20180301, 20:16  #1 
Feb 2018
140_{8} Posts 
The D.H.Lehmers problem. Find the mistake.
Never totient(n)  (n1) 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) = (21)(31)(51) = 1*2*4 (n1) = f( (21),(31),(51)) = f(1,2,4) = f(5,4) =29 If totient(n) (n1) , 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 
20180302, 03:48  #2 
Romulan Interpreter
Jun 2011
Thailand
10001011100100_{2} Posts 
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 20180302 at 03:54 
20180302, 11:32  #3 
Feb 2018
2^{5}·3 Posts 
Do you sincerely think are find a mistake on the proof ?
"2" are trivial cases, without any exit.
JM M 
20180302, 11:57  #4 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 

20180302, 12:03  #5 
Feb 2018
2^{5}×3 Posts 
yes, giving 4.4. I tested all with a computer, sir.
¿ Someone find a better mistake ?

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