So it is always a case k=1 when we take any prime number... Bcz prime no are odd(expect 2) and when we taking n-1 it is always give us even number... So n-1 = m*2^k... m and k in every case m=n-1/2 and k=1..
If k > 1, you may need to perform both tests. Primes will yield T == +-1 (mod n) or a successive square of T will be == -1 (mod n). Also, the M-R test is a compositeness test. n passing the test for base a means n is a strong probable prime to base a, but not a guaranteed prime. Base 2 is a strong liar for 2047 and infinitely many other composites.
@@CSEconceptswithParinita yes sister. btw , all the videos are so awesome. Yesterday i was watching them. Thank you so much for such an awesome explanation. Keep going 💯💯💯💯💯💯💯💯💯💯💯
There are up to k modular comparisons. T == +-1 (mod n) is still the first test, and if a prime does not pass that part, it will yield a remainder of -1 in one of the remaining k-1 checks. The remainder is squared and modded with n to get each next comparison.
@@CSEconceptswithParinita Hey can you make a video on Schoof's point counting algorithm tomorrow, because I have my exam on Computational Number Theory the day after tomorrow and I'm unable to find any proper explanation over that !
Thanks Apu for such nice explanation
Keep learning, keep sharing ✌ 😊
Very helpful tutorial...
Thanks for your feedback 😊
Mam how 25*8=234?
overall good work
Glad you liked it! Keep learning, keep sharing 😊
Thank you madam. Your videos made me understand all the concepts of cryptography 💥💥
I'm glad to know that 😊😊
Super explanation
Thank you for the acknowledgement 😊
Very clear👍
Thanks for supporting😊
Maam would we be getting value of k other than 1 as even numbers are always divisible by 2 and their is no logic to check prime for even numbers
Thank you :)
My pleasure 😊
So it is always a case k=1 when we take any prime number...
Bcz prime no are odd(expect 2) and when we taking n-1 it is always give us even number...
So n-1 = m*2^k...
m and k in every case
m=n-1/2 and k=1..
thankyou mam
My pleasure 😊 Keep learning, keep supporting 💯
If k > 1, you may need to perform both tests. Primes will yield T == +-1 (mod n) or a successive square of T will be == -1 (mod n).
Also, the M-R test is a compositeness test. n passing the test for base a means n is a strong probable prime to base a, but not a guaranteed prime. Base 2 is a strong liar for 2047 and infinitely many other composites.
nice explaination
Thanks for the acknowledgement 😊
sister atleast give the link of the previous video that you are refering in your description.
Sure, will take care next time 🙂
@@CSEconceptswithParinita yes sister. btw , all the videos are so awesome. Yesterday i was watching them. Thank you so much for such an awesome explanation. Keep going 💯💯💯💯💯💯💯💯💯💯💯
I'm glad to know that✌ Keep learning, keep sharing 😊
what if we get value of a other than 2?
Even if the value of a is anything other than 2, we have to carry out the same procedure
What when k=2
What is 'a' ?
if k=2 then how we find
There are up to k modular comparisons. T == +-1 (mod n) is still the first test, and if a prime does not pass that part, it will yield a remainder of -1 in one of the remaining k-1 checks. The remainder is squared and modded with n to get each next comparison.
That's Okay and everything...but 25 * 8 :))))
Bro u get ? 😅
Jisko 25*8 nhi aata usko kya hi bolna aur
I don't know how someone would write a number that's clearly not divisible by 5 or 8 lol
Prepare well before making the vedio. Overall performance is alright 👍
Thanks for your feedback!
@@CSEconceptswithParinita Hey can you make a video on Schoof's point counting algorithm tomorrow, because I have my exam on Computational Number Theory the day after tomorrow and I'm unable to find any proper explanation over that !