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Suppose $a,b,c\in\mathbb{N}$ and are each squarefree. Is there a general solution for this equation?

I found that for this equation to be soluble in integers there are three necessary and sufficient conditions which are $ab,bc$ and $ac$ should be quadratic residues $\bmod c,\bmod a$ and $\bmod b$ respectively. That is, the equations

$$ ab\equiv\alpha^2 \hspace{-0.8em}\pmod{c} \\ bc\equiv\beta^2 \hspace{-0.8em}\pmod{a} \\ ac\equiv\gamma^2 \hspace{-0.8em}\pmod{b} $$ should be solvable. These conditions can be easily derived from the given equation assuming that $a,b$ and $c$ are relatively prime in pairs, but I cannot understand how these are the only sufficient conditions and how the equation is solvable if these conditions are satisfied.

Mathew George
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  • The formula in the General form looks like this. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Sep 16 '14 at 14:59

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This is discussed in great detail here. See in particular:this theorem of Legendre. Those who don't want to read Ireland and Rosen can find a proof here.

Igor Rivin
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  • It would be very helpful to point out what the symbol $\Box$ in the referenced theorem of Legendre means. Is it just any constant? Or some kind of determinant, maybe? –  Dec 20 '13 at 15:20
  • Why is there a minus sign in the last congruence? – Mathew George Dec 20 '13 at 15:26
  • @pyramids the simple means "square" (a quadratic residue). – Igor Rivin Dec 20 '13 at 15:28
  • Thanks, Igor. I guess that was kind of obvious, but somehow I didn't get it, so thanks for clarifying. –  Dec 20 '13 at 15:30