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I'm trying to find the number of solutions to this congruence: $$x^2\equiv 121\pmod {1800}$$

I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:

$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\; x^2\equiv 121\pmod {2^3}$

$\Downarrow$

$ x^2\equiv 21\pmod {5^2} \;,\; x^2\equiv 4\pmod {3^2} \;,\; x^2\equiv 1\pmod {2^3} $

Now I'm pretty stuck with how to solve each of the above quadratic congruences, is there a fast way to do it?

Thanks

user401516
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2 Answers2

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Hint:

We need $\left(\dfrac x{11}\right)^2\equiv1\pmod{3^2,5^2,2^3}$

Now we can prove $y^2\equiv1\pmod{p^n}$ has exactly two solutions for prime $p\ge3$ and integer $n\ge1$

Finally we can apply Chinese Remainder Theorem to find the number of in-congruent solutions to be $$4\cdot2^2$$

See also: Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$

lab bhattacharjee
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No, there's not a fast way. To solve $x^2\equiv a \pmod{p^n}$ we first solve $x^2\equiv a \pmod{p}$ and then use Hensel's Lemma to "lift" the solution to $\pmod{p^2}$ and then $\pmod{p^3}$, etc. Sometimes the solutions lift uniquely, sometimes not.

In the case of your problem, there are two solutions modulo $5$ which lift uniquely to two solutions modulo $25$. The exact same thing happens modulo $9$. There is one solution modulo $2.$ It lifts to two solutions modulo $4.$ Each of these in turn lifts to two solutions modulo $8$, for a total of 4 solutions modulo $8.$

So in the end, there are $2\times 2\times 4 = 16$ solutions modulo $1800.$

B. Goddard
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    I should add that a "fast" square root algorithm for non-prime moduli would be an earth-shattering invention. Half of all cryptosystems would be compromised. – B. Goddard Jun 04 '18 at 16:20