proof that is prime for
We show that for , is prime. Of course, this can easily be seen by considering the cases, but the proof given here is illustrative of why the statement is true.
Recall that there is only one reduced (http://planetmath.org/IntegralBinaryQuadraticForms) integral binary quadratic form of discriminant ; that form is . The smallest prime that is represented by that form is . For suppose and . Then obviously , so , which is impossible. Since equivalent forms represent the same set of integers, it follows that any form of discriminant represents no primes less than .
Now suppose is composite for some . Then
and thus has a prime factor . Write ; then represents (); its discriminant is
Since there is only one equivalence class of forms with discriminant , is equivalent to and thus represents the same integers. But we know that cannot represent any prime , so cannot represent . Contradiction. So is prime for .
This proof works equally well for the other cases mentioned in the parent article, since for each of those cases, there is only one reduced form (http://planetmath.org/IntegralBinaryQuadraticForms) of the appropriate discriminant, which is .
|Title||proof that is prime for|
|Date of creation||2013-03-22 16:55:48|
|Last modified on||2013-03-22 16:55:48|
|Last modified by||rm50 (10146)|