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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – user99666
    Dec 20, 2013 at 15:20
  • $\begingroup$ Why is there a minus sign in the last congruence? $\endgroup$
    – Mathgrad
    Dec 20, 2013 at 15:26
  • $\begingroup$ @pyramids the simple means "square" (a quadratic residue). $\endgroup$
    – Igor Rivin
    Dec 20, 2013 at 15:28
  • $\begingroup$ Thanks, Igor. I guess that was kind of obvious, but somehow I didn't get it, so thanks for clarifying. $\endgroup$
    – user99666
    Dec 20, 2013 at 15:30

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