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Actually, I need to find if $4x^2 - 3y^2 - z^2 = 12$ is solvable. But I somehow feel that applying theory of integer representation by quadratic forms in three variables would yield quicker results... So far I haven't been lucky though.

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You can rewrite $4x^2-3y^2-z^2=n$ as $(2x+z)(2x-z)=3y^2+n$. – Next Nov 21 '13 at 13:48
Did you perhaps look for solutions? There are many. – Mike Bennett Nov 21 '13 at 19:42
up vote 1 down vote accepted

$(x,y,z)=(2,1,1)$ is a solution of $4x^2-3y^2-z^2=12$, so it is solvable. This is probably quicker than factoring or reviewing some theory -- although there is a rich theory about ternary quadratic forms, and the question which integers they represent. Ramanujan studied this for the quadratic form $x^2+y^2+10z^2$. It is associated to the elliptic curve $y^2 = x^3+x^2+4x+4$, and the question which odd numbers are represented by it is very difficult, and most results here depend on generalised Riemann hypotheses. A nice article can be found here:

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This case is rather easier as the form is indefinite (and can be easily shown to represent every $n \not\equiv 2 \mod{4}$ infinitely often). – Mike Bennett Nov 22 '13 at 1:03
@MikeBennett: yes, it is easier, of course. This solves the question completely, thank you. I wanted to say that this question in general can be quite interesting and difficult, and in fact Ramanujan considered such a case. – Dietrich Burde Nov 22 '13 at 9:27

If we consider the Diophantine equation: $qX^2+Y^2=Z^2+j$

If the root is a : $a=\sqrt{\frac{j}{q}}$

We use the solutions of Pell's equation: $p^2-(q+1)s^2=1$

Solutions can be written:




$L$ - any integer number given by us

number: $b,c,f,t$ - are solutions of the following equations



If we take the solutions of Pell's equation: $p^2-(q+1)s^2=k$

number $b,c$ - are solutions of the equation: $qb^2+c^2=f^2$

wherein: $c-b=k$

number $t,f$ - solutions of the equation: $t^2-(q+1)f^2=\pm{qk^2}$

These formulas allow us to find some solutions of Pell's equation using solutions of simpler equations. At least there will be another opportunity to find a solution to this equation. Later draw solutions with other factors.

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All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$

If the ratio is factored so: $a=(b-c)(b+c)$

Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$

where: $f=(q+1)k^2-2kt-(q-1)t^2$

Then the solutions are of the form:




All of numbers can be any character.

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