How to solve a quadratic congruence when the modulus is NOT prime

[@michaelz2270]

A little shortcut: if x works so does -x so you only need to figure out two of them.

[@blackpenredpen]

Ahhhh that’s right!!! For other viewers: I got 46 for x that means -46 will also work. I.e 9 mod 55

[@ffggddss]

@@blackpenredpen Yes, and the above insight means you can list the residue classes mod 55, as -27, ..., 0, ..., +27; and then you need check only 0, ..., +27. Fred

[@BrikaEXE]

Whoa thats a lot of damage

[@bikashchandraroy2481]

I have a formula for its solutions. It takes only three minutes to get all the four solutions.

[@bikashchandraroy2481]

There is no use of CRT.

[@xardasnecromancer590]

6:46 You can also do it this way: 5k Ξ 1 (mod 11) 5k Ξ -10 (mod 11) k Ξ -2 (mod 11) k Ξ 9 (mod 11)

[@suniltshegaonkar7809]

I did the same way

[@LordArrack1]

You can also always check yourself, since opposite pairs must add to 55. 46 (from 1,2) + 9 (from 4,9) = 55 31 (from 1,9) + 24 (from 4,2) = 55

[@surrealistidealist]

I'm trying to teach myself Mathematics after being out of school for a decade. Your enthusiasm and joy is so inspiring, and it's the only thing that makes me feel less alone as I struggle to establish my foundations. Thank you so much!

[@blackpenredpen]

Surrealist Idealist thank you!!!!

[@surrealistidealist]

@@blackpenredpen Thank YOU!

[@henryh.448]

371 likes, 0 dislikes...on a math video. I think you broke the record for youtube. Great job. The only downside? The ratio is undefined!

[@喵帕斯-大萝卜鸡]

Human fears something they don’t know or they don’t understand, and they respect that.

[@beiyi6933]

Just graduated from university and thank you for maintaining my interest in math 😂

[@blackpenredpen]

Bei Yi yay! Glad to hear!!

[@GenerationXSwag]

In* maths. Practise ur english as well😀

[@MrConverse]

My module arithmetic is nowhere near as strong as I want it to be. Keep these videos coming!!

[@ZectonplaysMC]

Yayy! I love modular arithmetics please do more of these! I'm learning RSA for cryptography and i really need to learn mod

[@JamesMatimela]

5 Years Have Gone By , And People Are You Winning

[@pilotomeuepiculiares3017]

I never had a class of number theory or studied by myself and I was able to get the video. You are awesome!

[@RexxSchneider]

At 6:00 "Or, you can just do it in your head", If you're going to do it in your head, you can just check each integer that's congruent to 2 *or* 9 (mod 11) and less than 55. Start with { 2, 9 } and keep adding 11. There are just ten of them: { 2, 9, 13, 20, 24, 31, 35, 42, 46, 53 } Reject any that are not congruent to 1 *or* 4 (mod 5), which is easy because you only need examine the final digit. Then you have the list of four solutions: { 9, 24, 31, 46 } in one pass. Note that they come in pairs, summing to 55, because if x works, then 55-x also works, so you can get all four solutions just by testing up to 27 "in your head".

[@anonthief]

You could solve it by rewriting x^2 Ξ 26 (mod 55) as 1= 26(x^-1)^2 (mod 55) this means all you have to do is find the inverse of 26 mod 55 and take the square root 26^-1 Ξ 36 (mod 55) √36 is 6 now, 6 is x^-1 so, the inverse of 6 (mod 55) will be x 6^-1 Ξ 46 mod 55 46 is a solution for x

[@tracyh5751]

By doing this you have assumed that 6 is the only square root of 6(this is not true) And that x is not divisible by the factors of 55(which is true in this example, but not in general).

[@bikashchandraroy2481]

I have a formula to find all the four solutions.

[@zwatotem]

I'm really excited about these modular algebra videos. Actually, I never had occasion to learn this section of mathematics. I hope for more.

[@blackpenredpen]

ɯǝʇoʇɐʍZ thanks!!!!

[@disguisedhell]

Ya, really they are not taught in schools but they come handy in Olympiad maths

[@blackpenredpen]

@Matthew Hutchinson An application is called the "cryptography" look it up!

[@disguisedhell]

@Matthew Hutchinson I just said that modular arithmetic is almost left unexplored in school mathematics. It will be really helpful if he posts more over this topic.

[@disguisedhell]

@@blackpenredpen one suggestion: how to choose correct modulus in proving no solutions

[@doctorscalling9479]

Learned much more from this video than the lecture and books I read combined.

[@jonathanspilhaus3165]

This has been brilliant: I have really struggled to get to grips with this. Thank you very much!

[@schrodingerbracat2927]

Note that if u = 11a + 45b, then u=a (mod 5) and b (mod 11). Plug in a=±1, b=±2 gives all the solutions (mod 55).

[@snnwstt]

4:09: Easier without the Chinese Reminder Theorem like this: eq. 1 X = 1 mod 5 eq. 2 X = 2 mod 11 Multiply each equation by the modus of the other equation, that is 11 * eq.1 and 5 * eq 2: 11 X = 11 mod 55 ( the mod is now 55, check it is ok using the definition of modulus is not convinced) 5 X = 10 mod 55 We can now add/subtract since they both share the same modus: 6 X = 1 mod 55 46 is the inverse of 6 in mod 55 ( check 1 == 46*6), so, (46*6) X == (1) X = 46 * 1 mod 55 => X = 46, as you found, ... later on, with much more computations.

[@rwex1]

Hi, you did i to the power of i, can you do i to the superpower of i? I hope you reply and make video about it please.

[@oledakaajel]

Does the maths for that even exist yet?

[@ffggddss]

Alas, AFAIK, superpowers are defined only for integer "superexponents." Fred

[@rwex1]

@@oledakaajel Yes.

[@marcushendriksen8415]

So you mean something like tetration, but with complex exponents?

[@rwex1]

@@marcushendriksen8415 Yes exactly.

[@wkingston1248]

Can you use linear algebra to solve system of congruences like normal systems of equations?

[@willnewman9783]

Yes, if you are working mod a prime number

[@snnwstt]

And be careful with the a mod equation equals to 0. 3X = 0 mod 6 has THREE solutions, X= 0 , X =2, X=4 (the gcd(3, 6) = 3 and divides 0, so 3 solutions. To get a first solution, divide by the gcd and then, solve it, here: X = 0 mod 2; add the new modus, here 2, pgcd -1 times to get the other solutions.). This way, you can also solve AX=B mod C if the gcd(A, C) divides B, for a general case and there is no solution if: B mod C / gdc(A, C) as in 2X = 3 mod 10. Sure, with B = 0, then there is always at least one solution, X=0. Note: I assumed that we used: (A*X) | C == (A|C) * (X|C) mod C that is, that the involved A is less than C.

[@LS-Moto]

The moment you stood still and stopped talking, I tought my video has stopped loading🤣

[@blackpenredpen]

Л.С. Мото hahahhaha that silence!!

[@LS-Moto]

@@blackpenredpen its the power of silence :)

[@aashsyed1277]

@@blackpenredpen he is lars

[@PMA_ReginaldBoscoG]

Create a Cayley's table for multiplication modulo 55; Take out the diagonal elements where you get 26; Ta da! You've got your integers.

[@Green_Eclipse]

Wow I thought you meant a system of 4 equations but its actually a system of 4 systems of 2 equations. Really cool.

[@blackpenredpen]

lol yea : )))

[@MoonLight-sw6pc]

I just finished ur 100 integrals video ! No 85 integral was hilarious ! Good job as always ! !!!!!!!

[@blackpenredpen]

Moon Light thank you!!!!!

[@RexxSchneider]

If you're going to do it in your head, you might as well deduce that x is of the form 5n +/- 1 *and* 11m +/- 2. You still get the four cases, but they are a little easier to hold in your head and to see the solutions. For example, when x = 5n+1 = 11m+2, you get n = (11m+1)/5 and you then see m=4 gives integer n=9, so x = 46. Knowing that all four solutions repeat at multiples of 55 immediately gives you -9 and another solution; hence x=9 is also a solution, and you're half-way there. Similarly taking x = 5n+1 = 11m-2 it becomes n = (11m-3)/5 yielding m=3, n=6 and x=31. Knowing x=31, you get x=-24 which implies x=24 as the fourth solution in the range 0 to 55. Of course, if you're not going to do it in your head, you can just use a computer anyway.

[@Daydreamer-h1t]

Sir, do you make tutorials on advance theory of numbers??

[@Tehom1]

Chinese Remainder Theorem FTW!

[@ffggddss]

IOW, 55k + 26 = x², that is, a square. To check x, we need only check 0, ..., 27, because the rest of the possible residues are ≡ -27, ..., -1, and when squared, these give the same values as the positive group. 0...5 are too small (x² < 26). The first "hit" is 9² = 81 = 55 + 26. To find another x, we could use x² - 9² = (x-9)(x+9) = 55k and so, the factors on the left have to have a 5 and an 11 between the two of them. And we need consider only 9 < x ≤ 27; i.e., 0 < x-9 ≤ 18. So we try x-9 = {5, 10, 11, 15}, and see whether x+9 = (x-9) + 18 = {23, 28, 29, 33} can provide the missing factor. The "hit" then, is x = 24. So there are four solutions: x ≡ {±9, ±24} ≡ {9, 24, 31, 46}, to put the residues back into the 0...54 range. If instead, we start checking k-values, we need check only as long as 55k + 26 ≤ 27² = 729. We get: k = 0: 26 ≠ ⧠ k = 1: 81 = (±9)² k = 2: 136 ≠ ⧠ k = 3: 191 ≠ ⧠ k = 4: 246 ≠ ⧠ k = 5: 301 ≠ ⧠ k = 6: 356 ≠ ⧠ k = 7: 411 ≠ ⧠ k = 8: 466 ≠ ⧠ k = 9: 521 ≠ ⧠ k=10: 576 = (±24)² k=11: 631 ≠ ⧠ k=12: 686 ≠ ⧠ and again, only x ≡ ±9 and ±24 work. Fred

[@RonaldoEuSi]

0 dislikes! people who hate math hate it so much they dont even bother comin and disliking hahah

[@vitalsbat2310]

x^2 congruent 11 (mod 96) will make you go crazy

[@jaspergodfreyzann5421]

My favourite math youtuber!

[@blackpenredpen]

Guiseppe Pizzaro thank you!!

[@aashsyed1277]

@@blackpenredpen same for me!!!!!!!!!!!!!!!

[@ramaprasadghosh717]

x^2 -26 is divisible by both 5 & 11. Hereby x^2 modulo 5 = 1 and x^2 modulo 11 = 4 chinese remainder theorem solves the remaining

[@SmileyMPV]

Here is a fun exercise: Find all x such that there exist y=5mod8 and z=2mod4 such that x=yz.

[@blackpenredpen]

SmileyMPV well, we can just solve for y and z first then multiply the results.

[@SmileyMPV]

@@blackpenredpen But can you find a modular epression for x? In general, for all a,A,b,B there exist c,C such that x=c mod C if and only if x=yz for some y=a mod A and some z=b mod B. Some other interesting examples are (y=6mod8, z=9mod12) and (y=5mod25, z=6mod10). The ultimate goal would be a general formula for c and C, which does exist!

[@blackpenredpen]

SmileyMPV hmmm I will have to think about this.

[@SmileyMPV]

​Hm, I must admit that I made a mistake. My formula actually gives the largest C such that x=yz => x=ab(modC), however we do not have x=ab(modC) => x=yz for some y,z. Unfortunately we can prove the if and only if to be impossible in some scenarios. However, the question of finding this largest C is still interesting. In the case (y=5mod8, z=2mod4) you actually get C=4, so x=2mod4. However, x can not be 2. But since x can be both 6 and 10, there is no C larger than 4. Exactly which x can be made by x=yz with y=5mod8 and z=2mod4 actually looks really interesting, as I can not find any patterns. Here are all x below 200. 6 10 18 22 26 30 38 42 50 54 58 66 70 74 78 86 90 102 106 110 114 118 122 126 130 134 138 150 154 162 166 170 174 182 186 190 198 Additionally, we can ask ourselves when the if and only if actually holds. I have no progress on this yet.

[@rcuster93]

@@SmileyMPV Really, I'm pretty sure that there's not a nice congruence for the example you gave. You could describe the solution set, though: x={32mn+16m+20n+10 | m,n are integers}. You could examine this in multiple mods, but I don't think any of those mods will account for those values 2 mod 4 that x cannot take. You could have x=0 mod 2, x=2 mod 4, x=4n+2 mod 8, x=4n+10 mod 16, and x=16m+20n+10 mod 32. I think that's the closest you could get to a nice answer.

[@MathNerd1729]

Dang it! I thought it was a new video, yet I clicked the link way earlier! Very funny, KZbin notifications!

[@skorpion0325]

So how about modulo with multiple divisors? Like 72. How can I deal with it? Thank you!

[@Yoshimaster96smwc]

I just noticed, 46+9=31+24=55, which is the modulus. I'm curious, will this always happen for mods which are a product of exactly two primes?

[@050138]

It is nothing to do with product of two primes or three primes.... It is because we have X² modulo something, and you have to solve for X modulo the same number.... For any X, which the congruence holds, it has to hold for -X as well.... So if 46 is a solution, as X = 46(mod 55), -46 is also a solution, but -46 modulo 55 is same as 9 modulo 55 (as 9 = 55 - 46)

[@deidara_8598]

This one was easier than it looked So i simply factor 55 = 5*11 Now let's check some roots 26 = 1 mod 5 sqrt(26) = 1 or 4 mod 5 26 = 4 mod 11 sqrt(26) = 2 or 9 mod 11 So we have our candidates: 1, 4, 2, and 9 Only one that squares to anything greater than 26 is 9, and indeed 9^2 = 26 (mod 55)

[@bird9436]

can someone plz explain what mod is support to mean? and the equal sign with 3 stripes?

[@paulfoss5385]

A - The equal sign is used when both sides of the equation are identical. When doing modular arithmetic though you're interested in when different numbers have the same remainder when divided by some number called the modulus. For example 8 does not equal 15, so we don't want to write "8=15", but they have the same remainder when divided by 7, and we can work out arithmetic properties 8 and 15 must share with respect to multiples of 7. Same thing happens in geometry, we only say two polygons/lines/circles/whatever are equal if they are literally the same thing in the exact same position. If they are the same size and shape but positioned elsewhere in the space they are said to be congruent so we don't open ourselves up to errors that could be derived from saying that two different things are the same thing.

[@ayush.kumar.13907]

mod means modulo. (mod b) means remainder when divided by b. The Triple sign stands for congruent which is sort of like equality in modular mathematics.

[@scarbotheblacksheep9520]

It's like if you had numbers mod 11, then you would count 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 etc.

[@faizahbegum5257]

love your energy! makes me love maths

[@janernn]

8:47 Think about this a little bit. Hm.. ... ... ... ... 36 WORKS!

[@si48690]

Which book to refer to for questions like these and more advanced Number theory?

[@philippelepilote7946]

A reference : An Introduction to the Theory of Numbers - by G.H. Hardy, E.M. Wright

[@si48690]

@@philippelepilote7946 Thanks Sir

[@swarnakshi_official8533]

Sir what is the answer of X^2 congruent 27(mod59)

[@bikashchandraroy2481]

The solutions are 26 & 33. As 59 is prime, the congruence has exactly two solutions. The middle-pair solution : 29, 30. Corresponding Congruence:X^2 congruent 15 (mod59). 27-15=12=3.4;Then Required solutions: 29-3=26; 30+3=33.

[@bikashchandraroy2481]

I think Use of Roy Formulation is the best option to find the solutions in a short time. it is time-saving.

[@aflah7572]

Something seems wrong video is shown to be uploaded 15 minutes ago and has comments shown to be 4 days ago. BTW love the vids!! Keep Up the good work

[@blackpenredpen]

Aflah 786 thanks!!!

[@aflah7572]

Are you planning on going up in a battle with Sophomore's dream integral?@@blackpenredpen

[@philkeyouz2157]

Little problem, how do you solve : X^2 = 14 mod 55 and Y^2 = 14 mod 55 with X + Y = 55 . Note that if X^2=55k+14 and Y^2=55l+14 then X-Y=k-l.

[@mateszabo5487]

Dear balckpenredpen!Can u help me solving a task?We search a function,f:R-->R,and this function have 2 attribution:1,if x1≠ x2 => f(x1)≠ f(x2) 2,exist a,b>0 constants wihch: f(x^2)-(f(ax+b))^2 ≥1/4 in every x(x ∈ R) f(x)=?(if exist this type function)

[@alexdemoura9972]

x^2 [C=] 26 (mod 55) the first one is: 55 + 26 = 81 ... x^2 = 81 ... x = 9; All numbers satisfying this congruence have the form: 55n + 26 , n = 1, 2, 3... but not all of them are perfect square numbers. n = 2 --> 136 , n = 3 --> 191 , n = 4 --> 246 , n = 5 --> 301 , n = 6 --> 356 , n = 7 --> 411 , n = 8 --> 466 , n = 9 --> 521 , And other perfect square: n = 10 --> 55 x 10 + 26 = 576 ... x^2 = 576 ... x = 24 ;

[@呂永志-x7o]

31+24=46+9=55應該不是巧合

[@blackpenredpen]

呂永志 沒錯 可以看看pinned comment.

[@meve5918]

(Sorry, I don't read Chinese but I did wonder about 31+24=46+9=55.

[@saatvik-agrawal]

Absolutely Gobsmacking content! Greetings from the middle East

[@borizoinamn8045]

Sir can u please do some videos for vector also . Please

[@ליאורו-ט3ו]

very interesting love your videos

[@thejoltmaster1231]

This is way easier than usual , solved it in less than 30 seconds

[@blackpenredpen]

Wanna share and show all your steps?

[@thejoltmaster1231]

@@blackpenredpen for some reason i keep responding but it doesn't appear Well i have commented it down in the comments how i solved

[@blackpenredpen]

THE JOLT MASTER MC5 your comment is prob too long so YT hides for now. I will have to check later on when I use my laptop.

[@Edwin-wn3ss]

What happens if x^2 congruent to 3 or something other than 1/4/9/16?

[@sergiogarofoli573]

How do I find X when it's on the (modX) part of the sentence? Answers pelase!!!

[@pyrosmantia8117]

thank you so much

[@davidseed2939]

ths is how i did it. x^2 = 55n+ 26 try n=1 a solution x^2 = 81 x=9 so now what other solutions. x^2 = 9^2 + 2.9. m + m^2 ie (18+m)m =55k try m=22 k=12

[@rexsybimatrimawahyu3292]

Can we say X=9? Since X²=26(mod 55), just add 55+26=81 then we square it?√81=9 so 9²=26(mod 55)?

[@milindsawant1895]

I have one question, in Euler's identity (e^iπ = -1), if I take logarithm on both sides, then I will end up with iπ =1/e. Is it true?

[@_P_a_o_l_o_]

It is not true: iπ is a purely imaginary, while 1/e is real. So they cannot be equal. The error in your calculation is that ln(-1) is NOT 1/e. You probably got confused with the fact that ln(1/e)=-1. Hope this helps

[@milindsawant1895]

Yeah I didn't notice it. Thank you very much

[@royformulation2161]

Hi, I think formulation of solutions will work efficiently. It will be time & labour saving.

[@sabriath]

missed opportunity to explain fermat's last theorem to find the multiplicative inverse over modulus in prime. the inverse of 'a' over 'mod p' is 'a^(p-2) mod p'.....so when you were doing 5k = 1 mod 11, you need to "divide by 5" so you have to find the multiplicative inverse of 5. 5^9 is 1953125, mod 11 is 9....yes it's the same answer you got, but you "used your head" on an extremely small prime modulus, it becomes extremely more difficult as the prime increases, so to know the formula is good. To work out on paper a^b for large values, you can also do this easily with a binary representation. For example, 5^9 can be done like this: 5 (mod 11) 5*5 = 25 = 3 (mod 11) 3*3 = 9 (mod 11) 9*9 = 81 = 4 (mod 11) 4*4 = 16 = 5 (mod 11) repeated 5*5 and so on from here (as you can see, each new row is the previous mod answer squared and reduced) binary valuation is

[@estebanrodriguez5409]

Now you are ready to play the 999 games...

[@vigneshsrinivasan9692]

why does he solve 4 linear congruences? what is the logic behind doing so?

[@aranyaroy]

What is happening ? I can't seem to understand, I know what congruency mean and what modulus mean(well not that well, my modulus is screwed little bit due to my coding background) so what mod world and congruenc means ? Ty :D

[@BrikaEXE]

Mod world is just what remains from a number mod another number eg: 9 in the mod 7 world is just 2 , bc 7×1 +2= 9

[@BrikaEXE]

Congruence is a writing that links the remainder of a number with the number itself

[@devvratbani5209]

I just learned about this today. How is this coincidence possible???

[@blackpenredpen]

Darth Vader hahaha nice!!!!

[@husklyman]

Can you solve for x? AΞB (mod x)

[@wesleydeng71]

Because 26+55 = 81, I saw 9 right away.

[@lionbryce10101]

(before I watch) x^2 = 26 mod 55, 26/55 = 0 R 26, so 26 mod 55 = 26 Therefore x = + or - sqrt(26) (after) wtf? (google) oh, notation differences I see why this is weird. I hate that notation (I program so I'm use to 5 % 7 = x)

[@alexanfung]

31 works means -31=24 works

[@amitavasarkar7922]

Why 45k is equal to k in 1st case

[@philippenachtergal6077]

For more generic case with a congruence that has many factors, search youtube (or elsewhere) for "chinese remainder theorem" kzbin.info?search_query=chinese+remainder+theorem

[@Patapom3]

Amazing!

[@milindsawant1895]

Nice video

[@thejoltmaster1231]

I have a way easier method X^2 =- 26 [55] We add 56 to 26 it doesn't change anything That become : X^2 =- 81[55] Yes that's what you are thinking right now That become X=- 9[55] or X=- -9[55]

[@blackpenredpen]

LOL, yes. But you would miss other solutions. I did that in the previous video on how to "produce a square"

[@thejoltmaster1231]

@@blackpenredpen aren't there only 2 solutions ?

[@thejoltmaster1231]

Oh wait you can add 55 to have other solutions

[@RonaldoEuSi]

do you think proofs are too long, takes too much time and gather to low views to make? man there are bout no good proof video series on youtube, i think that would be great, nobody has done that until now, new market, mb pick it up, although i understand it takes alot more prep that just examples, anyways, great work man!

[@blackpenredpen]

Thank you!!

[@ליאורמלכה-ע9ש]

can't you just add 26+55, see it is equals to 81 which has 9 as it's squre root, so this is the answer?

[@thedoublehelix5661]

Try doing a quadratic congruence in the mod 2310 world :)

[@Oskar-zt9dc]

cant i just say x=+-sqrt(n*55+26) for n element of Z

[@dxgiang.6]

The title can be called Quadratic Congruence Modulo a Composite? Am I right?

[@rahulchowdhury7635]

I have question plzz solve it

[@nimmira]

Wouldn't it be easier to write the formula as: X mod Y Ξ Z (or even better, invent a symbol for "mod" altogether). Flipping back and forth makes me lose the way to my mouth, and brain :/

[@toufikguenane]

x^3 mod 187 =13 i've got a problem with this can anyone solve it?

[@JoshuaHillerup]

I'm still unclear how I would do say x^2 congruent to 8 (mod 32), even with me writing programs on a computer.

[@someoneuppingdudetechnical6320]

x^2 % 32 = 8

[@alkankondo89]

HOW IN THE WORLD were you coming up with those answers at the end?? Did you just memorize them, or did you seriously just perform lightning-fast guess-and-check?

[@iabervon]

It's easy to come up with the numbers that are 1 mod 5; they end in 1 or 6. Numbers that are 2 mod 11 have the ones digit two more than the tens digit. The other remainders work similarly, with a bit of coping with carries on the 11.

[@blackpenredpen]

Kinda just guess and check. : )

[@blackpenredpen]

Except for 9 tho, notice since 46 works, so -46 would also work, which is the same as 9.

[@robertcotton8481]

Since one was mod 5 other mod 11 if he just adds 11 on the mode 11 one he only had 5 to check before they repeat

[@ssdd9911]

yay

[@SartajKhan-jg3nz]

Here is a calc question: Wtf is the derivative of x! ??

[@SmileyMPV]

The derivative of Gamma(x) is Gamma(x)Digamma(x) by definition of Digamma(x), so the derivative of x!=Gamma(x+1) is Gamma(x+1)Digamma(x+1). You can also use Leibniz integral rule on the integral expression of Gamma(x) to find an integral expression of its derivative.

[@johnariessarza3622]

Notice me sensei...

[@patemafang8588]

Chinese theorem

[@saatvik-agrawal]

Dammmm no dislikes

[@markcarranza2032]

❤

[@yrret1996]

its 'mod' not 'mawd'

[@mahmoodmawed4347]

Stop Asian hate

[@bustdooms2638]

does this comment deserve a like??

[@blackpenredpen]

bust dooms sure!!

[@bustdooms2638]

@@blackpenredpen thanks

[@MrRyanroberson1]

555 likes

[@yangleleo8333]

哥我咋感觉你一直在骂人啊😂

[@thetetrix4474]

Stop doing arithmetics ur bad at them just stick to calculus man. Really.

[@blackpenredpen]

The Tetrix Stop commenting ur bad at them just stick to Pokemon man. Really.