Hello everybody!

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

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

**Case 1**

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

**Case 2**

When \(a=b\)

**Case 3**

When \(a\) and \(b\) are both primes and , \(a>b\)

## Comments

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TopNewest\(Consider\quad f\left( n \right) =\phi (n)\quad \quad \quad Where\quad \phi (n)\quad is\quad Euler\quad Totient\quad Function\\ f\left( 5186 \right) =f\left( 5186+1 \right) =f\left( 5186+2 \right) =2592\quad \\ =>\quad f\left( 5186 \right) =f\left( 5187 \right) =f\left( 5188 \right) =2592\quad \\ 5186\quad is\quad the\quad only\quad number\quad which\quad satisfy\quad f\left( x \right) =f\left( x+1 \right) =f\left( x+2 \right) \\ and\quad is\quad less\quad than\quad { 10 }^{ 10 }.\) – Rishabh Deep Singh · 1 year ago

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