help me with this doubt
given a quadratic congruence that x^2 cong -1 mod p. Where p is an odd prime. Consulting burton's book I have found that (p-1/2)! satisfies the expression. Now for instance p is taken to be 13 . Hence the solution will be 6!=720. Which means 720 much satisfy the expression. But I am confused that there the congruence of 720 is taken to mod 13 yields 720 cong 5 mod13 and 5 is implied into the solution. Please help me understand this. @ganeshie8
Ref:
heyy
@ganeshie8
still stuck on this ?
yes !!! :(
@ganeshie8
consider the factors in \((p-1)! = (p-1)(p-2)\cdots 3\cdot 2\cdot 1\)
would you agree \[\large p-1\equiv -1 \pmod{p}\] ?
yes!
thats simply because \(p\equiv 0 \pmod p\)
so \(p-1\equiv 0-1\equiv -1\pmod p\)
I am confused with this fact that 6! is a solution to the equation but why is that 6! was again taken with modulo. I am getting confused with this part basically.
\(x = 6! \) is a solution to the congruence \(x^2\equiv -1\pmod {13}\)
plugin \(x = 6!\) into the conruence and reduec : \[6!^2 \equiv -1 \pmod{13}\]
\[720^2\equiv -1 \pmod {13}\]
but we know \(720\equiv 5 \pmod{13}\) so we have \[5^2\equiv -1\pmod{13}\]
\[25\equiv -1\pmod{13}\] which is true, so yes \(x=6!\) is indeed a solution to the congruence \(x^2\equiv -1\pmod{13}\)
Okay! I know my question is foolish, but why do I need to take the mod of 720 again. :(
I mean 720 itself is the solution. right ???
thats right, 720 itself is the solution we're just checking whether 720^2 reduces to -1 or not
Also notie, 720 is same as 5 in modulo 13 so essentially \(5\) is a solution to the conruence \(x^2 \equiv -1 \pmod{ 13}\)
oho. sorry for the turmoil... Thank you :)
\[\color{Red}{720}^2\equiv -1 \pmod{13}\] is same as \[\color{Red}{5}^2\equiv -1 \pmod{13}\]
np :) it is good to ask questions and make sure every little thing makes sense
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