Some additional stuff: In general, gcd(x^a - 1, x^b - 1) = x^gcd(a,b) - 1 for any a and b, which implies that x^n = 1 mod p has exactly gcd(p-1, n) solutions. (This will probably be also shown later when primitive roots are discussed, but I haven't seen that yet.) The proof basically only relies on the fact that if a >= b, then x^a - 1 = x^(a-b)(x^b - 1) + x^(a-b) - 1 which is thus congruent to x^(a-b) - 1 mod x^b - 1, so gcd(x^a - 1, x^b - 1) = gcd(x^(a-b) - 1, x^b - 1), which is the key step of Euclid's algorithm, so doing this repeatedly, we're basically doing Euclid's algorithm until one of the exponents becomes zero, in which case we have gcd(x^a - 1, x^0 - 1) = x^a - 1, which means the exponent must end up being the gcd. In general, gcd(x^a - y^a, x^b - y^b) = x^g - y^g where g = gcd(a,b), as long as gcd(x,y) = 1. (Note that if x and y are indeterminates---that is, we work on the PID k[x,y]---then this is automatically true.) The proof is rather similar to the above.
Surprise! If multiply denominator to right side get σ_p-2 = σ1*σ_p-1 mod p, but right side only divisible by p, i.e.σ_p-2 ≠ σ1*σ_p-1 mod p^2. So sometimes can get stronger result when look at original polynomial.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Some additional stuff:
In general, gcd(x^a - 1, x^b - 1) = x^gcd(a,b) - 1 for any a and b, which implies that x^n = 1 mod p has exactly gcd(p-1, n) solutions. (This will probably be also shown later when primitive roots are discussed, but I haven't seen that yet.)
The proof basically only relies on the fact that if a >= b, then x^a - 1 = x^(a-b)(x^b - 1) + x^(a-b) - 1 which is thus congruent to x^(a-b) - 1 mod x^b - 1, so gcd(x^a - 1, x^b - 1) = gcd(x^(a-b) - 1, x^b - 1), which is the key step of Euclid's algorithm, so doing this repeatedly, we're basically doing Euclid's algorithm until one of the exponents becomes zero, in which case we have gcd(x^a - 1, x^0 - 1) = x^a - 1, which means the exponent must end up being the gcd.
In general, gcd(x^a - y^a, x^b - y^b) = x^g - y^g where g = gcd(a,b), as long as gcd(x,y) = 1. (Note that if x and y are indeterminates---that is, we work on the PID k[x,y]---then this is automatically true.) The proof is rather similar to the above.
At 11:32, the coefficient of x^2 for x(x-1)...(x-6) should be -1764, instead of -176 (a typo, I believe).
At 36:11, (y^n - 1) should be divisible by (y - 1), instead of y, another typo.
Surprise! If multiply denominator to right side get σ_p-2 = σ1*σ_p-1 mod p, but right side only divisible by p, i.e.σ_p-2 ≠ σ1*σ_p-1 mod p^2. So sometimes can get stronger result when look at original polynomial.
7 | 176?
give or take :)
Typo it should be 175 If I am not mistaken
yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
This is not sigma, this is six.