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Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.

Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)

which implies that 1/f(r) = f(k+1-r)/N

Then 1/f(1) + 1/(f(2) + ... + 1/f(k)

= f(k)/N + f(k-1)/N + ... + f(1)/N

= [f(k) + f(k-1) + ... + f(1)] / N

= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N

Q: Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?

This answer is:

Related answers

Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.

Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)

which implies that 1/f(r) = f(k+1-r)/N

Then 1/f(1) + 1/(f(2) + ... + 1/f(k)

= f(k)/N + f(k-1)/N + ... + f(1)/N

= [f(k) + f(k-1) + ... + f(1)] / N

= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N

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Given the function f(x) = 2x + 3 and a = -1, we can find f(a) as follows:

f(a) = 2(-1) + 3

f(a) = -2 + 3

f(a) = 1

So, f(a) = 1.

To graph f(x) and 1/f(x), we can plot several points and connect them to visualize the functions. Here are some points for f(x):

For f(x):

When x = -2, f(x) = 2(-2) + 3 = -1

When x = -1, f(x) = 2(-1) + 3 = 1

When x = 0, f(x) = 2(0) + 3 = 3

When x = 1, f(x) = 2(1) + 3 = 5

When x = 2, f(x) = 2(2) + 3 = 7

Now, to find 1/f(x), we take the reciprocal of each f(x) value:

For 1/f(x):

When x = -2, 1/f(x) = 1/(-1) = -1

When x = -1, 1/f(x) = 1/1 = 1

When x = 0, 1/f(x) = 1/3 ≈ 0.333

When x = 1, 1/f(x) = 1/5 ≈ 0.2

When x = 2, 1/f(x) = 1/7 ≈ 0.143

Now, we can plot these points and connect them to obtain the graphs of f(x) and 1/f(x).

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The function (sequence generator) f(x) with x∈ℕ has been defined as a recursive function (sequence), with the initial value defined for some x, ie starting form some some natural number (in this case 1), the value of the function (sequence) is given (in this case f(1) = 1), and each successive value of the function (sequence) is defined in terms of the current value f(x+1) = f{x} + g(x) where g(x) is a function with g(x) = 3x(x + 1).

f(x + 1) = f(x) + 3x(x + 1)

f(1) = 1

→ f(2) = f(1 + 1) = f(1) + (3×1)(1 + 1) = 1 + 3×2 = 1 + 6 = 7

→ f(3) = f(2 + 1) = f(2) + (3×2)(2 + 1) = 7 + 6×3 = 7 + 18 = 25

I'll let you evaluate the rest.

Hint:

f(4) = f(3 + 1) = f(3) + (3×3)(3 + 1)

f(5) = f(4 + 1) = f(4) + ...

f(6) = f(5 + 1) = f(5) + ...

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To:

trantancuong21@yahoo.com

PIERRE DE FERMAT's last Theorem.

(x,y,z,n) belong ( N+ )^4..

n>2.

(a) belong Z

F is function of ( a.)

F(a)=[a(a+1)/2]^2

F(0)=0 and F(-1)=0.

Consider two equations

F(z)=F(x)+F(y)

F(z-1)=F(x-1)+F(y-1)

We have a string inference

F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1)

F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1)

F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2)

we see

F(z-x-1)=F(x-x-1)+F(y-x-1 )

F(z-x-1)=F(-1)+F(y-x-1 )

F(z-x-1)=0+F(y-x-1 )

give

z=y

and

F(z-x-2)=F(x-x-2)+F(y-x-2)

F(z-x-2)=F(-2)+F(y-x-2)

F(z-x-2)=1+F(y-x-2)

give z=/=y.

F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2)

F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1)

F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1)

So have two cases.

[F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1)

or vice versa

[F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1).

F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1).

We have

F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2.

=(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4).

=x^3.

F(y)-F(y-1) =y^3.

F(z)-F(z-1) =z^3.

x^3+y^3=/=z^3.

n>2. .Similar.

We have a string inference

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

we see

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=0+G(y)*F(y-x-1 )

give z=y.

and

G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2)

x>0 infer G(x)>0.

give z=/=y.

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

So have two cases

[G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1)

or vice versa.

[G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)].

G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).]

We have

x^n=G(x)*[F(x)-F(x-1) ]

y^n=G(y)*[F(y)-F(y-1) ]

z^n=G(z)*[F(z)-F(z-1) ]

x^n+y^n=/=z^n

Happy&Peace.

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Последнее Пьер де Ферма теоремы.

(x,y,z,n) принадлежать( N+ )^4.

n>2.

(a) принадлежать Z

F является функцией( a.)

F(a)=[a(a+1)/2]^2

F(0)=0 и F(-1)=0.

Рассмотрим два уравнения

F(z)=F(x)+F(y)

F(z-1)=F(x-1)+F(y-1)

непрерывный дедуктивного рассуждения

F(z)=F(x)+F(y) эквивалент F(z-1)=F(x-1)+F(y-1)

F(z)=F(x)+F(y) выводить F(z-1)=F(x-1)+F(y-1)

F(z-x-1)=F(x-x-1)+F(y-x-1) выводить F(z-x-2)=F(x-x-2)+F(y-x-2)

мы видим,

F(z-x-1)=F(x-x-1)+F(y-x-1 )

F(z-x-1)=F(-1)+F(y-x-1 )

F(z-x-1)=0+F(y-x-1 )

давать

z=y

F(z-x-2)=F(x-x-2)+F(y-x-2)

F(z-x-2)=F(-2)+F(y-x-2)

F(z-x-2)=1+F(y-x-2)

давать

z=/=y.

так

F(z-x-1)=F(x-x-1)+F(y-x-1) не выводить F(z-x-2)=F(x-x-2)+F(y-x-2)

так

F(z)=F(x)+F(y) не выводить F(z-1)=F(x-1)+F(y-1)

так

F(z)=F(x)+F(y) не эквивалентен F(z-1)=F(x-1)+F(y-1)

Таким образом, возможны два случая.

[F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1)

или наоборот

так

[F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1).

или

F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1).

у нас есть

F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2.

=(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4).

=x^3.

F(y)-F(y-1) =y^3.

F(z)-F(z-1) =z^3.

так

x^3+y^3=/=z^3.

n>2. аналогичный

непрерывный дедуктивного рассуждения

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) эквивалент G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

мы видим,

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=0+G(y)*F(y-x-1 )

давать

z=y.

G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2)

x>0 выводить G(x)>0.

давать

z=/=y.

так

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

так

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

так

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

Таким образом, возможны два случая.

[G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1)

или наоборот.

так

[G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)].

или

G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).]

у нас есть

x^n=G(x)*[F(x)-F(x-1) ]

y^n=G(y)*[F(y)-F(y-1) ]

z^n=G(z)*[F(z)-F(z-1) ]

так

x^n+y^n=/=z^n

Счастливые и мира.

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