mersenneforum.org The D.H.Lehmers problem. Find the mistake.
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 2018-03-01, 20:16 #1 JM Montolio A   Feb 2018 1408 Posts 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
2018-03-02, 03:48   #2
LaurV
Romulan Interpreter

Jun 2011
Thailand

100010111001002 Posts

Quote:
 Originally Posted by JM Montolio A 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

 2018-03-02, 11:32 #3 JM Montolio A   Feb 2018 25·3 Posts Do you sincerely think are find a mistake on the proof ? "2" are trivial cases, without any exit. JM M
2018-03-02, 11:57   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by JM Montolio A "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.

 2018-03-02, 12:03 #5 JM Montolio A   Feb 2018 25×3 Posts yes, giving 4.4. I tested all with a computer, sir. ¿ Someone find a better mistake ?

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