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I'm trying to solve for 188 and 131, how do we get 188 and 131 using CRT?

I can calculate these, its just I do not get how to get 188 and 131.

√23 (mod 209)

√a (mod p) = ±a(p+1)/4, if p ≡ 3 (mod 4)

√4 (mod 11) = ±23^3(mod 11) = ±1

√4 (mod 11) = ±23^3(mod 11) = ±17

Using the CRT we can therefore calculate √ 23 (mod 209) as ±188, ±131

hashsaltpepper
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  • You combine the solutions mod $11$ and $19$ using CRT exactly as described in the duplicate. If that is not clear then ping me here and I will elaborate. – Bill Dubuque Jan 15 '19 at 21:46
  • What are you trying to calculate, exactly? – Bernard Jan 15 '19 at 21:48
  • See also here and here for more worked examples. – Bill Dubuque Jan 15 '19 at 21:49
  • We seek solutions to $,x\equiv a\equiv \pm1\pmod{!11},\ x\equiv b\equiv \pm2\pmod{!19}.,$ Applying CRT wwe find a solution for $,a,b = -1,2,$ is $,x\equiv 21\pmod{!209}$ and its negative (corresponding to $,a,b = 1,-2)$ – Bill Dubuque Jan 15 '19 at 22:10
  • Similarly $,a,b = 1,2,$ yields $,x\equiv 78\pmod{!209},$ and its negative (for $,a,b = -1,-2)\ $ – Bill Dubuque Jan 15 '19 at 22:12
  • $!\bmod 209!:,\ 21 \equiv -188\ $ and $,78\equiv -131\ $ so these are the same as your solutions. Thos are all the possible combinatios of roots mod $11$ and roots mod $19$ so those are all the possible roots mod $11\cdot 19$ by the CRT argument in the linked threads. – Bill Dubuque Jan 15 '19 at 22:12
  • could you elaborate about the combination of solutions of mod 11 and 19? – hashsaltpepper Jan 16 '19 at 05:13
  • @hashsaltpepper That's explained in the linked dupe, see "By CRT each combination..." – Bill Dubuque Jan 16 '19 at 16:25

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