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Old 2018-03-01, 20:16   #1
JM Montolio A
 
Feb 2018

1408 Posts
Post The D.H.Lehmers problem. Find the mistake.

Never totient(n) | (n-1)
Proof 
Definition f(a,b) = (a*b)+(a+b).
With adecuate extension to any number of parameters.
Let n not prime, and square free.
For example, n= 2*3*5 = 30
totient(n) = (2-1)(3-1)(5-1) = 1*2*4
(n-1) = f( (2-1),(3-1),(5-1)) = f(1,2,4) = f(5,4) =29
If totient(n)| (n-1) , then a product of some numbers divides to the f() of them.
But any f() of any numbers is finally f() of only two numbers. f(a,b).
Then (a*b)| f(a,b)=[(a*b)+(a+b)]
Then (a*b)|(a+b). What never happens.
¿ Where is the mistake ?
Thanks.
JM M

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Old 2018-03-02, 03:48   #2
LaurV
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Quote:
Originally Posted by JM Montolio A View Post
Never totient(n) | (n-1)

Try 2.
Not only that you never heard of Wieferich primes, but is looks like you never heard of primes in general...

Last fiddled with by LaurV on 2018-03-02 at 03:54
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Old 2018-03-02, 11:32   #3
JM Montolio A
 
Feb 2018

25·3 Posts
Post Do you sincerely think are find a mistake on the proof ?

"2" are trivial cases, without any exit.


JM M
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Old 2018-03-02, 11:57   #4
science_man_88
 
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Quote:
Originally Posted by JM Montolio A View Post
"2" are trivial cases, without any exit.


JM M
One mistake is, that you defined a function with 2 inputs not three. a=8;b=10 gives a+b divides a*b.
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Old 2018-03-02, 12:03   #5
JM Montolio A
 
Feb 2018

25×3 Posts
Cool yes, giving 4.4. I tested all with a computer, sir.

¿ Someone find a better mistake ?
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