(x^p )+(y^p) ≡ (x+y)^p) why?

Question

If x,y,p belong N and p is prime number
(N is Natural Number )

Answer 1

If you use the Binomial Theorem on the right side, you get

x^p + (p choose 1) * x^(p - 1)y + (p choose 2) * x^(p - 2)y^2 + ... + (p choose (p - 2)) * x^2y^(p-2) + (p choose (p - 1)) * xy^(p - 1) + y^p.

It turns out that p divides (p choose k) for 1 <= k <= p - 1. This is true because (p choose k) equals p!/(p - k)!(k!). Neither (p - k)! nor k! is divisible by p (since all their prime factors are less than p), but p! is, so (p choose k) is divisible by p.

Hence, when you expand out (x + y)^p, all the terms except x^p and y^p are necessarily congruent to 0 (mod p). So the result you have holds.

Answer 2

I don't think this is right.
Suppose x is 1 and y is 1 and p is 3
then 1^3 + 1^3 does NOT equal (1+1)^3
the left side is 2 and the right is 8

Answer 3

Congruent mod (p)?

I think there is a theorem that states (but check your textbook to be sure this is true. It's been a long time since number theory):

x^p = x mod (p)

and so

y^p = y mod (p)

(x+y)^p = (x+y) mod (p)

Adding the first two together, we get:

x^p + y^p = (x+y) mod (p)

according to the third one, x+y = (x+y)^p mod (p), and so:

x^p + y^p = (x+y)^p mod (p)

qed.

Again, double check that my assumption is true. Hope this helps.

Answer 4

It is wrong the way it is stated...
U can't factor out indices...
ex: (x^2 + y^2) is NOT equal to (x + y)^2

Answer 5

No.
x^2+y^2 does not equal (x+y)^2
(x+y)^2=x^2+2xy+y^2
They are not the same.

Answer 6

its just factoring out the ^p

Answer 7

it just does.