I have this diophantine equation and i need the general solution form for it
the equation
$$x^2+qy^2-z^2=4n$$
some conditions $y\le 0,x\ge 0,z\ge 0$ x,z are even or odd together
$n=5y+x-3xy$
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http://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2/789929#789929 – individ Jul 11 '16 at 13:40
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$$z^2-x^2=(z-x)(z+x)=y^2-4n$$ There is always a solution for any numbers $y,n$ . It is only necessary to decompose the difference in the multipliers. – individ Jul 11 '16 at 13:52
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I don't think you need infinite descent; the given equation is equivalent to $x^2+y^2\equiv z^2\pmod{4}$, so it is enough to just check all possible values of x, y, and z mod 4 (of course, it is even faster if you only check the values of $x^2$, $y^2$, and $z^2$ mod 4) – alphacapture Jul 11 '16 at 13:53
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1- if i said that there is only one solution for this case while i know that if x is even then z is even too and odd if odd ,in addition to there is a relation between x,y and n where n= 5y+x-3xy . – Sherif Abdalmoniem Jul 11 '16 at 14:02
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2- #individ can you explain the solution $qX^2+Y^2-Z^2=a$ it is too hard to solve with this method in your link it have many variables – Sherif Abdalmoniem Jul 11 '16 at 14:05
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I don't understand what You need decision? Suggested a few options - not satisfied. So what need??? – individ Jul 11 '16 at 15:49
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#individ calm down ! -in my research i have a problem of this diophantine equation $n^2+4r^2-s^2=(4/3)(5-(3n-5)(r-1))$ the only value i have is n but i know some properties about r and s that's all and i knew that there is only one solution where n,s positive integers and r is negative integer – Sherif Abdalmoniem Jul 11 '16 at 17:10