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Alecsey [184]
3 years ago
14

Twin Primes (a) Let p > 3 be a prime. Prove that p is of the form 3k +1 or 3k – 1 for some integer k. (b) Twin primes are pai

rs of prime numbers p and q that have a difference of 2. Use part (a) to prove that 5 is the only prime number that takes part in two different twin prime pairs.
Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

(a) Let us recall the division algorithm: given two positive integers n and p there exist other two positive integers k and r such that

n = pk+r where r and r is called the <em>remainder</em>.

So, given any positive integer n and 3 we can write

n=3k+r where r=0,1,2. Thus, every n can be written as

  • n=3k
  • n=3k+1
  • n=3k+2

Now, notice that n=3k+2 = 3k+3-1 = 3(k+1)-1 =3k'-1. Hence, every number can be written as n=3k, or n=3k+1 or n=3k-1.

A number p is prime if and only if its only factors are 1 and p itself. So, a number of the form n=3k cannot be prime. Therefore, every primer number is of the form n=3k+1 or n=3k-1.

(b) Assume that there are three prime numbers such that p, p+2 and p-2 are prime.

By the previous exercise p=3k+1 or p=3k-1. Let us analyze both cases separately.

<em>First case</em>: p=3k+1. Then p-2=3k-1 that can be prime, and p+2=3k+3 that is not prime. Hence, there are not such three primes with p=3k+1.

<em>Second case</em>: p=3k-1. Then, p+2=3k+1 that can be prime, and p-2=3k-3=3(k-1) that cannot be prime. Hence, there are not such three primes with p=3k-1.

Therefore, there are no three primes  of the form p, p+2 and p-2, except for 3, 5 and 7.

Notice that this is only possible because 5=2*3-1 and 2*3-3=3, that is the only ‘‘multiple’’ of 3 that is prime.

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2.794

Step-by-step explanation:

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F(x,y,z) = (xy, yz, zx)

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\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

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\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

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\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

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