Hello :) i want to give a answer op the following question:
For which prime number $p$ can we give a solution of the diophantic equation given by $x^2-65y^2=p$.
I want to solve the question without a Computer Algebra system. I want to solve it algebraic. I thought about the following:
We can maybe use the quadratic reciprocity, because $x^2-65y^2=p$ is equivalent with $(x-y\sqrt{65})(x+y\sqrt{65})=p$ thus we search for primes which splits in the integers of $\Bbb{Q}(\sqrt{65})$. But how to give a general answer? Can someone help me? Thank you :)
Largely because $65 \equiv 1 \pmod 8,$ the form $x^2 - 65 y^2$ represents exactly the same odd numbers as $x^2 + x y - 16 y^2.$ That is to say, residues mod 5 and mod 13. By the Chinese Remainder Theorem. So, all primes $1,4,9,14,16,29,36,49,51,56,61,64 \pmod {65}.$ For example, $$ 1 \pmod {65}: \; \; \; 131 = 14^2 - 65 \cdot 1^2. $$ $$ 4 \pmod {65}: \; \; \; 199 = 28^2 - 65 \cdot 3^2. $$ $$ 9 \pmod {65}: \; \; \; 139 = 42^2 - 65 \cdot 5^2. $$ $$ 14 \pmod {65}: \; \; \; 79 = 12^2 - 65 \cdot 1^2. $$ $$ 16 \pmod {65}: \; \; \; 211 = 74^2 - 65 \cdot 9^2. $$ $$ 29 \pmod {65}: \; \; \; 29 = 17^2 - 65 \cdot 2^2. $$ $$ 36 \pmod {65}: \; \; \; 101 = 19^2 - 65 \cdot 2^2. $$ $$ 49 \pmod {65}: \; \; \; 179 = 58^2 - 65 \cdot 7^2. $$ $$ 51 \pmod {65}: \; \; \; 181 = 21^2 - 65 \cdot 2^2. $$ $$ 56 \pmod {65}: \; \; \; 251 = 106^2 - 65 \cdot 13^2. $$ $$ 61 \pmod {65}: \; \; \; 61 = 49^2 - 65 \cdot 6^2. $$ $$ 64 \pmod {65}: \; \; \; 389 = 83^2 - 65 \cdot 10^2. $$
EEEDDDIIIITTT: In case anyone is looking. The first thing of importance was that $8^2 - 65 \cdot 1^2 = -1.$ That is, (and this is unusual for composite numbers such as 65), we need pay no attention to $\pm$ signs anywhere. I think what I will do next is display the full Lagrange cycle for the principal and the secondary genus, first for discriminant 65 and then for 260.
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