all fibonacci numbers are coprime (math)
So there I was, trying to actually explain the proof of Fermat's Last Theorem. The whole 100 or 200 page pdf. (I haven't even gotten to that because of background information and previous papers.) (Hmm, based on the screenshot link there might actually be a good easy-to-understand for my level proof. I picked a randomly dense section to seem "omg unknown math". Looking at that screenshot... https://doi.org/10.1214%2Faos%2F1176343654 probably isn't the most dense thing out there. Scary.)
And this problem simply claims: there are no integers a,b,c > 0 and n > 2 where:
a^n + b^n = c^n
So simple. But it took 3 and a half centuries to prove, and even then the proof came sooner than expected.
Anyways, there's an important subproblem for n=4
n=4 is important because for reasons I won't explain, the problem can be reduced to proving n for n=2*2 or n=prime>2
It has something to do with cancelling out all factors.
Here, I'm visualizing factors as multiplication of primes.
For example,
12 = 2 * 2 * 3
34 = 2 * 17
45 = 3 * 3 * 5
Numbers are coprime if they have no factors in common (gcd is 1)
So 12 has 2 in common with 34, and 3 in common with 45.
So 12 is not coprime with either 34 or 45.
But 34 is coprime with 45.
And now my rediscovery:
If a and b are coprime, then a + b is also coprime!!!!!!!!!
Consider 34 + 45
34 = 2 * 17
34 + 45 = 2 * 17 + (number coprime to 2 * 17) = number not divisible by 2 or 17
45 + 34 = 3 * 3 * 5 + (number coprime to 3 * 3 * 5) = number not divisible by 3 or 5
This is the sort of "visual "proof"" where if you understand it, it's basically like "by definition" and not very impressive, and if you don't, it's not very helpful for understanding. There's definitely a better explanation somewhere online.
And yeah, this applies to the fibonacci numbers. All fibonacci numbers are coprime:
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
8 + 13 = 21
....
Random exploration
Also, any a is coprime with a+1
Therefore a is coprime with: a+a+1, a+a+a+1, a+a+a+a+1,
and also a+a+a+1+1, a+a+a+a+1+1+1, and a+a+a+a+1+1
Thus within (a, ka), there are at least 1+1+2+3+4+5+6+... coprime numbers
Now, counting numbers within (k, 2k) coprime to k I get:
2: 1 (prime)
3: 2 (prime)
4: 2 (5, 7)
5: 4 (prime)
6: 2 (7, 11)
7: 6 (prime)
8: 4 (9 11 13 15)
9: 6 (10 11 13 14 16 17)
10: 4 (11 13 17 19)
11: 10 (prime)
12: 4 (13, 17, 19, 23)
And the OEIS gets the euler totient function! Hmm
