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Solving Quadratic Congruences: 16 Solutions
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[QUOTE="talolard, post: 3341873, member: 226208"] [h2]Homework Statement [/h2] How many solutions does [tex]x^{2}\equiv9 mod 7700[/tex] have? So my question is if this solution is "legitimate" Solution First notice that[tex] 7700=7\cdot11\cdot2^{2}\cdot5^{2}[/tex] Thus we must solve the system [tex]\begin{cases} x^{2}\equiv2 & \left(7\right)\\ x^{2}\equiv9 & \left(11\right)\\ x^{2}\equiv\mbox{1} & \left(4\right)\\ x^{2}\equiv9 & \left(25\right)\end{cases}. [/tex] This system is equivelent to [tex]\begin{cases} x\equiv\pm2 & \left(7\right)\\ x\equiv\pm9 & \left(11\right)\\ x\equiv\pm1 & \left(4\right)\\ x\equiv\pm9 & \left(25\right)\end{cases}. [/tex] since gcd(1,p)=1 is always true, all of these equations are solvable by the fundamental theorem of modulo arithmetic. There are [tex]2^{4}=16[tex] disjoint equations and thus 16 distinct solutions. Each solution must be distinct, since if x solves to different equations, then for instance we would have [tex]x\equiv2\equiv-2\left(7\right)[tex][h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] [/QUOTE]
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Solving Quadratic Congruences: 16 Solutions
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