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I have the following question:

Find all solutions $x^2=1(mod 143)$

This is what I did :

11,13 are prime therefore :

$$(1)\space x^2=1(mod 11)$$ $$(2) \space x^2=1(mod 13)$$

using The Chinese Theorem I get that :

$$x^2=(11*x_1*1)+(13*x_2*1)$$

Finding $x_1,x_2$

$13x_1=1(mod 11) \rightarrow x_1=6+11k$

$11x_1=1(mod 13) \rightarrow x_2=6+11k$

Therefore : $$x^2=(11*x_1*1)+(13*x_2*1)=x^2=(11*6*1)+(13*6*1)=144+143k$$

But wolframalpha finds $4$ different solutions and I just found one, Any ideas what I did wrong?

Any help will be appreciated.

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    i have got $$1,12,131,142$$ as the solutions – 2017-01-29
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    $x^2 \equiv 1 \pmod{11}$ is equivalent to $x \equiv \text{plus or minus } 1 \pmod{11}$. Same for mod 13. – 2017-01-29
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    You found one solution for $x^2$, and that's all there is, but the question wants you to find all solutions for $x$, and there are four of those. – 2017-01-29

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There are two solutions to $x^2 \equiv 1\pmod{11}$, namely $1$ and $-1\equiv 10$. The same thing goes for $13$. Therefore, the Chinese remainder theorem gives you four solutions to the entire problem, namely: $$ \cases{x\equiv 1\pmod {11}\\x\equiv 1\pmod{13}}\\ \cases{x\equiv 10\pmod {11}\\x\equiv 1\pmod{13}}\\ \cases{x\equiv 1\pmod {11}\\x\equiv 12\pmod{13}}\\ \cases{x\equiv 10\pmod {11}\\x\equiv 12\pmod{13}}\\ $$

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    So generally speaking when I want to use the Chinese Theorem I need to get rid of the powers in order to use it? – 2017-01-29
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    @JaVaPG No, the CRT doesn't care about that. Quadratic equations have multiple solutions (or none) in modular arithmetic, just as they do for the regular integers / ratioanls / reals / complex numbers, and you forgot to take that into account. – 2017-01-29