Anonymous at Mon, 27 May 2024 13:47:16 UTC No. 16195790

>>16195713
Copy paste is not a gift.
>They only focus on one thing.

Anonymous at Mon, 27 May 2024 14:05:12 UTC No. 16195810

>>16195713
there's only one solution, p=5, q=2, r=3. since p = (r+q)*(r-q), r-q has to equal 1. no other set of primes aside from what's mentioned above satisfies this since r - q = 1 implies one of these is even and 2 is the only even prime.

Anonymous at Mon, 27 May 2024 14:15:41 UTC No. 16195831

>>16195810
Also 3 + 1^2 = 2^2.

Anonymous at Mon, 27 May 2024 14:19:26 UTC No. 16195834

>>16195831
1 isn't a prime number.

Anonymous at Mon, 27 May 2024 14:22:54 UTC No. 16195845

>>16195834
>but its divisible by one and itself.
its not prime by convention >:(

Anonymous at Mon, 27 May 2024 14:25:02 UTC No. 16195850

>>16195810
great. thread closed.

>>16195713
you can delete it now OP

Anonymous at Mon, 27 May 2024 14:38:14 UTC No. 16195878

>>16195845
It's not prime by definition. Elementary schools teaches that numbers only divisible by one and itself are prime, but that's never been the case. Prime numbers are the building blocks of composite numbers. You can't multiply 1 with anything to get a a new number. While all prime share the PROPERTY that they are only divisible by 1 and itself not all numbers that have this are prime. It's like how all hoovers are vacuums but not all vacuums are hoovers, even though some people use them interchangeably.

Anonymous at Mon, 27 May 2024 14:39:22 UTC No. 16195882

>>16195834
> You meet the qualifications but we still choose to exclude you.
That's not mathematics, that's racism.

Anonymous at Mon, 27 May 2024 14:51:49 UTC No. 16195907

which would you rather have
>the extremely useful property that every natural number greater than 1 can be represented as a product of primes in a unique way up to the order of multiplication
or
>hehe 1 prime :DDD
this is apparently an impossible decision for some

Anonymous at Mon, 27 May 2024 15:03:47 UTC No. 16195929

>>16195713
Write the possibilities out modulo 2:
A) 0+0=0
B) 1+1=0
C) 1+0=1
D) 0+1=1
There are clearly no solutions for A or B since 2 is the only even prime.
For D, you have 2+q^2=r^2. However, there are no such primes since, at a minimum, two unique primes are (p+2)^2-p^2 = 4p + 4 > 2 apart from each other.
For C, you have p+4=r^2. Using a prime gap of d between p and r (so r=p-d), we get the equation (p-d)^2-p = p^2 - (2d-1)p + d^2 = 4. We can solve the quadratic equation for d here; one example solution is (p,r)=(5,3).

Anonymous at Mon, 27 May 2024 15:35:54 UTC No. 16195985

>>16195882
>all even numbers share the property of being integers, therefore all integers are even.
Mathematics is all about exclusion.

Anonymous at Mon, 27 May 2024 17:02:48 UTC No. 16196125

>>16195713
The answer is sex with Kirisu

Anonymous at Tue, 28 May 2024 07:36:28 UTC No. 16197322

>>16195929
inspecting the solution of the quadratic all solutions are of the form
q = 2
r = sqrt(p+4)

but it does not rule out further solutions besides (5,2,3)