Case Study

Hello everybody!

as you know that if , aba\geq b ,then it is not necessary that ϕ(a)ϕ(b)\phi(a)\geq\phi(b) . Since ϕ\phi not an increasing function.

But there are some cases in which above inequality holds true.

Case 1

When either of aa and bb , suppose lets take aa as an arbitary prime , then b=a+1b=a+1 or b=a1b=a-1. Same follows if bb is prime.

Case 2

When a=ba=b

Case 3

When aa and bb are both primes and , a>ba>b

#NumberTheory

Note by A Former Brilliant Member
5 years, 3 months ago

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Comments

Considerf(n)=ϕ(n)Whereϕ(n)isEulerTotientFunctionf(5186)=f(5186+1)=f(5186+2)=2592=>f(5186)=f(5187)=f(5188)=25925186istheonlynumberwhichsatisfyf(x)=f(x+1)=f(x+2)andislessthan1010.Consider\quad f\left( n \right) =\phi (n)\quad \quad \quad Where\quad \phi (n)\quad is\quad Euler\quad Totient\quad Function\\ f\left( 5186 \right) =f\left( 5186+1 \right) =f\left( 5186+2 \right) =2592\quad \\ =>\quad f\left( 5186 \right) =f\left( 5187 \right) =f\left( 5188 \right) =2592\quad \\ 5186\quad is\quad the\quad only\quad number\quad which\quad satisfy\quad f\left( x \right) =f\left( x+1 \right) =f\left( x+2 \right) \\ and\quad is\quad less\quad than\quad { 10 }^{ 10 }.

Rishabh Deep Singh - 5 years, 3 months ago
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