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Decide which of the following congruences are solvable, and if so, find all solutions:

a) $x^2 \equiv c \bmod 363$, where $c= 1,5,31$

b) $x^2 \equiv 54 \bmod 125$

Now, I know how to find which of those are solveable, but I have no idea how to find the solutions.

(For example b) is solvable).

Any ideas?

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    At $x=1$, $c=1$ is trivial. – 2017-01-01
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    @SimpleArt Yes, but for the nontrivial congruences?Is there any method generally to find it? Or a method for specific cases? – 2017-01-01
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    Note that $x^2\equiv0,1\pmod3$. Thus, we can already see $c=5$. – 2017-01-01
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    this is the modular log and not trivial, but it can be done depending on how many calculations you can afford. – 2017-01-01
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    for $c=1$ i have found $$x=1,122,241,362$$ – 2017-01-01
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    Yes, use the symmetry of $(+x)^2=(-x)^2$ to find solutions faster. – 2017-01-01
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    @Dr.SonnhardGraubner How? – 2017-01-01
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    @ChakaChaka From my last hint, one only needs to check the first half of the values from $1$ to $363$. – 2017-01-01
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    @SimpleArt you can get something cheaper by using CRT, and when attacking mod $11^2$ first solving $\bmod 11$ to avoid a ton of cases. – 2017-01-01
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    @SimpleArt But it's still 181 numbers to check... – 2017-01-01
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    @JorgeFernándezHidalgo Why $mod \ 11$? – 2017-01-01
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    well, $363=11^2 \times 3$, you only have to check $22$ cases. – 2017-01-01
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    $11^2\times3=363$. Same idea as I used to show $c=5$ is not possible. – 2017-01-01

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Perhaps these general ideas can help you:

Suppose that we want to solve $x^k \equiv c \bmod p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$

It suffices to find the solutions $\bmod p^a$ for each prime power and then combine them using the chinese remainder theorem.

So how can we solve $x^k\equiv c \bmod p^a$?

One way is to do it inductively over $a$.

First find all the solutions to $x^k \equiv p^{a-1}$, so when we check for the solutions to $x^k\equiv p^{a-1}$ we only have to check congruences which also solved the previous equation.