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Prove that the congruence equation $ax^2 + bx + c \equiv 0 \pmod{m}$, with $(2a, m) = 1$ is equivalent to a congruence of the form $x^2 \equiv r \pmod{m}$.

I really have no idea where to start. Is there a theorem which relates to this problem? What do we have to show? A hint would be sufficient.


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up vote 2 down vote accepted

Multiply it by the unit $\rm\:4a\:$ yielding the equivalent equation $\rm\ (2\:a\:x+b)^2\: =\ b^2 - 4\:a\:c\ \ (mod\ m)$

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Great hint! Thank you. – Chan Apr 2 '11 at 7:27

Remember the quadratic formula to solve $ax^2 + bx + c = 0$: $x = \tfrac{1}{2a}(-b \pm \sqrt{b^2 - 4ac})$.

This works for the congruence too if we just interpret it right.

Suppose $r^2 \equiv b^2 - 4ac$ then it's easy to check that $x \equiv (2a)^{-1} (-b \pm r)$ solves the quadratic. The condition 2a coprime is m is exactly what's needed to invert 2a.

if x solves the quadratic congruence then multiply by 2a and add b to get the other direction.

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Many thanks. – Chan Apr 1 '11 at 6:20
That only works for one direction of the equivalence. – Bill Dubuque Apr 1 '11 at 6:26
It works for both ways. – quanta Apr 1 '11 at 6:48
My point was that your first version did not say anything about the other direction, which, of course, is necessary to prove an equivalence. The equations are equivalent because scaling by a unit is an invertible transformation. – Bill Dubuque Apr 1 '11 at 7:06

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