Maths Olympiad Prep

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Number theory Difficulty 6.3 National olympiad Prove it

10. Consider the quadratic congruence ax2+bx+c0(modp)a x^{2}+b x+c \equiv 0(\bmod p), where pp prime and a,ba, b, and cc are integers with p\ap \backslash a.
a) Let p=2p=2. Determine which quadratic congruences (mod2)(\bmod 2) have solutions.
b) Let pp be an odd prime and let d=b24acd=b^{2}-4 a c. Show that the congruence ax2+bx+c0(modp)a x^{2}+b x+c \equiv 0(\bmod p) \quad is equivalent to the congruence y2d(modp)y^{2} \equiv d(\bmod p), where y=2ax+by=2 a x+b. Conclude that if d0(modp)d \equiv 0(\bmod p), then there is exactly one solution xx modulo pp, if dd is a quadratic residue of pp, then there are two incongruent solutions, while if dd is a quadratic nonresidue of pp, then there are no solutions.

Solution

None

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.