2:32 PM
@EllenEllen have you tried assuming that n²-3 is of form 4k+1 (not 4k, as then n² becomes of form 4k+3 which isn't possible)
Hence n²-3 is odd and as 3 is odd too, n² must be even
n is of form 4k, as it's a perfect square
Let divisor be d². It divides an odd 4k+1. So d² is of the form 4m+1 itself.
Assuming quotient Q, substitute the formula d².Q=n²-3 by terms k,m.
This will yield a formula for generating m in terms of Q,k. i.e. There'll be definite m and hence definite d² for each Q,k, i.e. Q, n².
Group Theory
