Former McDonald's Worker Does a Number Theory Proof
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- Опубліковано 6 жов 2024
- Number theory classic problem with Sophie Germain's identity. A few days ago I was in my storage room and found my good old McDonald's uniform with the name tag. So many memories rushed back to me after I put it on again after 10+ years. Of course, I will never forget about the teacher who once helped a hopeless kid to get his first job.
Thank you, Mr. Sergio Salas.
In this video, I showed if 2019^4+4^2019 is a prime number or not. Be sure to watch my previous video on "add then subtract to make things work" • How to factor x^6-64 (... I also introduced the Sophie Germain identity for you guys in this video.
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bprp
Future McDonald's worker watches a number theory proof
Lol
best comment in this section
Future McDonalds Worker watches a number theory proof by a former McDonalds Worker explaining the number theory proof
Lol
Lmao
A job at McDonald's is respectable, honest work. If you work there, or anywhere else, be proud of yourself!
@CogitoErgoCogitoSum I would at least use the word "some" to replace "the" in your sentence. Not to say at I would say anything similar to your sentence at the first place.
I work at a prostitution ring. Thank you for giving me encouragement
I am a hit man, this is great motivation to keep doing what I love
Yeah, there's even special word for this - McJob.
@@smortboi1320 Hitting on minors doesn’t count.
Nice puzzle. I solved a bit differently myself before I watched your solution. You can notice that the last digit of 2019^n for odd numbers n is always 9, and for even n is always 1. For n = 4 it will end with 1. Similarily, 4^m end with a 4 for odd m and ends with a 6 for even m. For m = 2019 it ends with a 4. This means that we add two numbers, one that ends with 1 and one that ends with 4. The last digit of the sum is thus 5, and from this we conclude that the result is divisible by 5, and that it's not a prime
How life changes drastically always amazes me. Good job man
Thank you!!!!
Can we have fries with that?
what size?
Supersize me!
@@blackpenredpen size of a definite integral
blackpenredpen a number six with extra dip
The size of i^i
There's also brute force where 2019^4 % 5 = 1 (using p-1 prime rule), and 4^2019 % 5 = 4 (using the pattern of the last digit) and when you add the mods, you get 5, which means the number is divisible by 5.
Awesome! I too was thinking of calculating the first digits of each number, but I don't think I would've come up with this solution.
It isn't really a brute force way
Today I was released from a heavy session week of Chemo. I'm trying to follow what you're doing, but I feel absolute miserable and I can't even follow anything. Maybe in the days to come I can actually make sense of math again. But your T-Shirt has arrived. I'm actually wearing it right now, and the first thing that happened was I fell asleep in it. Thank you again for the shirt. I'll send you a photo of it when I feel better. Thanks again bprp :)
I hope that things start looking up for you. Good on you for watching math videos, even if you can't follow along completely. That's what I'm doing, too.
@@chinesecabbagefarmer Thank you :). By chances of beating cancer are at 95%, but I'll need to withstand 3 heavy chemo sessions. I got one done, starting to feel somewhat better. Next session is in 2 weeks. Luckily my tumor is one of the least aggressive ones which makes it easier to treat. But its still a battle. But I'll stay strong and positive, because its all up from here.
@@williamdavis2505 Thank you :)
Best wish to you, Lars
@@blackpenredpen Thank you very much Steve. Means a lot :)
When n = 1 , n^ 4 + 4^n is trivially prime. Yay
@Emperor pussy pounder For no other n, n^4 + 4^n is prime.
@Emperor pussy pounder that was not the question xD
Edit: As Un canal UA-cam points out I wrote nonsense
Emperor pussy pounder haven't covered these options yet
I'm a french high schooler, so you know, I'm not into these level of thougts yet. It just intrigues me, I love mathematics.
I really love this guy! He's so intentional about passing on the help he received as a student by helping, encouraging, and inspiring other students and viewers! BlackPenRedPen: thank you for all you do to inspire us who want to learn math! I would love to meet you one day and tell you that in person and give you a hearty handshake!
alkankondo89 awwww thank you thank you!!!
I would definitely love to meet you in person too!!
Alkans music is 🔥
2019^4 end with 1, 4^2019 end with 4, 1+4 is 5, a number that end with 5 in not prime.
A number that end with 9 squared end with 1, that squared end with 1.
4 at the odd power end with 4, at the even power end with 6 (4, 16, 64, 256, 1024, ...) so 4^2019 end with 4.
Yes but you'd have to prove why 2019^4 ends in 1.
@@jibran8410 9*9*9*9=6561
@@jibran8410 it isn't hard. 2019^4 = (2020*2018 + 1*1)^2
5 is prime and it end with 5 hehe
Ivan Burilichev The entire point of the proof is to not have to do complicated calculations such as that. Otherwise, one would simply actually calculate 2019^4.
Omg i just learned Sophie Germain and you just uploaded a video about it. Exactly two days ago. Thanks for saving me
Great work! The first factor A^2-2AB+2B^2 can be written as
(A - B )^2 + B^2 > B^2 > 1 .
Actually, 8:30 not really need a "convincing work" 😃 'cos L=MN while L and N is positive(all positive term), therefore M will automatically be positive as well.
@@gagadaddy8713 That's not sufficient. We need to prove that the factors are greater than 1, not just greater than 0.
It is your dedication and hard work what resulted the fruit!!! We always love your content keep up the awesome work!!!
P.S. And yeah solution to that problem is n=1 because it is trivial that n cannot be even and if it is odd the expression is factorable so the only prime generated is 5.
Another approach: Do the calculation modulo 5.
2019^4 == (-1)^4 (mod 5) == 1 mod 5)
4^2019 == (-1)^2019 (mod 5),
and since x^4 == 1 (mod 5) we have:
4^2019 == (-1)^(504*4) * (-1)^3 == -1 (mod 5)
Adding both terms,
2019^4 + 4^2019 == 1 + -1 == 0 (mod 5)
So the sum is divisible by 5 and is not 5 itself, so 2019^4 + 4^2019 is composite.
(Really I tried 2 and 3 first, but they give non-zero answers)
Yeah. I noticed that 2019^4 ends in 1 and 4^2019 ends in 4, so the sum ends in a 5.
Truly the simplest solution!
Could also have said 4^2019 == (-1)^2019 == -1 (mod 5)
Very simple solution
@@TomasIngi00 Yes, you are right. I saw that I could use Fermat's Little theorem, but it would have been simpler to just use (-1)^(2n+1) = -1 as you did.
It is not true that x^4 == 1 mod 5 in all cases. For example, put x = 5.
The claim fails when x is any multiple of 5.
Everyone gangsta til the minimum wage AP Calc student starts using proofs at the cashier
Thanks for all your videos and your work here, you're really great and your explanations are very clear and fun. I really love what you do now and now I admire you even more, and finally (as someone else said) A job at McDonald's is respectable.
This Sophie Germain theorem brings back a memory for me. Namely, of seeing, in my dad's book of math tables and formulae, from his college days (1930's?), the following factorization of a sum of two fourth powers:
x⁴ + y⁴ = (x² + √2 xy + y²)(x² - √2 xy + y²)
Just replace y with √2 y, and your SG factorization pops out!
I recall this because it was so striking to me at the time - I thought such a factorization impossible until I saw that!
This also reminds me of one of the basic theorems of polynomial algebra; namely, that a real polynomial (i.e., real coefficients) of any degree, can always be written as a product of real polynomials, all of degree 2 or less. It is this theorem that guarantees that every such polynomial, p(x), has exactly n (= the degree of p) zeroes in the complex plane, and that the non-real zeroes always occur in conjugate pairs. (Conjugacy follows from the quadratic solution formula.)
Fred
I have a similar experience that my former algebra teacher wanted me to factor x^4+x^2+1. I was like how was that possible!!
n^4+4^n
If n=2m,
(2m)^4+4^(2m)
16•m^4+16^m
=16(m^4+16^(m-1))
Not prime!
If n=2m+1
(2m+1)^4+4^(2m+1)
(2m+1)^4+4•4^(2m)
(2m+1)^4+4•(2^m)^4
And then apply the theorem featured in the video.
Except for when m=0 and thus n=1
Yup : )
Nice!
This was uploaded on my birthday! Thank you for making my day better with the inspiration you gave us all. You and a few others on youtube are probably the best teachers in the world!
It’s not a prime number.
(2019)^4=.......1 because 9^4=6561
(4)^2029=.......4 because:
4^1=4
4^2=16
4^3=64
4^4=256
4^5=1024
....
So only “4” and “6” can be at the end of those number
When n-odd in 4^n then it will be “4” at the end of the number.
(2019)^4=.......1
(4)^2029=.......4
Add this two number and we will get number with “5” at the end.
So this number isn’t prime
(2019)^4+(4)^2019=......5
Beautiful
Алексей Багров Beautiful thinking. Thank you Алексей!
Wait, but 2^4 is 16, which ends with 6...
@@arnavpvpkumar5591 but the base of exponentiation is 4 not 2, why did you write 2^4?
@@backyard282 no, but the person who commented it suggested that it's always ends with 1 and 4, but with my example, it clearly doesn't.
Sophie Germain is my fav mathematician throughout history. I wrote an essay on female mathematicians throughout history and she was my fav part
Mad respect to you lad, this is goes to show that it doesn't matter wherever you're working as long as you're actually contribute something useful to the society, salutations to you sir
Absolutely inspiring figure!!!!!
Respect for you 👏👏👏
Thank you!
😊😊😊
We need to appreciate the geniuses in our lives
Incidentally, a simpler way to show that both factors are bigger than 1 is to pull a b^2 out of both, leaving a square-of-binomial result, so that the factors can be described as (a+-b)^2+b^2; this should make it much more clear that the factors are much bigger than 1.
I proved by finding the last digit of their sum which is 5 to prove that it is not a prime number. Hope that is correct.🤣
That's what I would have done if I had a piece of paper
Hmm, for n=2 we get 32 for instance tho....or did I misunderstand you?
@@slenderman478 Did I miss out something? I was wondering that you mentioned n=2 as I don't see any unknown in the question.
How do you know it is 5? Can you prove it?
@@srpenguinbr I think we have a same thought. The last digit of 2019^4 is 1, and the last digit of 4^2019 is 4, which gives the sum of the last digit is 5, and disproved the number is prime.
More Number Theory videos please! I love this channel :)
8:26 you can also prove that the left part is greater than one using the fact that
A^2 - 2AB + 2B^2 = (A - B)^2 + B^2
which is greater than one!
Mu Prime Math ah very nice. It was too obvious that I ended up don’t know the best way.
Thank you BPRP your awesome!
ur vids r gr8!
Thank you, Mr. Sergio Salas. for what you did ;) Now we can watch bprp
Jardel Kaique yay!
It's comforting to know that even though smart math teachers have humble beginnings
At the end, the left factor can be shown to be positive much easier than what you did. Just take it as (a-b)^2+b^2, and as b^2>>1, done
Great video!
Would you please consider doing a video on how to use the Heaviside Step Function to rewrite the limits of integration of double/triple integrals? I know it’s slightly off topic with the videos you’ve been making but it’s a super cool technique!
Man, your respect in my eyes got doubled, don't know your struggle story, but great efforsta nad hardwork in getting up till here
Best of Luck for future
Thank you!!
Have to love how there are so many commenters yelling out "BUT MY APPROACH IS SIMPLER" despite the fact that their approaches actually involve directly calculating fourth powers and what not.
It all depends on the definition of "simple" that one uses. General Relativity is "simpler" than Newtonian Gravitation in a sense.
This is a very great video. Thank you bprp for making this informative video about sophie germain identity in a very nice way.
Yes you prove that everything is possible on earth .....And now you are a math teacher and famous in all world
For anyone wondering, the number in question is 3623451821273106591605158955925970682490509265109095775271008260956357951888670068832469558835892986320727597296523443813577499688450395146308139232055896543173285131110954165671456072195188013709929685010113589124698392632948187686265979366525372884130811973141957444103103571989422275171036014833329908675248680952347232010573119358847821657591754309610336083982094408800289035060730735245523219187672432500415027217526834893543743791904279241132187921668007251314204468026474912038573975538977609307347667218167420915455923716200897866924993436765964180475552469823275361476229422184923231689763336008423109311448623660577453324627597003940475436305569797539724915405288833336876231760987788626810061976582113315875577404617660424515575257447401498311198128297991087923919209815986462573828643365671015448091577222174807328082857961908681123124688624141805842774527962861923987595332155496451244096960921957036442151318626571561562246969953507084246747139537419593141771359186956144897427224699832606974869881706123803285287985726213497970784989414439684363752011604989672735667173277877154000554324849138991497110625950738443188719122628638571758117822162537138380686725207315319577146283255778470976887541869265
Very nicely done, sir.
Thank you for your labors.
I tried by binomial theorem
(5-1)^2019 + (2020-1)^4
=(2019C0 . 5^2019 - 2019C1..... +2019C2018 . 5^1 - 2019C2019) + (4C0 . 2020^4 - 4C1 . 2020^3 +....- 4C3 . 2020^1 + 4C4)
Since 2020 and 5 are multiples of 5 and nCn = 1.
Therefore the equation reduces to (5k-1)+(5b+1)
=5a i.e. multiple of 5
Therefore not prime
I understand though that the question couldn't be solved by this method had the number been prime
Thank you for your service (to Mathematics and fast food).
2019^4 -> Let’s take a look at our last digit if we do it on the fourth power: 9*9=81; 1*9=9; 9*9=81. (We’re only looking at the last digit). We are dealing with a 1 on the unit’s place.
Now let’s look at four’s powers.
:
4^0=1; 4^1=4; 4^2=16; 4^3=64; 4^4=256; 4^5=1024; 4^6=4096; 4^7=16384, and so on.
See if you can spot the pattern. If you can’t, it’s basically that for all odd powers, we have a four in the unit’s place, and for the even powers, (apart from 0), we have a six in the unit’s place.
So we can say that 4^2019 has a four in it’s unit’s place, plus the one from the unit’s place on 2019^4 equals some number ending in 5.
As we know, any number besides five itself, that ends on a five, cannot be prime, because it’ll be divisible by five. So 2019^4+4^2019 is clearly not a prime.
This video put a smile on my face.
I believe that Yitang “Tom” Zhang also did some time in fast food (Subway?) at one point in his life before his recent seminal proof regarding bounded prime number gaps at 50 years of age.
At 3:24 “If you’re willing to just think about how can you make things happen then you can actually make good things happen.”
A powerful theorem in mathematics that generalizes to everything in life worth doing.
Thank you for the video and your words of wisdom and perseverance. Steve K.
For the last question, n^4+4^n is prime if the left part of the factorization is equal to one. It means n²-n*2^((n+1)/2)+2^n=1
This equation has 2 solutions : n=0 and n=1 (according to WolframAlpha, it's also possible to demonstrate it by derivating it).
We still have to prove that the other factor is a prime factor, which would imply :
X = n²+n*2^((n+1)/2)+2^n is prime.
For n=1 : X = 5, which is prime.
For n=0 : X = 1, which isn't prime by convention.
So n^4 +4^n is prime if and only if n=1.
1, conventionally, is not a prime. So, n=0 doesn’t work. So the only solution is n=1, which produces the prime number 5.
J4YB1RD 1 is by definition not a prime number.9
Indeed I'm sorry. Thanks for the correction@@j4yb1rd
Love number theory, awesome as always friend!
SpikePowerHD thank you!!
Great proof! Who knew a McDonald’s employee could be so good at math? 😊
He was probably in school, just like millions of other mcdonalds workers...
Your channel is awesome, keep up the good work!
The moment when i realised the a2-b2 form I practically lost my shit, you are a genius.
The easiest way is:
2019 to the even power ends with "1"
4 to the odd power ends with "4"
So, the sum of this numbers ends with "5", and it can be divided by 5.
______________
At 8:35 in the first parentheses you have full square plus something:
2019^2 - 2•2019•2^1009 + 2^1009 + 2^1009 = (2019 - 2^1009)^2 + 2^1009 > 1
This video made me happy. That Sophie Germain identity is really cool. Never seen it before.
I was expecting deeper story about your life. A glimpse of it is itching!
Anyway, thumbs up for you! You reminds me of Yitang Zhang. He used to be a pizza guy before getting a mere lecturer position.
The roots to x^4+4y^4 is given by y(+-1+-i), which pairing up conjugate gives the factorization x^4+4y^4 = (x^2-2xy+2y^2) (x^2+2xy+2y^2).
Also (x^2 - 2xy + 2y^2) = (x-y)^2+y^2 = 1 if and only if x=y=1. (where x,y are positive integer)
I have nothing but respect for you my man :)
Goku17yen thank you Goku! Btw, please do me a big favor of never change your profile pic and name!
Another, more general approach to the question: Note (A^2-2AB+2B^2) = (A-B)^2+B^2. So the only way the Sophie Germain number can fail to be composite is if (A-B)^2+B^2 = 1. The only way this can happen is when A=B=1. In this case the other factor is 5, a prime. Conclusion: n^4+4^n is prime only when n=1.
There is a way to motivate people to think about factoring a sum of fourth powers without reference to a sum of squares.
Consider viewing the roots of x^4 + 1 = 0 (i.e., the 4th roots of -1) in the complex plane. There are two pairs of complex conjugate roots and each pair corresponds to a quadratic polynomial with real coefficients. These polynomials are factors of x^4 + 1.
The preceding contains the key ideas -- what follows is an overview of the details.
The first pair of conjugate roots lie at angles 45 and -45 degrees; second pair at 135 and 225 degrees. The corresponding polynomials are, respectively,
x^2 - (sqrt(2))x + 1 and x^2 + (sqrt(2))x + 1.
It's easy to verify that the product of the polynomials shown is x^4 + 1. A minor modification gives a two-variable identity
x^4 + y^4 = [x^2 - (sqrt(2))xy + y^4] . [x^2 - (sqrt(2))xy + y^4]
from which the assignment x = A, y = (sqrt(2))B gives the Sophie Germain identity cited in the video presentation.
The approach here probably would be unsuitable (too time-consuming) to be part of the proof presented in the video.
Also see comments from *tomogames* apparently posted before this one, and *ffggddss* posted after this one.
.
Is a Sophie Germain identity, so it can be factored as a difference of a simple square with a perfect square
But you will never know if the two resulting factors ARE NOT ALWAYS the prime number and one. It has to be proved but most people choose to resolve and check.
To prove that IS A PRIME all you need to do is to prove than the negative member equals the othe rtwo positive members, AND OR prove than the all three positive members are equal to the main number itself.
Awsome. I love Sophie life and work. Thanks for the video. THIS ONE It was a VERY GOOD ONE
in the last part when we were trying to prove that it isnt one, couldnt we just on the left hand side do:
(2019^2 - 2*2019*2^1009 + (2^1009)^2) + (2^1009)^2
that will become:
(2019-2^1009)^2 + 2^1009
of which we know the bracket is going to end positive because of the square, and not zero because of different parity ( doesnt matter but had to mention ), and we are adding a positive integer greater than 1 to a number that is >= 0 which is another proof of it not equalling one
The smaller of the two factors is sqr(a) - (2*a*b) + 2* sqr(b) which can be written as sqr(a - b) +
sqr(b) which will be equal to 1 only if a = b = 1. This makes the first factor = 1 and the second factor = 5. Hence the only possible prime value of the expression is 5.
10:04 n=1 => n^4+4^n=5 and 5 is prime
Thanks for your videos blackpenredpen!
I speak spanish but i try to understand your videos because love math
Érebo14 thanks!
I love pure math and physics!!!!!!
I was read about sophie yesterday !!! She was smart
... she is writing to gaus in france and he is answer her
n^4+4^n is prime for infinitely many real numbers as it is continuous and increasing constantly for positive n. In fact all primes can be generated by n^4+4^n. n=1 is, however, the only natural number which gives a prime.
I would actually love to see you treat x^4+4^x=y for all real x in another video. Finding an inverse function (for positive x), or showing the minimum, or anything like that really.
Edit: clarification on domain of the function to be inverted (oops)
f(n) = n^4 + 4^n cannot be continuous by definition as the notation specifies that f(n) has f : N -> N, not f : R -> C.
Also, x^4 + 4^x is not invertible.
Angel Mendez-Rivera Both of the things you’ve said would be true if I was talking about natural numbers or all real numbers, but I specifically extended/restricted the domain to positive real numbers, at which point the expression is, in fact, continuous and (to the best of my knowledge!) invertible. Why would it not be invertible? It’s a one to one mapping of real numbers to real numbers.
Joël Ganesh okay, again, I’ve restricted the domain to the positive real numbers. I am fully aware that the function x^4 + 4^x is not invertible over the entire real domain
Although I really appreciate the responses!
An easy way to show that your factors are not 1 is by looking back at the identity. Notice that
a^2 - 2ab +2b^2 = (a^2 - 2ab + b^2) + b^2 = (a-b)^2 + b^2
Since (a-b)^2 is at least 0, this expression is at least b^2. Therefore, with your b=2^1009, there's definitely no problem,
2019^2 - 2(2019)2^1009 + 2(2^1009)^2 >= 2^2018 >> 1.
U are amazing!!!! Love how u proof statements!
Alternatively, you could look at the expression mod 5, but sophie germain identity is really nice ; )
Why modulo 5?
Unit digit?
Thank you for existing !
A much simpler proof:
4 ≡ -1 (mod 5) and (-1)^2019 = -1 so 4^2019 ≡ -1(mod 5)
2019 ≡ -1 (mod 5) and (-1)^4 = 1 so 2019^4 ≡ 1(mod 5)
Therefore, sum of these two will be equivalent to 0 on mod 5.
Therefore, it is divisible by 5, and it is not prime
Also, it's congruent mod 5 to (-1)^4 + (-1)^2019 = 1 - 1 = 0, so it's divisible by 5.
You can also say that, since 2019^4 will always end in a 1 (9^2 = 81, 9^3 = 729, 9^4 = xyz1), and 4^(uneven number) will always end in 4, their sum will always end in 5, which is never prime, because all whole numbers that end in 5 can be divided by 5.
Factorizing (a^4+4b^4) is just a Grade 8 exercise problem
but applying this in such a problem is also quite interesting
Condition for it to be prime is
n = 2^((n-1)/2) (+ or -) root(1-2^(n-1))
since the root part cannot be negative, n
n^4 + 4^n is even for even n, hence composite
also, (n^4 + 4^n) = (n^2 + 2^n)^2 - 2^(n+1).n^2
for odd n, n+1 is even and the thing becomes a difference of squares, which is factorable hence composite (except 1)
hence the given number is composite for all n except 1
hence it is prime only at 1
The interesting thing about this is... For any n>1 where n is a Natural number n⁴+4^n is not a prime. It can be easily showen by using Sophie Germain's Identity.
😃😃
Nope try it on paper if you do so then for every n= even it's will be irrational prime
harshit bansal You are incorrect. 1. Any number that is an integer is by definition not irrational. For n even, n^4 + 4^n is an integer, so it is not irrational. 2. If n is even, then n = 2m for some integer m, so n^4 + 4^n = (2m)^4 + 4^(2m) = 16m^4 + 16^m. 16m^4 is even since 16m^4/2 = 8m^4 is an integer, and 16^m is also even since 16 is dividible by 2, and therefore any power of 16 is also divisible by 2. However, 16m^4 + 16^m is the sum of 2 even numbers, and we know the sum of even numbers is also an even number. Therefore, 16m^4 + 16^m is an even number and is thus divisible by 2. Therefore, it is not a prime number.
@@angelmendez-rivera351 I said if you he will use Sophie identity then for n= even he will end up with 2 factors which are irrational
@@harshitbansal7908 No, we wont. Integer multiplication, squaring and adding can't produce irrational numbers, the result will always be an Integer.
@@harshitbansal7908
You're right. You shouldn't use Sophie Germain's identity to factor the number for even n, because with an even n you can't get it into the right form to factor it into integers with SG. You shouldn't have called it prime though, because it's pretty obvious that for even n, the sum is even, and so not a prime.
You can use SG for odd n larger than 1, and it's obvious that the number isn't prime, so the proof that n=1 is the only prime does require SG and the trivial fact with n being even.
I had a problemwho looks like to this one this year in a math competition, we have to find all prime numbers a and b such that a^b+b^a is also prime (the only solution was a=2 b=3 and a=3 b=2)
You motivate me to become a great mathematician such as yourself ! 💯
Wonderful. Well done!
Respect=Infinity
Quick answer: It is not a prime, there has to be a quick way of showing it as divideble, if it is a prime, we don't have enough time to check every prime from 2 to roughly (2019^2 + 4^1009) so we can prove it is a prime.
He evolved from BigMacs to Calculus 🔥🔥
So what is the Prada Lu for McDonald's? 😂😂
Great video , thank you ❤️
you should clarify when discussing the Sophie Germain identity that a and b have to be different from each other. for example if a=b=1 the identity falls apart.
Why does it fail? A=B gives 5A^4 = (A^2)(5A^2) which is true.
Why does it fail? A=B gives 5A^4 = (A^2)(5A^2) which is true.
Sophie Germain has worked for McDonald's?
I like this factorization.
I love how you use doraemon music at the start. Don't know why.
I figured this out without factoring like that. I used 9^4 ends in a 1, and 4^2019 ends in a 4, and 4+1 is 5. Therefore, it is divisible by 5
n^4 + 4^n will be prime wherever the graph intersects prime values of y, infinitely many values. For example: -1.158, 0.481, 1
I got a minor in math in college, because I wanted to understand more math. But it was always challenging to do a lot of algebra, because I make mistakes. This was high school level, or pre calc level algebra, but as it was a lot of it, it's a bit hard to keep it all clear. I suppose it helps to practice this. I remember one step was like 2^2^x, and I had forgotten the rule (a^b)^c = a^(b*c).
All he does is help society,,,, legend
Simple - Just watch both last digits and their sum. 2019^4 ends in 1 (since 9*9=81, 1*1=1) and 4^2019 ends in 4 (since odd power of 4), so this large number sum ends in 5, hence multiple of 5, hence composite, hence non-prime.
Sophie Germain Factorisation
Inspirational bro!
CHALLENGE:
Find Y
Y"×Y'=Y
Y is not equal to 0
@@mikhailtatmyshevskiy2666 honestly I don't know this thing came in my mind like uuuhhh.. wtff maan
Got an answer but it's in the terms of an non elementary integral. The integral is du/((u^2+c)^(1/3)) where c is any constant. This is trivial for c=0 but i believe the anti derivative is not expressable in terms of elementary functions.
I won't deny the existence of an elementary general solution, though.
A simpler way to check if the first term is greater than one is to notice that A^2 - 2AB + 2B^2 = (A - B)^2 + B^2. B^2 is obviously greater than 1, and (A - B)^2 is greater or equal to 0.
The only odd number n for which 4^n+n^4 is prime is n=1.
This because if n is even it is divisible by 2 if n is equal to 2m+1 then 4^n+n^4=(2m+1)^4+4*(2^m)^4.
If we set A=2m+1 and B=2^m, we have that 4^n+n^4=(2*A^2+B^2-2*A*B)*(2*A^2+B^2+2*A*B).
Since 2*A^2+B^2-2*A*B is the smallest it has to be 1.
2*A^2+B^2-2*A*B=A^2+(A-B)^2, since A and B are natural we have only two options:
1) A=1, A-B=0 => m=0.
2)A=0, A-B=1 => no natural m.
Then the only m is 0, then the only n=2*0+1, then n=1, THEN Q.E.D.
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Riccardo Costantini yes! That was great!!
Respect teacher
4^n mod 5 is (-1)^n is -1 when n is odd.
n^4 mod 5 is 1 when n is not divisible by 5.
So 4^n + n^4 can only be prime when n is divisible by 2 or 5 (and otherwise has a factor of 5).
for the last question. through fermats last theorem, k^5=k mod 5, k^4=1 mod 5 (you could also prove this through exhaustion), 4^n=(-1)^n= -1 mod n if n is odd or 1 mod n if n is even. if n is odd, n^4+4^n=-1+1=0 mod 5, so it is divisible by 5. checking n = 1, we can see it actually is prime, so n=1 is prime. for any other odd n, n^4+4^n is divisible by 5 and so not prime. when n is even, n^4 is also even, and since 4^n is even as well the entire thing is divisible by 2, so it isnt prime (since it is more than 2). therefore the only case where n^4+4^n is prime is n=1
There's a pretty "easy" prime checking method where as long as you can calculate the last few digits, you can look at the last digit after converting to several bases and if it remains a valid ending after the first few bases that you check, you know it is probably prime