The Prime Problem with a One Sentence Proof - Numberphile
video description
1) Every prime number >2 is odd.
2) odd+odd = even+even = even number & ODD + EVEN = ODD number
3) odd-even = EVEN - EVEN = EVEN & odd-odd = odd
Because of 1) we're looking for two natural numbers x-2 and y-2, so that their sum is odd, if the prime p=x-2+y-2 is >2. With 2) we know, that one of those numbers is an odd number and the other is an even number. The numbers, which we're looking for, are squares, what means, that each of them is a product of two times the same number x-2=x-x (or y-2=y-y. 3) let's know, that the even number has to be calculated by two even numbers. Let's say, that x-2 is the even number and so is x=2-a. That's why x-2=x-x=2-a-2-a=4-a-2. To proof the conjecture now, we have to set y=1, so y-2=1 too, which conclutes to p = 1 + 4-a-2. This is why p-1=4-a-2 is divisdable by 4. But this is also true for any other other odd y-2, where p-y-2=x-2.
Note that this is not only the case, if the number p is an odd prime number. Instead every odd number, which is modulo 4 equals 1, has this property.
Please tell me, if I made a mistake or anything like that.
Have a nice day, dear reader.
Date: 2022-04-08
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Comments and reviews: 9
ED
Is this conjecture true:
a = any whole number
x = number of digits in a
b = multiple of a
b > 2a-x
n = -b-(a-1)-x-
n = prime or divisible by a-x
Example 1:
a = 4
x = 1
b = 1000
n = -1000-(4-1)-1-
n = -1000-(3)-1-
n = -1000-3-
n = -397-
n = 397
n is prime
Example 2:
a = 11
x = 2
b = 253
n = -253-(11-1)-2-
n = -253-(10)-2-
n = -253-100-
n = -153-
n = 153
n is divisible by a-x or 9
Example 3:
a = 7
x = 1
b =49
n = -49-(7-1)-1-
n = -49-(6)-1-
n = -49-6-
n = -43-
n = 43
I have tested this conjecture a number of times on paper. I think it might be false, but I have failed to find a counter example. May someone please find one?
reply
Is this conjecture true:
a = any whole number
x = number of digits in a
b = multiple of a
b > 2a-x
n = -b-(a-1)-x-
n = prime or divisible by a-x
Example 1:
a = 4
x = 1
b = 1000
n = -1000-(4-1)-1-
n = -1000-(3)-1-
n = -1000-3-
n = -397-
n = 397
n is prime
Example 2:
a = 11
x = 2
b = 253
n = -253-(11-1)-2-
n = -253-(10)-2-
n = -253-100-
n = -153-
n = 153
n is divisible by a-x or 9
Example 3:
a = 7
x = 1
b =49
n = -49-(7-1)-1-
n = -49-(6)-1-
n = -49-6-
n = -43-
n = 43
I have tested this conjecture a number of times on paper. I think it might be false, but I have failed to find a counter example. May someone please find one?
reply
PhilBagels
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, .
reply
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, .
reply
Renato
There's a much easier proof imo: To write a prime number bigger than 2 as two squares, you allways need an odd and an even square number. The even square is obviously dividable by 4 since its root has a 2 in it which is squared. To prove that the odd square -1 is allways dividable by 4 you take the fact that the difference between two squares allways increases by 2:
Let's start with 1. 1-1 equals 0 which is dividable by 4. The difference between 1 and 4 is 3, and the difference between 4 and 9 is 5. 16-9=7 and 25-16=9. So the difference between two odd squares is allways dividable by 8 and therefore dividable by 4.
reply
There's a much easier proof imo: To write a prime number bigger than 2 as two squares, you allways need an odd and an even square number. The even square is obviously dividable by 4 since its root has a 2 in it which is squared. To prove that the odd square -1 is allways dividable by 4 you take the fact that the difference between two squares allways increases by 2:
Let's start with 1. 1-1 equals 0 which is dividable by 4. The difference between 1 and 4 is 3, and the difference between 4 and 9 is 5. 16-9=7 and 25-16=9. So the difference between two odd squares is allways dividable by 8 and therefore dividable by 4.
reply
T. C.
Ahhh. I knew a method to get the value of x and y. I made it by number theory. I just swirl the equation and got a square to be of the form 4n+1 satisfy that and subtracting 1 we get 4n+1-1 giving p-1 to be divisible by 4. Then, I equated the second term of the equation to be =1 as p is a prime and a wonderful relation between x and y which can easily give their value either by a Diophantine but for very large primes it becomes hard by hand.
reply
Ahhh. I knew a method to get the value of x and y. I made it by number theory. I just swirl the equation and got a square to be of the form 4n+1 satisfy that and subtracting 1 we get 4n+1-1 giving p-1 to be divisible by 4. Then, I equated the second term of the equation to be =1 as p is a prime and a wonderful relation between x and y which can easily give their value either by a Diophantine but for very large primes it becomes hard by hand.
reply
John
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.
reply
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.
reply
Ismael
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!
reply
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!
reply
Jayashri
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: )
reply
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: )
reply
Z-mZ-m
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? :)
reply
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? :)
reply
Jacob
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-
reply
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-
reply
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