let $a,b,c,x,y$ be non-zero positive integers such that $$\gcd(x,y,z)=1$$ find all the non-trivial integral solutions of the diophantine equation:$$ax^2+by^2=cz^2$$ I know that the Legendre's theorem can be helpful in solving this equation but beyond the insight that this theorem provides into the criteria of solvability, it did not help me get any further. Please help.
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I find it easier to use the formulas. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Oct 22 '14 at 12:23
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@Individ, Thanks. But there must be a method to get to the formulas. That's what I am interested in. – Oct 22 '14 at 12:28
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The theorem of Legendre does not give the formula of the solution. It gives a criterion that says when this equation can be solve. – individ Oct 22 '14 at 15:13
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That's my point. How do I get to your formula? – Oct 22 '14 at 15:23
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No! It is necessary to wait when the article prints. – individ Oct 22 '14 at 15:29
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ok. it's worth the wait. Is your paper archived yet on arxiv.org? – Oct 22 '14 at 15:38
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Let us continue this discussion in chat. – individ Oct 22 '14 at 15:40
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Give more this topic will not be discussed. – individ Oct 22 '14 at 15:43
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There is at least one integer solution $(U,V,W)$ to $a U^2 + b V^2 = c W^2$ with small variables, $a U^2 + b V^2 + c W^2 < 4abc.$
Once you have a single such integer solution, all primitive integer solutions are given by taking $\gcd(m,n) = 1,$ then $$ X = -aUm^2 - 2 b V mn + b U n^2, \; \; Y = aVm^2 - 2 a U mn - bV n^2, \; \; Z = (a m^2 + b n^2)W. $$ Then take $$ g = \gcd(X,Y,Z), $$ finally $$ x = X / g, \; \; y = Y/g, \; \; z = Z /g. $$ Depending upon the actual coefficients $a,b,c,$ it may be possible to predict $g,$ resulting in a finite number of slightly different formulas to give all primitive solutions.
Will Jagy
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Like i said I found this form except that small integer solution. That's my challenge. Can you write down one for me? – Oct 24 '14 at 01:46
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@num, I am missing something...why would you think there was a way to predict a solution to $ax^2 + by^2 = cz^2?$ – Will Jagy Oct 24 '14 at 02:22
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Because Euler found an identity for equations of this form. I just cannot find his approach. – Oct 24 '14 at 02:39
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@num Sigh. Do you have a reference for the idea that Euler said something about this? – Will Jagy Oct 24 '14 at 02:50
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Once you divide both sides by $z^2$, you have an equation of the form$ aX^2+bY^2=c$. How did he get his identity? I just wish I could find the reference. That'd help me tremendously. – Oct 24 '14 at 03:04
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@NumThcurious, you have misunderstood. The identity of Euler says merely how to multiply values of the binary form $a X^2 + b Y^2.$ It is essentially Gauss composition. – Will Jagy Oct 24 '14 at 03:05
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We need to write generally speaking the more General equation:
$$aX^2+bXY+cY^2=jZ^2$$
Although I formula solutions recorded, but I see it is of interest expression solutions using any one of the known solution.
If we know what any one solution: $(x,y,z)$ - then you can write a formula for the solutions of this equation.
$$X=jxt^2-cxk^2+2(cyk-jzt)s+(by+ax)s^2$$
$$Y=jyt^2-2jztk+(cy+bx)k^2+2axks-ays^2$$
$$Z=jzt^2-(bx+2cy)kt+czk^2+(bzk-(2ax+by)t)s+azs^2$$
$k,t,s$ - any integer asked us.
individ
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It's interesting that it is easier to find the general form than one actual solution. Thanks Individ. – Oct 24 '14 at 10:30
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