Congruences| Part 11| Wilson's theorem and its converse

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This video is about Congruences| Wilson's theorem and its converse

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6:50 how can you definitely assume that there exists a unique a' such that aa'=1 (mod p)
Without proving ?

ayushdoglaaurharamihai

which writing interface software do u use?

PathFindersAcademy

You are great onctly the best video i have watch explain this theory

tonyhaddad

Proof of Wilson theorem. Let us consider numbers
2x, 3x, ...., (p-2)x, where x is natural number from 2 to (p-2) and p>=3 is prime. First I claim that all these numbers give different remainders when dividing by p. Indeed, if at least two numbers give equal remainders then their difference should be divisible by p., i.e. i*x - j*x = (i-j)*x should be divisible by p., Where i, j are from 2 to p-2. But 1<= i-j < p and x< p, thus (i-j)*x is not divisible by p. we got contradiction. Second, non of these remainders is x or p-x. It is clear that x not equal to p-x. Indeed, if remainder is x, then
i*x -x= (i-1)*x should be divisible by p, but 1<=i-1<=p-3, and 2<=x<=p-2 are not divisible by p, so we got contradiction. If remainder is p-x then i*x -(p-x)= (i+1)*x -p should be divisible by p, but 3<=i+1<=p-1 and 2<=x<=p-2 are not divisible by p, we got contradiction again. Therefore we obtained the following, all above mentioned remainders are distinct numbers from 1 to p-1 and not equal to x or p-x. So it means total quantity is p-3 remainders. But quantity of our numbers 2x, 3x, .. (p-2)*x are also p-3. Which means that there is one and only one remainder is 1 for each x from 2 to p-2. So we proved that 2x for only one x from 2 to p-2 has got remainder 1 when divided by p, 3y for only one y from 2 to p-2 has got remainder 1 - where x not equal y and so on the same for (p-2)*z for only one z from 2 to p-2 where x, y, ..., z are all distinct. So we got that all numbers from 2 to p-2 is partitioned into pairs of numbers which product gives remainder 1 when dividing by p. It means that all their products also gives remainder 1 when dividing by p. So 2*3*4*... (p-2) gives remainder 1 when dividing by p.i.e 2*3*4*(p-2) = A*p +1 for some number A. multiply both parts by p-1 we get (p-1)! = A*p*(p-1) +p-1 then (p-1)!+1 = (A*(p-1)+1)*p divisible by p. That is et!

omrquliyev

Continue doing videos on Number Therory and its applications

narendrareddyr

Explained so well but I have doubt in converse part how d/1 ? I am not getting this. Is it because of d /{ (n-1)! + 1 - (n-1) } => d /1 ??

chusrangagasirmarak