A better method: divide it in a system of congruences mod 5, 7 and 9. Then rewrite in these forms: For mod 5: x^2+3x+2=(x+2)(x+1)=0, x=3 or 4 For mod 7: x^2+3x-4=(x+4)(x-1)=o, x=1 or 3 For mod 9: x^2+3x-10=(x+5)(x-2)=0, x=2 or 4. Then solve then system of congruences in these 6 cases. With a bit cleverness we have not to use 6 times chinese method, it’s much simpler
You said there weren't really any tricks, but your polynomial for 0 (mod 5) was already factorable into (x+2)(x+1), which give you -2, -1 --> 3, 4. Then for the mod 7 polynomial you could have changed 17 to -4 then factored into (x+4)(x-1) giving you 1, -4 --> 1, 3.
I noticed the error at 07:00 f'(2) = 4 ?? but luckily it resolved itself. ;-) - It is very important for your overall aura and reputation that you don't watch at the cheat-sheets all the time. ;-)
Think x^2 = 2 (mod 3). It does not have any solutions, x^2 is only congruent to either 0 or 1. We should be able to make a table and show that there are no solutions where we are working mod a small number. For working mod a large prime, I can't think of a better solution, though.
2. Find all values of x in the congruence x² = q(mod p), where p is a prime number, given (i) q = 15 p = 967 (ii) * q = 14, p = 941 . how to solve this problem
Chinese remainder theorem is lengthy and uneasy to handle. Instead use the following formula for this problem. Let x=x(1) mod(5), x=x(2) mod(7) and x=x(3) mod(9). x(1)=3,4 x(2)=1,3 x(3)=2,4 Formula:- x=126x(1)+225x(2)+280x(3) mod(315) All the 8 solns follow for different values of x(1), x(2) & x(3).
"mod" stands for the modulo operation ( en.wikipedia.org/wiki/Modulo_operation ). I'd highly recommend starting at the beginning of this playlist and watching the videos in order ( ruclips.net/p/PL22w63XsKjqwAgBzVFVqZNMcVKpOOAA7c ).
This is an amazing problem/solution. Especially when you consider all the 2^3 solutions. Me, I would just brute force it in python.
at 12:47 you say 5*45*7 but it's actually 5*45*1 (multiply by the choice of x, not the mod base). 1163 is the right result though.
i know I'm pretty randomly asking but does anyone know a good place to watch newly released movies online?
Professor Michael Penn, I really understand this lecture from start to finish. This is solid mathematics.
A better method: divide it in a system of congruences mod 5, 7 and 9. Then rewrite in these forms:
For mod 5: x^2+3x+2=(x+2)(x+1)=0, x=3 or 4
For mod 7: x^2+3x-4=(x+4)(x-1)=o, x=1 or 3
For mod 9: x^2+3x-10=(x+5)(x-2)=0, x=2 or 4.
Then solve then system of congruences in these 6 cases. With a bit cleverness we have not to use 6 times chinese method, it’s much simpler
8 cases
At 13:49 you say there are 7 other solutions but only show 6. Isn't 164 also a solution? 164^2+3 x 164 + 17 = 27405 = 87x 315 = 0 mod(315)
yes
Thanks Very much, but I think in the last operation of calculating the sum , instead of the last 7 we should write 1.
You are totally right, thanks for catching that!
you, thank you, I understood from your video and I did well in today's exam. good luck.
@@MichaelPennMath you, thank you, I understood from your video and I did well in today's exam. good luck.
You said there weren't really any tricks, but your polynomial for 0 (mod 5) was already factorable into (x+2)(x+1), which give you -2, -1 --> 3, 4. Then for the mod 7 polynomial you could have changed 17 to -4 then factored into (x+4)(x-1) giving you 1, -4 --> 1, 3.
Yeah, but the modulo you are working with is so low it much of a muchness
@@tomatrix7525I don't understand your comment
Any more information on gcd(3,f'(1))? Thank yoy very much.
I did 1 after I realized you had 2^3 combinations and thought fuck it for the other 7... too much work.
Let us ponder the non-UFDness of Z_(315) [X] :
x^2 + 3x + 17 is the product of 4 distinct couples of polynomials
Merci!
Awesome, dear Michael! 🥳🥳🥳
I noticed the error at 07:00 f'(2) = 4 ?? but luckily it resolved itself. ;-) - It is very important for your overall aura and reputation that you don't watch at the cheat-sheets all the time. ;-)
My theory is, that he omitted 3 (mod 3) and just calculated 2 * 2 (mod 3) for the derivative. But it doesn't matter (mod 3), if you take 4 or 7!
@@rainerzufall42 Man, I waited a year for this. ;-)
When the polinomail congruence has no solution؟
Think x^2 = 2 (mod 3). It does not have any solutions, x^2 is only congruent to either 0 or 1. We should be able to make a table and show that there are no solutions where we are working mod a small number. For working mod a large prime, I can't think of a better solution, though.
2. Find all values of x in the congruence x² = q(mod p), where p is a prime number, given
(i) q = 15 p = 967
(ii) * q = 14, p = 941 . how to solve this problem
It's amazing. I tried another calculation using Matlab.
Chinese remainder theorem is lengthy and uneasy to handle.
Instead use the following formula for this problem.
Let x=x(1) mod(5), x=x(2) mod(7) and
x=x(3) mod(9).
x(1)=3,4 x(2)=1,3 x(3)=2,4
Formula:-
x=126x(1)+225x(2)+280x(3) mod(315)
All the 8 solns follow for different values of x(1), x(2) & x(3).
Biceps and now quads!? jk but in all seriousness, thank you so much for making all of these wonderful videos!
Awesome thanks
What is mod
"mod" stands for the modulo operation ( en.wikipedia.org/wiki/Modulo_operation ). I'd highly recommend starting at the beginning of this playlist and watching the videos in order ( ruclips.net/p/PL22w63XsKjqwAgBzVFVqZNMcVKpOOAA7c ).
If we see board clear than may be this video will helpful but unfortunately wording which shows at screen is much disturbing by seen board
This video is for 3 years ago now quality of his video is so much better.