Let $x,y,a,b,m$ be non-zero integers where $m$ is square free. I am trying to find all the integer solutions of $$x^2-y^2=a^2+mb^2$$ I tried to factorize both sides but the righthand side equation is not too friendly. Any hints?
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do you want the hints as answers ? – Sep 22 '17 at 22:41
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Reference, answers, you name it. This problem appears so elementary but yet I cannot figure it out. Thanks. – Sep 22 '17 at 22:45
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If $x=1, y=0, m<0$, finding this part of the solution set is equivalent to Pell's equation... – Daniel Schepler Sep 22 '17 at 22:49
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$$a^2+mb^2=(zc+mvq)^2+m(zq-vc)^2=(z^2+mv^2)(c^2+mq^2)=(x-y)(x+y)$$
$$x=\frac{1}{2}(c^2+z^2+m(q^2+v^2))$$
$$y=\frac{1}{2}(c^2-z^2+m(q^2-v^2))$$
individ
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Here we have an answer someone can follow, not like some of your other answers – MaximusFastidiousIrreverence Oct 07 '17 at 22:26