Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

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: http://www.stanford.edu/~rjlo/papers/09-TernaryQuadratics.pdf.

share|improve this answer
    
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:

$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$

$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$

$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$

$L$ - any integer number given by us

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

$qb^2+c^2=f^2$

$t^2-(q+1)f^2=\pm{q}$

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.

share|improve this answer

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:

$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$

$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$

$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$

All of numbers can be any character.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.