I'm trying to solve $13j^2 \equiv 1 \pmod{264}$. Now I've seen the form $x^2 \equiv n \pmod{m}$ but the form $ax^2 \equiv n \pmod{m}$ although it looks like a quadratic congruent equation, I don't seem to be able to follow exactly the same steps as $ax^2 +bx+c \equiv 0 \pmod{p}$. Some help would be appreciated (with detail and explanations). Thank you in advance.
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I agree but how do I solve that then? – Yoni Mar 25 '18 at 06:12
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Chinese Remainder Theorem? – Angina Seng Mar 25 '18 at 07:17
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Hint because $\gcd(13, 264)=1$ then $13x+264y=1$ has $(61,-3)$ as a solution, thus $13\cdot 61 \equiv 1 \pmod{264}$ and $$13 j^2 \equiv 1 \pmod{264} \iff 13\cdot61 j^2 \equiv61 \pmod{264} \iff\\ j^2 \equiv61 \pmod{264}$$
rtybase
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Thank you for the hint. To find j from j squared, I can't find a perfect square by adding 61 + 264, then 61 +2(264)....61+n(264). I didn't try 264 numbers...is there a way to do that easier and faster? – Yoni Mar 25 '18 at 15:12
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There is no solution, see the last theorem in this chapter and $264$ is divisible by $8$. – rtybase Mar 25 '18 at 18:46