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 2013-01-04, 00:29 #1 Raman Noodles     "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts Representation by quadratic forms Representation by quadratic forms After research by writing with my own program, I got with this following result, great to be had them proved, been! $\begin{tabular}{|l|l|l|l|l|} \hline {If\ N\ can\ be\\ written\ as\\ (with\ a\ \ge\ 0,\ b\ \ge\ 0)} & {Prime\ factors\ of\ N\\ which\ have\ no\\ restriction} & {Sum\ of\ exponents\ of\\ all\ these\ prime\ factors\\ of\ N\ should\ be\ even\\ when\ combined\ together} & {Prime\ factors\ of\ N\\ whose\ exponent\\ should\ be\ \ge \ 2} & {Each\ of\ these\\ prime\ factors\ of\ N\\ can\ only\ occur\\ in\ pairs\ individually} \\ \hline a^2+b^2 & {2\\ 1\ mod\ 4} & ~ & ~ & 3\ mod\ 4 \\ \hline a^2+2b^2 & {2\\ 1\ mod\ 8\\ 3\ mod\ 8} & ~ & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline a^2+3b^2 & {3\\ 1\ mod\ 3} & ~ & ~ & 2\ mod\ 3 \\ \hline a^2+4b^2 & 1\ mod\ 4 & ~ & 2 & 3\ mod\ 4 \\ \hline a^2+5b^2 & {5\\ 1\ mod\ 20\\ 9\ mod\ 20} & {2\\ 3\ mod\ 20\\ 7\ mod\ 20} & ~ & {11\ mod\ 20\\ 13\ mod\ 20\\ 17\ mod\ 20\\ 19\ mod\ 20} \\ \hline a^2+6b^2 & {1\ mod\ 24\\ 7\ mod\ 24} & {2\\ 3\\ 5\ mod\ 24\\ 11\ mod\ 24} & ~ & {13\ mod\ 24\\ 17\ mod\ 24\\ 19\ mod\ 24\\ 23\ mod\ 24} \\ \hline a^2+7b^2 & {7\\ 1\ mod\ 14\\ 9\ mod\ 14\\ 11\ mod\ 14} & ~ & 2 & {3\ mod\ 14\\ 5\ mod\ 14\\ 13\ mod\ 14} \\ \hline {a^2+8b^2\\ (for\ odd\ numbers)} & 1\ mod\ 8 & 3\ mod\ 8 & ~ & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+8b^2\\ (for\ even\ numbers)} & {1\ mod\ 8\\ 3\ mod\ 8} & ~ & 2 & {5\ mod\ 8\\ 7\ mod\ 8} \\ \hline {a^2+9b^2\\ (for\ non-multiples\ of\ 3)} & 1\ mod\ 12 & {2\\ 5\ mod\ 12} & ~ & {7\ mod\ 12\\ 11\ mod\ 12} \\ \hline {a^2+9b^2\\ (for\ multiples\ of\ 3)} & {2\\ 1\ mod\ 12\\ 5\ mod\ 12} & ~ & ~ & {3\\ 7\ mod\ 12\\ 11\ mod\ 12} \\ \hline a^2+10b^2 & {1\ mod\ 40\\ 9\ mod\ 40\\ 11\ mod\ 40\\ 19\ mod\ 40} & {2\\ 5\\ 7\ mod\ 40\\ 13\ mod\ 40\\ 23\ mod\ 40\\ 37\ mod\ 40} & ~ & {3\ mod\ 40\\ 17\ mod\ 40\\ 21\ mod\ 40\\ 27\ mod\ 40\\ 29\ mod\ 40\\ 31\ mod\ 40\\ 33\ mod\ 40\\ 39\ mod\ 40} \\ \hline \end{tabular}$ Is it possible in general to give by an explicit formula, for the general representation of the quadratic form a²+kb² form, like this? The necessary condition for a prime p to be written as a²+kb² form, is being that (-k/p) = 1. How to determine the class number, in general, but what it is being, first of all, in properly defined sense? It seems to me that it is being an abstract concept... For this example, consider with the representation by using the next quadratic form, namely, a²+11b² form, in general. The necessary condition for a prime p to be written as a²+11b² form, is being that (-11/p) = (p/11) = 1. i.e. p ≡ 0, 1, 3, 4, 5, 9 (mod 11). It is not being a sufficient condition as since every prime of these residue classes cannot be written, in general, in the form, a²+11b² form. But if we extend it to include numbers of the form (a/2)² + 11(b/2)², whereby a, b are being integers ≥ 0, then certainly every prime of these residue classes can be uniquely written of the form (a/2)² + 11(b/2)². But how do we know that, what primes of the form 0, 1, 3, 4, 5, 9 (mod 11) do require with odd a, b values, for the representation of the form (a/2)² + 11(b/2)²? In general, the square-free numbers 1, 2, 3, 7, 11, 19, 43, 67, 163 have got with class number 1. Excluding thereby the non-square-free numbers 4, 8, 12, 27. Of course, for these k values, in a²+kb² form, (or thereby (a/2)²+k(b/2)² form - if k ≡ 3 mod 4), if a composite number C can be written in that form, then every prime factor P of it can be written in that form, thereby (a²+kb²)(c²+kd²) = |ac+kbd|² + k|ad-bc|² = |ac-kbd|² + k|ad-bc|² Last fiddled with by Raman on 2013-01-04 at 00:45

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