The Prime Problem with a One Sentence Proof - Numberphile
Вставка
- Опубліковано 11 вер 2024
- More links & stuff in full description below ↓↓↓
A prime number problem posed by Fermat that has been proved multiple times - including a famous proof using one just sentence.
See the next video (proof) at: • The One Sentence Proof...
This video features Professor Matthias Kreck from the University of Bonn.
See him discuss wobbly tables: • Fix a Wobbly Table (wi...
Prime Number Videos: bit.ly/primevids
Support us on Patreon: / numberphile
NUMBERPHILE
Website: www.numberphile...
Numberphile on Facebook: / numberphile
Numberphile tweets: / numberphile
Subscribe: bit.ly/Numberph...
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumb...
Videos by Brady Haran
Brady's videos subreddit: / bradyharan
Brady's latest videos across all channels: www.bradyharanb...
Sign up for (occasional) emails: eepurl.com/YdjL9
Numberphile T-Shirts: teespring.com/...
Other merchandise: store.dftba.co...
Hey, why did you make this two videos and not just one?
Because two videos would have been very long and deterred casual viewers from even learning about the problem.
Makes more sense to make a video which describes the problem and the solution, plus its history and the fact a one sentence proof exists... then have a more technical second video for people who want to see the proof itself. I have made links to the second video as clear as possible and I suspect more technical fans subscribe to Numberphile2 anyway.
Correction: At 4:08 the number at the bottom should be the prime 23, not 22... - Наука та технологія
"3 = no" is now my favorite formula.
Docteur Zeuhl -3=yes
3 = no
7 = no
3 = 7
Q.E.D.
😂
@@MinusPi-p9c 🤣🤣
@@MinusPi-p9c 7+7=14
3+7=10
3+3=6
14=10=6
10+10=20
14+6=20
20=20
If 3 was not equal to 7,
20 would not be equal to 20.
Since 20=20
3=7.
5:14
"Too complicated for this video"
So you have a proof but it's too complicated for that video? That's so Fermat-like
Brilliant! (Note: I had a much longer reply, but it could not fit within the margins of the UA-cam comment box. Hint hint wink wink).
Video is the new notebook
"i have a proof but wont make it for the rest of the video..."
But this proof fits in 1 sentence.
The professor's accent is wonderful!
I don't think an accent can get any more german than that.
Didn't know that there are people who actually like the german accent! :D I always thought that people hated it. Maybe because most Germans do.
I think it's more likely most Germans just hate the English language.
Nah, we Germans just really love rolling dem ssss sounds.
I don't think Germans hate the English language, it's just really hard for most people to get the pronounciation right, especially the rather soft sounds, because it's so different to what we're used to from our own language. Also I can imagine that some people simply aren't that interested in getting rid of their accent, which is ok too.
Stop commenting his English. He speaks well and understandable. Try to imagine yourself explaining difficult problem in foreign language. Most you you, who complain, are not even within range of sight knowledgewise.
And btw rather german accent than french one.
+chris mclaughlin Best comment so far! Maths are indeed a language out of this world. And I have huge respect for those who can actually speak mathematics, sometimes when I look at it I'm like whaaaaat 😂 but anyway his accent is quite cute, definitely more understandable than the French one lol
+
You're totally right with that. Also, he has the advantage of pronouncing german names the right way, what I think to be an important thing.
I admit, his accent means that working out what words the speaker is saying is a little harder with him than usual. But that's also true of a lot of people with American accents who present UA-cam videos.
As a frenchman I feel offended
Because arbitrarily large numbers are mentioned, just a note that because 4 cleanly divides into 100, you only have ever to check the last two digits for divisibility.
Cool! Question for computer science folks: is this strategy employed for the modulo function?
I doubt it. mod is made to work on any possible number so there shouldn't be any special cases, at least in the modulo function/operator itself. it would be easy to program in your own code though.
I was hoping someone already mentioned this. I noticed the same thing, so I checked the largest known prime and it ends in 51. Therefore, theoretically, 2^74207281-1 can be written as the sum of two squares.
I'm an idiot. Change that "can" to "can't".
+MultiPaulinator If you make a mistake in your comment it is better to delete it completely and write it again correctly, rather than making a new comment correcting the previous one. It makes things clearear and much straight forward. Try it!
The moment Euler popped out I started laughing like crazy. Of course it's Euler!
@lessur dgreat Euler will use up all of his appearance quota, so that when they do a video about him, Gauss will be the main focus.
At 4:08 : It's supposed to be 23 = x²+y², not 22 = x²+y² isn't it ?
Yes, it should be 23
yes, it's now mentioned in description (in the end)
3 = 2^2 + i^2
1337!
last time I checked i wasnt a positive natural number
i^2 is negative (-1)
+Phalax: You are absolutely right. Silly rule we have stumpled over and were smashed to the bottom. (unhappy + Sorry, when the grammar is unintentionally wrong)
+Theplayer1209 The Player Still, it's not a natural number
they didnt even say the scentence... i call click bait
It's shown in the video - it has a HUGE RED BOX around it... and the linked extension video is 12 minutes all about the proof.
See video description for more explanation.
at least watch the video before whining on the comments section.
Tsk tsk tsk, duzn't watch the whole video or read the description box before commentating. U gave Brady another papercut.
Should try looking up how to spell "sentence" before calling someone else out on UA-cam, haha.
This channel is incredible. I saw many 18-minute long videos on my recommendations, by I thought "I don't have time for that". then I saw this 6-minute long video and decided to watch it. Now I NEED to watch the 12-minute sequel. You are brilliant.
At 4:07, it should say 23 at the bottom, not 22. (23 is the prime)
indeed
I was going to like your reply, but your likes for that reply are at 23 right now so I'm replying instead. :-)
Ta-daaa~
It's back
He added an Annotation to cover it up, unless you're on mobile and you cant see it.
It isn't an annotation (in the video, that is); it's in the Description notes.
2:27 "Who did this first" - Probably Euler, or something. It's always Euler... oh, darn, it was Fermat :-/
5:08 Aha! Redemption! It _is_ always Euler!
Last time I was this early, this joke didn't exist.
Last time I came this early, last time I came this early jokes were funny.
+Mariux - So you were never early?
hello poirot
damn grazer Hi
"Joke"
Euler strikes again!!!
I laughed so hard at the unintended laugh-like sound at 3:17 :D
The actual, full statement is that if & only if p ≡ 1 or 2 (mod 4), is a prime p expressible as a sum of 2 squares.
Of course, p ≡ 2 (mod 4), is the case only for p = 2.
And 2 = 1² + 1²
This is one of the most beautiful 'simple' theorems in number theory .....
Zagier's proof is very famous. Probably every mathematician has seen/heard of it. Glad to see it on numberphile.
I love how he calls himself ordinary compared to Zagier when he actually was friends with him and they published very interesting papers together
Optimus prime numbers, prime number sums in disguise
this guy sounds like such a wizard, i'd love to be taught by him
slayerficated i blocked that crazy person
What is the formula for the random zoom setting?
Fermat: *Writes Theorem Without Proof*
Andrew Wiles: Am I a Joke to You?!
tbh, he was almost going to be.....
That 2 is a special case is less trivial than it might seem. All prime numbers EXCEPT 2 can be expressed in the form p = 2m + 1, that is, they're all odd. Exactly half the odd numbers (but not necessarily half the primes) can be expressed in the form 2(2k) + 1, that is, 4k + 1. So if p is one more than an even number, the prime can be written as x^2 + y^2 iff that number is doubly even. If p is itself even, the formula doesn't apply, but it doesn't need to since 2 is the only even prime number.
A way to write the formula to include the special case: ∃(x, y) | p = x^2 + y^2 iff p 4k - 1 (where p is prime and x, y, and k are all positive integers).
... "finically"? ...
@chris mclaughlin what?
2-1=1 and 1 isn't divisible by 4 and 2 works
The test only holds for primes > 4. It was only designed to prove the pattern for primes that we can't easily compose a sum for. Math is a weird thing, and there are other tests with cases where the result of the test doesn't tell you anything about the property of the number you are testing. For example, the Ratio Test will tell you if a series converges (test comes back 11) based on the result of the test, but if the result of the test is 1, then you don't find out anything.
Which part are you not understanding?
what about 50=1^2+7^2.
50-1=49. 49 isn't divisible by 4
50 is no prime
This can be fixed by saying that, for p = m + 1, m is either a multiple of 4 or it's odd. In other words, if p = 2n + 1 is a solution, then n = 2k. Which is a long way to go to just say "2 is an exception". ;)
It could be more easily fixed by limiting it to odd p (there's no need to limit it to primes > 4, as RekiWylls does, because there are only two primes less than 4, and only one of these is an exception).
Fermat's theorems remind me so much of the way Ramanujan worked. He instinctively knew he was right and did not feel the need to explain himself.
Professor Kreck is back! I'm frequently reminded of his wobbly table.
Hey Brady,
It would be cool to see some applied mathematics - as in following a mathematician doing some real life work using math. It's just difficult for me to understand exactly how these theorems and primes factor into real world applications. So, I'm asking for perspective I suppose :)
(Still looking forward to that Rachel Riley video)
Heeey! When I saw x^2 + y^2, my first thought went to the 3b1b video about pi hiding in prime regularities, where he mentioned the factorisations of primes in the complex plane into gaussian primes, and that primes 1 above a multiple of 4 do further factor into gaussian primes, while primes 3 above a multiple of 4 are already gaussian primes themselves :)
Thanks, that was a very interesting video.
@@danielbarnes3406 np. His whole channel is wonderful.
And that is very closely connected to the result discussed in this video.
Anyone else reminded of Robert Guillaume's cartoon voice performances? Like Rafiki or that time he did a Land Before Time character. That's what his accent reminds me of.
Just wanted to tell you guys that you made a mistake in the video description. You say you made two videos because two videos are very long and deter viewers from learning about the problem. Pretty sure that's what would happen if you made one long video. ^^
If there exists an x and a y such that p = x^2 + y^2, we can conclude for all non-even p that x must be even and y must be odd or vice versa. There exists a j and k such that x = 2j and y = 2k + 1. This means that p = (2j)^2 + (2k + 1)^2 = 4j^2 + 4k^2 + 4k + 1 = 4(j^2 + k^2 + k) + 1. j^2 + k^2 + k is a natural number, thus there must exist an n such that p = 4n + 1. That's not complicated.
Yeah, I did the exact same thing, and I'm 15. I kept wondering if I did something wrong, but everything seems right, so is this a proof of it?
I hope you understand you just proved the following statement:
p = x^2 + y^2 => there is a n in N with p = 4n +1
However, that was not all we wanted to prove. You didnt show:
there is a n in N with p = 4n +1 => p = x^2 + y^2
It's always a good thing to see a new upload from any of Brady's channels.
So the proof is not actually in the video? Bummer
he does it here - ua-cam.com/video/yGsIw8LHXM8/v-deo.html
Cheers
Oh dude the proof is super simple, its only 1 sentence long.
The involution on the finite set S = {(x,y,z) E rkJ3: X2 + 4yz = p } defined by
((x + 2z, z, y-x-z) if x 4 (2y - x, y, x - y + z) if y - z < x < 2y
I(x - 2y, x -y + z, y) if x > 2y
has exactly one fixed point, so ISI is odd and the involution defined by (x,y,z) -
(x,z,y) also has a fixed point. O
Morning everybody
is that a typo at 4 minutes? shouldnt it be 23 not 22 for the x squared plus y squared line?
Yes, it should be 23
At 4:10 ,shouldn't it be: 23=x^2+y^2 not possible?
What about 2? 2 is prime, And can be expressed as 1^2+1^2=2. additionally 2-1=1 And 1 is not divisible by 4.
@Numberphile: Does that mean the other primes (those that can be written as 4k - 1) can be written as x^2 + y^2 - 2? More generally, does the 4k + 1 rule apply to all odds, or only to primes?
The answer to my second question is no: Unless we allow x to be 0, it doesn't apply to 9; even we do allow x to be 0, it doesn't apply to 21.
@@PaulHartzer A number can be written as the sum of two squares if and only if there is a prime p = 3 mod which divides the number an odd number of times. 21 = 3 *7, so indeed it cannot be written as the sum of two primes even though 21 = 1 mod 4.
Every even number squared is a multiple of 4, and every odd number squared is one more than a multiple of 4. Therefor every sum of an even square and an odd square is 4k+1 for some natural number k. Every sum of two even squares or two odd squares will be an even number, and therefor can't be prime (except 1+1). In addition, every square of a number 4k+1, must also be one more than a multiple of 4.
Is there any investigation of the non-primes that are the sum of two squares and are one more than a multiple of 4? That is all non-primes that are the sum of an even square and an odd square. 25, 45, 65, 85, 117, 125, 145, 153, ...
Set up a spread sheet and decompose the first 289 numbers. 0 to 289 will cover it sufficiently for you to get a start on it. Yes, all WHOLE (W) numbers can be expressed in terms of sums of powers of 2. I say Whole in order to break the limit requirement of "sum of two square numbers" and allow "sum of as many squares as are needed". Some are like 9, which is simply 3^2. Some will pop up with two squares repeatedly like 10 = 3^2 + 1^2. Some are like 21 which has three solutions(4^2+ 2^2 + 1^2) or (3^2 + 3^2 + 1^2 + 1^2 + 1^2) or (3^2 + 2^2 +2^2 + 2^2). To create the decomposition in the first order, do this: first find the Integer Value of the Square Root of W. Then, square that irW= (int(sqrt(W)))^2) Then subtract W-(I(W)^2). Repeat with the remainder until the remainder is 3 or less. So, INT(SQRT(21)) = 4. 4*4 = 16 so 21-16=5. INT(SQRT(5)) = 2. 2*2=4, 5-4=1. This tells you that 21 = (4^2+ 2^2 + 1^2), Doing this as a first step will bring out the generalized pattern to how all the numbers can be decomposed to a summation of squares. Not just two but as many as are needed. You can repeat this process with using a seed value of INT(SQRT(W))-1, and there by gain or not any additional solutions to each number W (as in the case of 21 (3^2 + 3^2 + 1^2 + 1^2 + 1^2) or (3^2 + 2^2 +2^2 + 2^2) ). You then set up test fields and put out a notifier if you have a solution that is in the form W=X^2+Y^2. You can then determine if there is any other patterns to that system of solutions or not. (Edit: there comes a point where there is a great many ways of decomposing, not just with this linear sequential approach, and the only way to handle multiple outcomes is with set theory and the one rider that only 1,2 and 3 can be 1^2, 1^2+1^2 and 1^2+1^2+1^2)
It's a cool pattern, but I have a small point.
when he says the rule he says "if and only if" then says the 'minus 4, divisible by 4' rule.
2 breaks this rule.
2-1 is not divisible by 4. so either the rule needs the extra rule including the special case or he shouldn't say "if and only if"
Let p be an odd prime. Very common to start off number theory proofs with that statement, he simply forgot to add it or did so purposefully for simplicity.
I wish numberphile posted certain pdf paper links of proofs or what not that explained their videos into greater details for those interested.
Harry potter called, he wants his glasses back
Aren't all natural numbers positive?
Depends on whether you include 0 or not.
Nope. -1 is also a natural number.
yes - I think he was just wanting to emphasise the point for the layman - saying positive natural numbers may be redundant to someone who already knows, but I don't think the meaning is lost!?
+Nareth Erakian
natural numbers are a subset of the integers which are all positive (or 0 idk)
No, natural numbers are just positive integers.
{P is sum of 2 squares "if and only if" (P-1) % 4 = 0} suggests double implication.
How about the exception:
Exception : P = 2: P - 1 = 1; 1% 4 != 0; but 2 = 1 ** 2 + 1 ** 2;
Either the video does not mention that or I missed it. One cannot supply "patches" for exceptions in maths. There also exists a "Proof" video; trying to see where "2" (was or got) excluded in the proof.
So 1 is devideable by 4? As 2-1 = 1 and 2 = 1^2+1^2? Or is it the only exception? Are there others? What makes it special?
All primes with more than 2 digits are equally divisible by 6 if you first subtract or add 1. Found that from doing a variant ulam spiral in python. Not all numbers that are like this are primes, but all primes are. Easier to detect. The last digit of the numbers is always 1,3,5,7,9 and you can toss out the 5 values because those are all divisible by 5. It's really an easy test so you don't have to filter through everything else.
A prime that is not 2 or 3 will have to can't be equal to 0, 2, 3 or 4 mod 6. So it will be 1 or 5 mod 6, which means either subtracting 1 or adding 1 will make it 0 mod 6 and thus divisible by 6.
Incredible! Also, stop fetishizing his accent as either positive or negative. Yeah, we as an anglicized audience might find it aesthetically pleasing or dreadfully unintelligible (in which case, educate yourself), but why not appreciate his eloquence and the fact that he's educating us about a brilliant mathematical feat instead? :)
The proof of Zagier shows that p can be written as the sum of two squares if it is of the form p=4k+1.
But the other half of the claim, namely that other odd primes can *not* be written as the sum of two squares, was not mentioned anymore.
It is very simple to prove, in fact any number of the form 4k+3 (for nonnegative integer k) cannot be written as the sum of two squares. Here's my one sentence proof:
Assume it could, it would obviously be the sum of an odd and an even square, but for integer a, b,:(2a+1)^2+(2b)^2 = 4(a^2+a+b)+1.
4k + 1 where K =5 results in 21. I realize that's not prime, but it seems this proof should work for all natural numbers that can be written in the form 4k + 1, but it doesn't not work with 21. 5^2 by itself = 25 and is too high. The lower combos don't work.
It turns out you the fact that p is prime to show that the involution in has only 1 fixed point (1,1,k),. A fixed point (x,y,z) must have x=y, so p = x^2 + 4yz = x^2 + 4xz = x(x + 4z). Since p is prime, this means x=1, so the only fixed point is (x,y,z) = (1,1,k) where p=4k+1. If p=21, then you can take x=3, z=1, and you get another fixed point (3,3,1).
One sentence proof
not once
cheers
tnx
Hi numberphile I big fan!
@Matthias: Kann es sein, dass sein deutscher Wikipedia-Artikel den "One-Sentence-Proof" nicht erwähnt?
Is it possible his german Wikipedia entry doesn't feature this proof?
When a prime can be written as a sum of two squares if the prime minus 1 is divisible by 4, that would mean every prime of the form 2^n + 1 can be written as a sum of two squares. Right?
Yes, if n>1 :)
Cool, a professor from Uni Bonn! That's the university where I'm studying (in a different field though) :) I know a few people studying mathematics there, should ask them if they know Prof. Kreck!
the proof i remember is as follows, assume p is odd if p=3(4) then it clearly cant be a sum of two squares as the squares are 0,1 mod 4.
If p=1 mod 4
then we take some primitive root g mod p then g^(p-1)=1 modp and as g is a primitive root g^(p-1/2)=-1 modp as it cant be 1 as has to have order p-1 but also must square to 1. as p-1 is a multiple of 4, p-1/2 is even so we can rewrite this as a^2=-1 mod p for some a
we may now use the following theorem
if in n dimensions we have a symmetric convex set S and a lattice L such that Volume(S)>=2^n det(L) then S contains a non zero point of the lattice. Here det(L) essentially refers to the fundamental volume of the lattice eg a parralelogram formed by the lattice, in this sense the theorem is intuitive
if now we fix p and let S={x,y integers| x^2+y^2
Is 2 an exeption? 2-1=1, 1 isnt divisible by 4, but 2 can be written as 1^2+1^2
Nice to see some German professors here.
I don't know how to express this, but I think I might have found the pattern of primes... I can almost create an algorithm, but, maybe someone has tried this before and seen this pattern, because it's too obvious that someone hasn't seen it...
There are 2 problems with it... which a programmer or a mathematician could fix though.
Basically I make a single assumption and then ask what breaks that assumption. This means that what I'm doing is exclusionary. The problem is that as you go higher, this becomes problematic because we're looking for which are primes and not which are not and as numbers get higher there are more excluded rather than included. I assume if I showed this pattern to a programmer or mathematician they'd be able to flip this.
The other problem is that there is a secondary issue in the pattern that I don't know how to account for in a single algorithm... Think of it like this... I'm getting a list where the pattern is AAAAAAA where A tells me what to exclude, but then I'm geting AAAABABAAABAAAAAAAB where both A and B tell me what to exclude, but B would never come up in an algorithm, because the number of As is infinite I don't see a pattern for when B ( and C and D for that matter) interjects.
I would just tell the pattern to see if someone else has seen this, but I understand this answer is worth millions and I need moneys so >.> yeah. I think I might figured out half of the prime problem, but then so have a number of other people in the past.
The shape of his glasses is positively hilarious.
'He wrote down a theorem without a proof'. Gödel would be proud of him.
when you're showing how 23 doesn't work using the theorem, you use 22 at the bottom... is this an editing mistake or how it's supposed to be done. If you watch the first example with 7 you use 7 again at the bottom, so I figure it's just an editing mistake, but I wanted to ask anyway 😅
That accent though. For one moment I thought he was going to say "here is what I came up with, let me show you its features" :))
To help with finding out if a number is divisible by 4:
Only look at the last two digits; if those two are divisible by 4, then the whole number is.
Does not work for 2. 2-1 is not divisible by 4. So the divine plan is not working in small scale?
Hmm. 2-1 is not divisible by 4, but 2 can be written as a sum of the squares of natural numbers. He should have specified that either x or y can be 1, but not both. Or am I missing something?
fantastic video, very well explained.
If X^2 and Y^2 have any common factors other than 1, then X^2 + Y^2 will be composite. Therefore, if X and Y have common factors other than 1, X^2 + Y^2 will be composite. So we know that the solutions are a subset of all pairs of coprime numbers. Also, since all primes greater than 2 are odd, X and Y must have opposite parities, except for the pair 1, 1. So the trick is to check all pairs of numbers with one odd, one even that are coprime.
I thought: "He sounds german." And the description reveald my assumption as true. :D
i don't understand why it's supposed to be true that only primes p such that (p-1)mod4 = 0 can be written as the sum of two squares considering that 2 = 1^2+1^2 and (2-1)mod4=1
but... prime 2, 2-1=1, 1 not divisible by 4... is it just because 2 < 4?
So 2 is an exception to that rule (p-1 must be divisible by 4), right?
Why does he say that you will not be able to find x and y? It is easier (takes less time) to find x and Y than to verify that p is a prime at all.
Due to his accent, it was a bit hard to follow. And I'm from Germany myself. But that actually doesn't matter, since it was explained well enough that it's understandable even though the accent barrier exists. Well done. ;)
That means if I have a twin prime pair, only one of the primes can be written as the sum of two squares, right?
Yep
what is a twin prime pair?
+Dosluke two prime numbers that are 2 apart, such as 5 and 7, 29 and 31, etc.
And then sauteed like shrimp! :D
Every odd integer can be written as the difference between two square integers, excluding 1.
reading the book "Fermats Enigma" right now it's awesome
Makes so much sense now! Thanks for the video!
How about 2? 2 is a prime and 2 minus 1 is 1. But 1 is not divisible by 4. So why can the prime 2 get in two square numbers?
there is a mistake on the video:
here is the one example with number 23
23 -1 =22
now 22 is not divisble by 4 so we know that 23 won't satisfy the addition of two squares.
however the video goes on to tell that is 22 instead the number that won't satisfy the addition of two squares.
I know this could be a video editing mistake but it could be enough to throw some people off.
cheers!
thank you for this very comprehensible video !!!
What happened to Lagrange between Euler and Gauss @ 5:21?
I really like the way he talks. Or accent
ITS MY GRANDPA!!!
O really. If yes then you'll be so lucky.
At 4:10, it should say "23=x^2 + y^2 X"
I wish Numberphile had an another channel where they would show Euler and Gauss' proofs, I can actually understand those topics, I just wish I could watch them in neat, Numberphile form... I suppose audience retention would be too low... Unfortunate.
This is one of my favorite topics!
Any number that is the same when squared is a bit suspect. 1 is not what I call a genuine working number. The first in any mathematical sequence is often anomalous in it's behaviour.
Don Zagier basically proposed a system of equations that, when solved, gives you the primes you want. Clever.
5:22 why lagrange was not mentioned??
But 2-1 is not divisible by 4, is it an exception?
Hey Numberphile can you proof that there is a primenumber between (x)squarede and (x+1)squared
Dude, Euler is everywhere.
I think the theorem should have said "Let p be an odd prime"
So for any number A, if A-1 is divisible by 4, then A is a prime number??? that's a strong theorem
21
Nice troll.
Did you watch the whole video?
I love Numberphile videos omg
+
I thought the theorem was that the p = x² + y² only if p is congruent to 1 (mod 4), and the theorem was not that all primes (except 2) could be solved by x² + y².
And if I am correct, then unless I misunderstand what Fermet’s theorem on sums of two squares means, which is very possible, isn't the solution simple as I show below? I'm a lawyer, not a mathematician. And this is also why my proof follows no standard format.
The sum of any two integers where one is even and one is odd will always sum to an integer which can be expressed as 4K + 1.
Because they can never sum to 4K + 3, any 4K + 3 prime could never be solved by x² + y².
Except for 2, all prime integers are odd integers. The sum of any two integers is only an odd integer when one integer is even and one is odd. An even integer is always divisible by 2. These are fundamental and do not require proof. Similarly, the square of any even number is divisible by 4: [N² = (2x)² = 2²x² = 4x²].
As for the square of the odd number, if M is 1, then the sum of N² + M² = 4x² + 1.
But other than 1, all odd integers are an even integer +1. Thus, if M is an odd integer not equal to 1, then M = 2y +1. So M² = (2y + 1)² = 4y² + 4y + 1.
Thus, the sum of two squares, where one is even and the other is odd is: 4x² + 4y² + 4y + 1. And this can be factored to 4(x² + y² + y) + 1, or 4K + 1. This shows that the sum of two squares where one is an even number and one is odd will always take the form of 4K + 1. Thus only an odd prime which is in the form of 4K + 1 might be solved by the sum of two squares. But any prime number which is in the form 4K + 3 can have no sum of two squares solution.
"The sum of any two integers where one is even and one is odd will always sum to an integer which can be expressed as 4K + 1"
No, 4 is even, 3 is odd, and 4+3=7 cannot be expressed as 4k+1.
"This shows that the sum of two squares where one is an even number and one is odd will always take the form of 4K + 1. Thus only an odd prime which is in the form of 4K + 1 might be solved by the sum of two squares. But any prime number which is in the form 4K + 3 can have no sum of two squares solution."
But what you need to show is that an odd prime of the form 4K+1 is ALWAYS the sum of two squares. This you did not do.
for anyone wondering, the biggest prime known to humankind is NOT the sum of two squares
Subtract 1, divide by 4. Why 4?
4 is the only even square of a prime
The theorem is actually a prime p is the sum of two squares iff p=2 or p=1[4].
Please make one more video to explain the assumption, seems like the whole argument is based on that