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I want to find how many solutions the congruence $x^2 \equiv 121 \mod 1800$ has.

What is the method to find it without calculating all the solutions? I can't use euler criterion here because 1800 is not a primitive root. thanks!!!

CnR
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2 Answers2

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As $(121,1800)=1,$ $$x^2\equiv121\pmod{1800}\iff(x\cdot 11^{-1})^2\equiv1$$

Now as $1800=3^2\cdot2^3\cdot5^2$

as $y^2\equiv1\pmod{p^n}$ has exactly two in-congruent solutions for odd prime $p$

and $y^2\equiv1\pmod8$ has four

the number of solutions $$(x\cdot 11^{-1})^2\equiv1\pmod{1800}$$ will be $2\cdot4\cdot2$

lab bhattacharjee
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  • hi, thank you but I didn't understand 2 things: 1. how did you get that: $ x^2\equiv121\pmod{1800}\iff(x\cdot 11^{-1})^2\equiv1 $ and 2. why do you multiply to get the number of solutions? – CnR May 01 '16 at 09:19
  • @CnR, See http://www.johndcook.com/blog/quadratic_congruences/ – lab bhattacharjee May 01 '16 at 09:35
  • it doesn't explain my questions :( – CnR May 01 '16 at 09:53
  • @CnR, See http://math.stackexchange.com/questions/345668/solutions-of-the-congruence-x2-equiv-1-pmodm – lab bhattacharjee May 01 '16 at 09:55
  • I really don't understand how what I asked and these links are related.. I appreciate your help but would like more elaborated explanation so I can undersand the steps of your proof – CnR May 01 '16 at 14:17
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Here is a different approach to find the number of solutions to the congruence $$ x^2 \equiv a \space(\bmod n)$$ If $$ n = 2^kp_1^{k_1}\cdots p_r^{k_r}$$ where $p_1, \dots ,p_r$ are odd different primes, $k \ge 0$ and $k_1,\dots,k_r \ge 1$.

If the congurence has solutions, the number of the solutions will be equal to $2^r \cdot 2^m,$ Where $$ m= \begin{cases} 0, & k=0,1\\ 1, & k=2\\ 2, & k \ge 3 \end{cases} $$

In our case, $$ x^2 \equiv 121 \space(\bmod 1800)$$ $$n = 1800 =2^3\cdot 5^2 \cdot 3^2 \Longrightarrow k=3,r=2$$ And the number of solutions is $2^2 \cdot 2^2 = 16$.

NirF
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