apparent magnitude of -8.0?

The article is already out of date. The round 1 candidates were announced back on December 11. Since that time, 11 candidates have been broken. For the latest information, I recommend visiting the SHA-3 Zoo.

Also, the article suggests that candidates will continue to be broken quickly, but I doubt this will happen. The weak hashes will be broken quickly, but there are likely to be many strong candidates which will not be broken during the contest. Other factors (speed, simplicity, etc.) will determine the ultimate winner.