Mathematics, 27.08.2020 22:01 kaylastronaut

Let p be a prime number. The following exercises lead to a proof of Fermat's Little Theorem, which we prove by another method in the next chapter. a) For any integer k with 0 ≤ k ≤ p, let (p k) = p!/k!(p - k)! denote the binomial coefficient. Prove that (p k) 0 mod p if 1 ≤ k ≤ p - 1. b) Prove that for all integers x, y, (x + y)^p x^? + y^p mod p. c) Prove that for all integers x, x^p x mod p.

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Answer from: niki209

Hello,

a) We know the binomial coefficients are all integers, so

$\dfrac{p!}{k!(p-k)!}$

is an integer.

And we can notice that the numerator p! is divisible by p.

If we take $1\leq k\leq (p-1)$

It means that k! does not contain p, and we can say the same for (p-k)!

So, we have no p at the denominator so the binomial coefficient is divisible by p, meaning this is 0 modulo p.

b) We can write that

$\displaystyle (x+y)^p=\sum_{i=0}^{p} \ {\dfrac{p!}{i!(p-i)!}x^{p-i}y^i$

We use the result from question a) and the binomial coefficients are 0 modulo p for i=1,2 , ... p-1 so there are only two terms left and then,

$(x+y)^p=x^p+y^p \text{ modulo p}$

c) Let's prove it by induction.

step 1 - for x = 0

This is trivial to notice that

$0^p=0 \text{ modulo p}$

Step 2 - we assume that this is true for k

meaning $k^p=k \text{ modulo p}$

and we need to prove that this is true for the k+1

We use the results of b)

$(k+1)^p=k^p+1^p=k^p+1 \text{ modulo p}$

and we use the induction hypothesis to say

$(k+1)^p=k^p+1^p=k^p+1=k+1 \text{ modulo p}$

And it means that this is true for k+1

Step 3 - conclusion

We have just proved by induction the Fermat's little theorem.

p a prime number, for for all x integers

$\Large \boxed{\sf \bf x^p=x \textbf{ modulo p}}$

Thank you

Answer from: Quest

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

Answer from: Quest

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

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Let p be a prime number. The following exercises lead to a proof of Fermat's Little Theorem, which w...

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