Suppose you a number N (128 bit encryption , 4096 bit encryption etc)

Suppose you had a fake solution K = prod_i p_i^a_i

Now take
lim G->infinity
(N/K)^(1/G) = exp( (log(N) - log(K))/G)

Both N and K are finite
= exp(0)
= 1

(N/K)^(1/G) = 1

now, power both sides by G
(N/K) = 1^G
= lim G->infinity 1^G = 1
N/K = 1
N = K

A random solution indistingusable from actual

If friend, K is the number we want, if enemy, K is an algebraic variable, which changes.

Suppose K1 is solution, then K2 is solution also. If K2 is solution, then K3 is solution, so on.

See most pgp messages are this is a test
https://www.google.com/search?q=%22this+is+a+test%22+pgp&sxsrf=ALeKk02t96NP4ywXsPPp51sT71O3104CYQ%3A1622305243365&ei=22myYOvlFdK4ggew-6GADQ&oq=%22this+is+a+test%22+pgp&gs_lcp=Cgdnd3Mtd2l6EAMyBwgjEK4CECc6BwgAEEcQsANQ3QtY-yNg9yVoAXACeACAAX2IAcgDkgEDNS4xmAEAoAEBqgEHZ3dzLXdpesgBCMABAQ&sclient=gws-wiz&ved=0ahUKEwirv_39pe_wAhVSnOAKHbB9CNAQ4dUDCA4&uact=5

-suresh

On Saturday, May 29, 2021 at 12:22:05 PM UTC-4, me wrote:
Suppose you had a **finite** number N (128 bit encryption , 4096 bit encryption etc)

> Suppose you had a fake solution K = prod_i p_i^a_i
>
> Now take
> lim G->infinity
> (N/K)^(1/G) = exp( (log(N) - log(K))/G)
>
> Both N and K are finite
> = exp(0)
> = 1
>
> (N/K)^(1/G) = 1
>
> now, power both sides by G
> (N/K) = 1^G
> = lim G->infinity 1^G = 1
> N/K = 1
> N = K
>
> A random solution indistingusable from actual
>
> If friend, K is the number we want, if enemy, K is an algebraic variable, which changes.
>
> Suppose K1 is solution, then K2 is solution also. If K2 is solution, then K3 is solution, so on.
>
> See most pgp messages are this is a test
> https://www.google.com/search?q=%22this+is+a+test%22+pgp&sxsrf=ALeKk02t96NP4ywXsPPp51sT71O3104CYQ%3A1622305243365&ei=22myYOvlFdK4ggew-6GADQ&oq=%22this+is+a+test%22+pgp&gs_lcp=Cgdnd3Mtd2l6EAMyBwgjEK4CECc6BwgAEEcQsANQ3QtY-yNg9yVoAXACeACAAX2IAcgDkgEDNS4xmAEAoAEBqgEHZ3dzLXdpesgBCMABAQ&sclient=gws-wiz&ved=0ahUKEwirv_39pe_wAhVSnOAKHbB9CNAQ4dUDCA4&uact=5
>
> -suresh