If $k(a^2+mb^2) = c^2+md^2$, what can be said about the form of $k$?

Asked Apr 28 '16 at 15:43

Active Apr 28 '16 at 15:43

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Let $k,a,b,c,$ and $d$ be integers, and let $m \ge 2$ be a non-square integer, such that $$ k(a^2+mb^2) = c^2+md^2. $$

QUESTIONS:

What can be said about the form of $k$ with no further restrictions?
What can be said about the form of $k$ when $m$ has further restrictions (e.g., $m$ is a prime $p$ of a given form)?
In particular: Under what conditions is it guaranteed that $k=u^2+mv^2$ for some integers $u$ and $v$?

asked Apr 28 '16 at 15:43

Kieren MacMillan

The formula in General as follows. http://math.stackexchange.com/questions/1127654/parametrization-of-solutions-of-diophantine-equation Next, it is necessary to consider the equivalent form and find out the required coefficients. – individ Apr 28 '16 at 15:58
Can we say $k $ is a square? – N.S.JOHN Apr 28 '16 at 16:17
@N.S.JOHN: If that's one of the restrictions added, that's fine — but please make that explicit. – Kieren MacMillan Apr 28 '16 at 16:26
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@N.S.JOHN If $k$ square. Then, for any $m$ the solution is always there. – individ Apr 28 '16 at 16:38
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For $m =1,2,3$ and $5$, point 3 is satisfied. I think a general solution has been given by Gauss using Quadratic Forms in his book "Disquisitiones Arithmeticae". – MathGod May 18 '16 at 15:14
@KierenMacMillan Releated – Henry May 20 '16 at 13:49
@IshanSingh Would you mind writing a solution to this question and my related question? Thanks. – Henry May 20 '16 at 13:50

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