This question probably proves my ignorance when it comes to factoring: How did you find factors with thousands of digits?

[QUOTE=Puzzle-Peter;332975]This question probably proves my ignorance when it comes to factoring: How did you find factors with thousands of digits?[/QUOTE]

Algebraic factors.
2^(2n)-1 = (2^n-1)*(2^n+1)

[QUOTE=henryzz;332977]Algebraic factors.
2^(2n)-1 = (2^n-1)*(2^n+1)[/QUOTE]

Haha, give me the right side of the equation and I give you the left side in no time at all. Doing it in reverse never crossed my mind... thanks!

(Re: It is a [URL="http://www.primenumbers.net/prptop/searchform.php?form=%2830%5E78857-1%29%2F29&action=Search"]known PRP[/URL], found by R.Price. )
Thank you Batalov. I looked for but couldn't find this kind of info so I had to roll it.
Many computer cycles saved!

[QUOTE=ishkibibble;333242](Re: It is a [URL="http://www.primenumbers.net/prptop/searchform.php?form=%2830%5E78857-1%29%2F29&action=Search"]known PRP[/URL], found by R.Price. )
Thank you Batalov. I looked for but couldn't find this kind of info so I had to roll it.
Many computer cycles saved![/QUOTE]

I think you are confused. You still have not proven its primality so no computer cycles have been saved. Just because it is a PRP does not mean that it is prime. There is not currently an easy way to test a > 100,000 digit number for primality unless n-1 or n+1 can be factored to 33%.

When you made the bold and clearly incorrect statement that "I was able to complete one test where the value is shown as prime.", Batalov was being sarcastic when he said "And what test would that be?" knowing that it is not easily possible to prove its primality.