The solution
Here's the solution.
We have:
k^2-a^2=b^2-1
(k+a)(k-a)=b^2+1
Then choose b such that b is even.
This implies b^2+1 is odd. If b^2+1 is a prime, then put k+a=b^2+1 and k-a=1. You will get k&a. If b^2+1 is not a prime, then choose k+a and k-a as its two odd factors. Solving...
Hey, I read about arbitrary elements sometime ago. Could anybody tell me what affects the values of these? And I also read that some say that the values of these are fine-tuned by the Creator. What is the general take on that?
Hey guys, u know how Euclid proved that primes r infinite. Now knowing that primes r infinite, if we take some primes p1, p2, p3,...,pn then will p1*p2*...*pn(+/-)1 always be prime?
Sorry for the confusion!
Sorry, a*b= k^2, k belongs to the naturals. Not that a*b=a. Really sorry!
Thanks for the hint, Shmoe! I may be clser to the solution by an another method. I'll let you know if I get it. BTW, TPT is "to prove that". :wink:
I somewhere read that Euler proved that primes are infinite by proving that the series 1/2 +1/3 + 1/5 +... diverges. Can anybody tell the proof?
Aditya