How to solve a non-factorable quadratic congruence

  Рет қаралды 50,054

blackpenredpen

blackpenredpen

Күн бұрын

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@Ni999
@Ni999 5 жыл бұрын
"We can do it because we have this powerful green marker..." _Perfect!_ Seriously, perfect.
@wongbob4813
@wongbob4813 5 жыл бұрын
Do a mod 6 next time!
@thedoublehelix5661
@thedoublehelix5661 4 жыл бұрын
Method #3: Check all the numbers from 0-6 to see if any of them work
@jeremy2719
@jeremy2719 5 жыл бұрын
And i just started learning modular arithmetic today, thank you!!!!
@holyshit922
@holyshit922 5 жыл бұрын
x_{1}=3+7k_{1} x_{2}=2+7k_{2} where k_{1} and k_{2} in Z Calculated mentally and used 2nd version - the factorization
@JM-us3fr
@JM-us3fr 5 жыл бұрын
This is pretty cool! I hope you make more number theory videos
@petermhart
@petermhart 2 жыл бұрын
Very cool! Thanks for posting!
@balancedactguy
@balancedactguy 5 жыл бұрын
Can you give an introduction to TENSORS??
@dectorey7233
@dectorey7233 5 жыл бұрын
I'd love to see BPRP tackle tensors but if you want, Andrew Dotson has a great ongoing series on the topic right now
@balancedactguy
@balancedactguy 5 жыл бұрын
@@dectorey7233 Thank You!
@dwaipayandattaroy9801
@dwaipayandattaroy9801 5 жыл бұрын
Tell honestly that hasnt you mugged up this thing , you are patient , kudos
@jonathanhanon9372
@jonathanhanon9372 3 жыл бұрын
I just did it as x^2 + 2x - 1 = x^2 - 5x + 6 = (x-3)(x-2) mod 7
@Daydreamer-h1t
@Daydreamer-h1t 4 жыл бұрын
your vidoes are so useful
@ericthegreat7805
@ericthegreat7805 2 жыл бұрын
You can also do X2+2x=-1 mod7 = 6 mod7 X2+2x-6=0 mod7 (X-3)(x-2) = 0 mod7 X = 2 mod 7, 3 mod 7
@salmamuhammad5416
@salmamuhammad5416 4 жыл бұрын
Thank you very much, i was stuck in a quadratic congruence like that and you helped me .. all respect
@satoruai3475
@satoruai3475 2 жыл бұрын
I hope there are more of number theory videos. Thanks for the videos😁
@theactualbowmonk
@theactualbowmonk 5 жыл бұрын
So using mod 7 means we can add as many 7s as we want to the righthand side because there is no remainder?
@Domestofobia
@Domestofobia 5 жыл бұрын
yes, in congruences you can replace anything with anything as long as they are im the same /remain group/(not sure if it is called this in english but for example x :=: y (mod n) then you can replace x with y
@ahmedbenlahrech5352
@ahmedbenlahrech5352 3 жыл бұрын
@@Domestofobia Exactly , thats what differentiates modular arithmetics from standard math but the type of problems and questions asked in this field is kinda hard ngl
@jzanimates2352
@jzanimates2352 5 жыл бұрын
Can you try to differentiate x! using the definition of factorial as pi or gamma function plz!!!
@srpenguinbr
@srpenguinbr 5 жыл бұрын
Once I did it and the result was in terms of psi(x) a special function I had never heard about
@maxwongpt2channel328
@maxwongpt2channel328 5 жыл бұрын
@@srpenguinbr That's the digamma function.
@mathforbem
@mathforbem 5 жыл бұрын
Nice😍😍thx
@SmileyMPV
@SmileyMPV 5 жыл бұрын
In general you can actually just use the quadratic formula. For the square root you just have to calculate the quadratic residue. Since the quadratic residues of 2 mod 7 are 3 and 4, we find x=-1+3=2 and x=-1+4. Interestingly, some numbers don't have a quadratic residue mod 7, for example 3. This means that some quadratic equations, such as x^2+2x-2=0mod7, do not have any solutions.
@robertlozyniak3661
@robertlozyniak3661 5 жыл бұрын
Because it is "mod 7", there are only seven possible cases that you need to look at. You could plug in x=0, x=1, and so forth, up to x=6, and see which values of x actually work.
@marcushendriksen8415
@marcushendriksen8415 5 жыл бұрын
Excellent video and thought-provoking material as usual! It's got me wondering about cubic congruences now...
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Marcus Hendriksen It’s just crazier. : )
@andywright8803
@andywright8803 5 жыл бұрын
Are you only restricted to integers in modular arithmetic?
@willnewman9783
@willnewman9783 5 жыл бұрын
Modular arithmetic refers to only doing things in the integers, I believe. You can extend it to all real numbers, but only addition makes sense in this realm.
@WilliamLeeSims
@WilliamLeeSims 5 жыл бұрын
Not unless your solution requires it. For example, 12 mod 6.28 = 5.72. In that example, you have an angle of 12 radians which is one turn around the circle (2 pi) plus another 5.72 radians. Another example is the sawtooth wave function y = x - floor(x); I like to think of it instead as y = x mod 1.
@WildAnimalChannel
@WildAnimalChannel 5 жыл бұрын
I don't know why but I don't like modular arithmetic. It just feels like counting gone wrong!
@helloitsme7553
@helloitsme7553 5 жыл бұрын
Another way: substitute x=7k+c, then (7k+c)^2+2(7k+c)-1=49k^2+14kc+c^2+14k+2c-1=7(7k^2+2kc+2k)+c^2+2c-1 thus we have to find c between 0 and 7 for which c^2+2c-1 is congruent to 0 mod 7. And then check for all those numbers
@reeeeeplease1178
@reeeeeplease1178 Жыл бұрын
Good one! 😂
@amaliacoughlan7071
@amaliacoughlan7071 4 жыл бұрын
Thanks man! That was so cool.
@Quadratic4mula
@Quadratic4mula 5 жыл бұрын
I'm trying to click on the in video, "Video" Link that says, "Chen Lu" and Dr. ¿Peyam? ... where is that video.
@deidara_8598
@deidara_8598 3 жыл бұрын
x^2 + 2x - 1 = 0 (mod 7) x^2 + 2x = 1 mod 7 x + 2 = inv(x) mod 7 15 = 3*5 = 1 mod 7 3 + 2 = inv(3) mod 7 x = 3 mod 7 8 = 2*4 = 1 mod 7 x = 2 mod 7 x = 2 or 3 mod 7 Of course that only works here because we easily found 15 and 8. A more general solution would require the quadratic formula, of course keepin in mind that division would have to be replaced with multiplying with the inverse, and the square root would be calculated with Tonelli-Shanks
@nicholasleclerc1583
@nicholasleclerc1583 5 жыл бұрын
If all this is 0 modulus 7, then that means the answer is a *multiple* (i.e. “k”) of 7, right ? Then: x^2+2x-1=7k x^2+2x-(7k+1)=0 x=-1+/- sqrt(-28k-3) x=-1 +/- i*sqrt(28)*sqrt(k+3/28) So: k
@lonigaming5880
@lonigaming5880 5 жыл бұрын
Great video!
@thetetrix4474
@thetetrix4474 5 жыл бұрын
Juste make a congruence table and take values from 1to7
@Jordan-zk2wd
@Jordan-zk2wd 5 жыл бұрын
x^2+2x+1=2 mod 7 x^2+2x+1=9 mod 7 Doesn't that also mean: x^2+2x+1=16 mod 7 (because 9+7=16). Therefore: (x+1)^2=4^2 mod 7 x+1=4 mod 7 and x+1=-4 mod 7 x=3 mod 7 and x=-5 mod 7 Adding 7 to the last one, we get... Wait x=3 and x=2, same solutions opposite order. Huh. Is it always as easy as checking one case (9 is enough and 16 is unnecessary) for all quadratic equations mod some prime?
@KatzKitz10
@KatzKitz10 5 жыл бұрын
Where i can get this subject? I dont get it in high school
@98danielray
@98danielray 5 жыл бұрын
its anything related to modular algebra. usually given in abstract algebra or number theory
@balancedactguy
@balancedactguy 5 жыл бұрын
You will find this in basic NumberTheory. See if you can find an "Introductory Number Theory" book.
@Rafael-oq9vu
@Rafael-oq9vu 5 жыл бұрын
yep, just go for number theory.
@superjugy
@superjugy 5 жыл бұрын
Oh shit, green pen! This is getting serious now
@gordonchan4801
@gordonchan4801 5 жыл бұрын
5:02 "indi-k"
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Gordon Chan hahhahaha
@mathdiscover
@mathdiscover 5 жыл бұрын
Really good,,, sir
@solomonbirhane3648
@solomonbirhane3648 5 жыл бұрын
you areThe Brilliant one keep it up
@أبوبراء-ظ5ي
@أبوبراء-ظ5ي 3 жыл бұрын
Very good
@AaronHe
@AaronHe 5 жыл бұрын
You should do geometry.
@BeauBreedlove
@BeauBreedlove 5 жыл бұрын
You could also change it to x^2+2x-15 and factor that to (x-3)(x+5) to get x=3 or x=2 Or you can factor the original x^2+2x-1 to (x-2)(x-3) knowing that -2 * -3 is congruent to -1 (mod 7) and -2 - 3 is congruent to 2 (mod 7)
@Sesquipedalia
@Sesquipedalia Жыл бұрын
IS THIS ALLOWED??
@radouaniabdelhadi332
@radouaniabdelhadi332 5 жыл бұрын
Very nice
@DarrenMcStravick
@DarrenMcStravick 5 жыл бұрын
Bro that intro was so gangsta my supreme boi
@LS-Moto
@LS-Moto 5 жыл бұрын
Goes to show how cool math really is
@shandyverdyo7688
@shandyverdyo7688 5 жыл бұрын
What about some integrals again?
@swarnakshi_official8533
@swarnakshi_official8533 5 жыл бұрын
Sir what is the answer of X^2 congruent 27(mod59)
@hassanalihusseini1717
@hassanalihusseini1717 5 жыл бұрын
If you have x^2+2x-1==0 mod 7 it is much easier to try all possible reminders: 0: 6 1: 2 2: 0 This is a solution 3: 0 This is also a solution. 4: 2 5: 6 6: 5
@DeepakKumar-qv1zb
@DeepakKumar-qv1zb 5 жыл бұрын
Good bro keep on
@thebloxxer22
@thebloxxer22 5 жыл бұрын
From Algebra I to Either an advanced form of Calculus or Pre-Calc.
@98danielray
@98danielray 5 жыл бұрын
advanced?
@xcalibur6482
@xcalibur6482 5 жыл бұрын
How can I send you a problem?🤔
@srpenguinbr
@srpenguinbr 5 жыл бұрын
Maybe you could that the quadratic is equal to 7k and use the quadratic formula. Then, use those formulas for the pithagorean triples and find what values for k give an integer solution to x
@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ
@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ 3 жыл бұрын
or just complete the square lol. x^2+2x+1-2=0 x^2+2x+1= 2 (x+1)^2 = 2 x+1 = sqrt(2) x sqrt(2)-1 OOOOR x+1= -sqrt(2) x = -sqrt(2) - 1. or just use the determinant lol
@dwaipayandattaroy9801
@dwaipayandattaroy9801 5 жыл бұрын
1 step seems forced logic like nah ? But solvable a+b whole sq , what training impact on mind
@hero947
@hero947 5 жыл бұрын
@Blackpenredpen you can solve this : ∫1/(e^x + e^-x +1) dx ?
@christiansmakingmusic777
@christiansmakingmusic777 3 жыл бұрын
You can’t take the square root unless it is a quadratic residue. Only half the non-zero residues are quadratic residues. 1,2,4 are the quadratic residues, 3,5,6 have no square root over the finite field of integers modulo seven.
@wojtek9395
@wojtek9395 5 жыл бұрын
What about non prime numbers in place of 7?
@ivan1793
@ivan1793 5 жыл бұрын
That's a very general question. And my very general answer would be: use the Chinese remainder theorem.
@呂永志-x7o
@呂永志-x7o 5 жыл бұрын
可能加了別的倍數也是解,怎麼證明這就是全部的解?
@muhammadqasim7056
@muhammadqasim7056 5 жыл бұрын
Video on cycloids next?
@nicolassamanez6590
@nicolassamanez6590 5 жыл бұрын
but what about (x+1)^2=16mod7? since 16mod7=2mod7
@skallos_
@skallos_ 5 жыл бұрын
That will give you x+1=+-4 (mod 7). x=3,-5 (mod 7). x=2,3 (mod 7).
@Theraot
@Theraot 5 жыл бұрын
Ends up the same, even if you take (x+1)^2 = 2+7*14 (mod 7) or any other square of the form 2+7n. Note: 2+7*14 = 100. Proof? Hmm... This is what I have been able to come up with: Any other square you can build form adding 7 multiple times to 9 will be 2+7+7m where m is positive integer... And must be a square, at least equal to 9 = 3^2. Thus, 2+7+7m = 9 + 7m = (3 + k)^2 => 9 + 7m = (3 + k)^2 => 9 + 7m = 3^2 + 2(3)k + k^2 => 9 + 7m = 9 + 6k + k^2 => 7m = 6k + k^2 Given that 6 = 2 * 3, 6 does not share factors with 7, thus this factor must come from m.... 7a +7(6)b = 6+k^2 => 6 = 7(6)b 7a = k^2 Given that 7a must be a square, therefore a must be a multiple of 7, in fact, it must be a square by 7: a = 7c^2 => 7(7c^2) = k^2 => 7c = k Or simply, given that we know that 6k = 7(6)b => k = 7b We conclude that k must be a multiple of 7, in other words k = 0 (mod 7) Alright, go back to (x+1)^2 = 2 (mod 7) knowing that 7m = 0 (mod 7) and 7m = 6k + k^2 => (x+1)^2 = 2 + 7 (mod 7) => (x+1)^2 = 2 + 7 + 7m (mod 7) => (x+1)^2 = 2 + 7 + 6k + k^2 (mod 7) => (x+1)^2 = 9 + 6k + k^2 (mod 7) => (x+1)^2 = 9 + 2(3)k+k^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => x + 1 = 3 + k (mod 7), x + 1 = -3 - k (mod 7) => x = 2 + k (mod 7), x = -4 - k (mod 7) And since we know that k = 0 (mod 7): x = 2 + k (mod 7), x = -4 - k (mod 7) => x = 2 (mod 7), x = -4 (mod 7) => x = 2 (mod 7), x = 3 (mod 7) That is the same result on the video (4:36), this shows that the selection of square is irrelevant for this case.
@conorbrennan5838
@conorbrennan5838 5 жыл бұрын
Can you integrate cos (x^2) next , hint : use taylor series
@drpeyam
@drpeyam 5 жыл бұрын
That intro 😂
@newtonnewtonnewton1587
@newtonnewtonnewton1587 5 жыл бұрын
Nice subjct thanks a lot for u
@snigdhasahoo9952
@snigdhasahoo9952 4 жыл бұрын
If there is any coefficient in x² then what should to do ???
@ZipplyZane
@ZipplyZane Жыл бұрын
Then you divide all the terms by the the coefficient. Since you divide by the same thing on both sides, the equation remains valid, and thus it doesn't mess up the value of x. So, say, 3x²+4x=-3 becomes x²+(4/3)x=-1
@lakhwinderhairtransplantco6803
@lakhwinderhairtransplantco6803 5 жыл бұрын
I have some problems in 3 dimensional geometry
@mryip06
@mryip06 4 жыл бұрын
Amazing
@scimaniac
@scimaniac 5 жыл бұрын
What if it is even?
@onlystudy5733
@onlystudy5733 5 жыл бұрын
Even means not prime so...
@josephhtoo1
@josephhtoo1 5 жыл бұрын
Can you do a day in my life?
@thomaswilliams9320
@thomaswilliams9320 5 жыл бұрын
You should run in 2020
@UrViridescentLeaf
@UrViridescentLeaf 5 жыл бұрын
👍
@ssdd9911
@ssdd9911 5 жыл бұрын
doraemon in the middle of the video
@BigDBrian
@BigDBrian 5 жыл бұрын
so... what do you do if it isn't prime?
@blackpenredpen
@blackpenredpen 5 жыл бұрын
mrBorkD I have that video in the video already. It’s unlisted so be the first few to watch!! : )
@BigDBrian
@BigDBrian 5 жыл бұрын
@@blackpenredpen oh cool, thanks
@diederickfloor4261
@diederickfloor4261 4 жыл бұрын
why doesn't this work: x^2 + 2x - 1 cong. 0 mod 7 x^2 + 2x cong. 1 mod 7 x(x + 2) cong. 1 mod 7 x cong. 1 mod 7 x + 2 cong 1 mod 7 x cong. 6 mod 7 edit: I know line 4 is not justified you can only split the factors with 0 mod n.
@johnariessarza3622
@johnariessarza3622 5 жыл бұрын
Make a video of mod is even..
@blackpenredpen
@blackpenredpen 5 жыл бұрын
John Aries Sarza I have another video when mod is composite. See description
@johnariessarza3622
@johnariessarza3622 5 жыл бұрын
blackpenredpen thanks bprp.. hope i can meet you personally. Always watching your tutorial blog, thanks.
@blackpenredpen
@blackpenredpen 5 жыл бұрын
@@johnariessarza3622 : ))))) Thank you!!
@rogerkearns8094
@rogerkearns8094 5 жыл бұрын
Up the mods.
@RexxSchneider
@RexxSchneider 3 жыл бұрын
At 4:32 "Two answers - that's it, right?" Not really. You failed to consider that 4^2 is also congruent to 2 mod 7. Now, it turns out that we will end up with the same values for x, but how is the viewer to know that will always be the case? It is in fact true that quadratic residues (mod p) occur in pairs in Z/pZ, if p is an odd prime, but that also isn't obvious to the viewer, and you haven't proven it. You can also use the quadratic formula directly to find solutions by changing from a congruence to an equality with an extra (integer) variable: x^2 + 2x - 1 ≡ 0 mod 7 is equivalent to x^2 + 2x - 1 = 7n where n is an integer. So x^2 + 2x - (1 + 7n) = 0 and we can use the quadratic formula: x = (-2 ± √(4 + 4(1+7n)))/2 = -1 ± √(1 + 1+7n) after diving top and bottom by 2 x = -1 ± √(2+7n) but we require integer solutions, so the term inside the square root must be a perfect square, which we can call m^2 x = -1 ± m where m^2 = 2 + 7n. The condition m^2 = 2 + 7n is equivalent to m^2 ≡ 2 mod 7 and the only values of m that have quadratic residues of 2 mod 7 are m=3 and m=4 mod 7. So either m = 3 + 7k or m = 4 + 7k where k is an integer. That gives x = -1 ± (3 + 7k), producing x = 2 + 7k or x = -4 - 7k; or x = -1 ± (4 + 7k), producing x = 3 + 7k or x = -5 - 7k. But -4 - 7k gives the same set of numbers as 3 + 7k does, and -5 - 7k gives the same set of numbers as 2 + 7k does, as k takes on all positive and negative values. The complete set of x values may therefore be summarised as 2 + 7k and 3 + 7k.
@isaacmojica3463
@isaacmojica3463 4 жыл бұрын
Dude you're very awsome. I was looking this kind of way s¡to solve that equiation for hours an nothing: Ty. This is a beautiful simple way and easy to understand :D
@رياضياتتوجيهيوجامعةأ.عبدالرحمن
@رياضياتتوجيهيوجامعةأ.عبدالرحمن 4 жыл бұрын
When the congruence has no solution ؟
@garavelustagaravelusta9717
@garavelustagaravelusta9717 4 жыл бұрын
Better Solution: x^2 + 2x = 1mod7 x(x+2) = 1 mod7 (x+2) = (1/x)mod7 Find pair elements in mod7 such that their multiplication is equal to 1 in mod7. (Hence they form a pair of x and 1/x) Pair 1: 1 1 x+2 = 1mod7 and x = 1mod7 ? No solution for the system! Pair 2: 2 4 (8 = 1mod7) Option 1: x = 2mod7 and x+2 = 4mod7 x = 2mod7 (Valid) Option 2: x = 4mod7 and x+2 = 2mod7 -> x=0mod7 (Not Valid!) Pair 3: 3 5 (15 = 1mod7) Option 1: x = 3mod7 and x+2 = 5mod7 x = 3mod7 (Valid) Option 2: x = 5mod7 and x+2 = 3mod7 -> x = 1mod7 (Not Valid!)
@jabir5768
@jabir5768 5 жыл бұрын
How to be good at maths: 1) Be crazy about maths 2)There is no number 2 3)There is no number 3 ... n)There is no number n Pls validate my worthless existence
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Jabir Fatah I like it!!!
@carftinginshakthi5720
@carftinginshakthi5720 5 жыл бұрын
BRO PLZ SPEAK LITTLE BIT LOUDER
@carftinginshakthi5720
@carftinginshakthi5720 5 жыл бұрын
SORRY BRO MY EAR PHONE HAD A PROBLEM BY THE WAY YOU ARE A BI..CH
@gauravsuyal4456
@gauravsuyal4456 5 жыл бұрын
1st like
@ttaullah5317
@ttaullah5317 5 жыл бұрын
I think 2nd
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