# Prove for all $m \in \mathbb{N}$ exists $n \in \mathbb{N}$ such that $\varphi(n)-\varphi(n+1)>m$ and $\varphi(n)-\varphi(n-1)>m$

Number Theory :

Prove for all $$m \in \mathbb{N}$$ exists $$n \in \mathbb{N}$$ such that :

$$\varphi(n)-\varphi(n+1)>m$$ and $$\varphi(n)-\varphi(n-1)>m$$

Attempt:

For $$m\in \mathbb{N}$$ let $$q_m$$ be prime number form of $$4k+3$$ for some $$k\in \mathbb{N}$$ such that $$2m+3

• Get some ideas from WA. Dec 17, 2020 at 22:40
• @NeatMath Thanks , I will check it out . maybe it will give me a ways to solve
– ATB
Dec 17, 2020 at 22:43

Hint: Suppose $$n$$ is an odd prime. Then $$\varphi(n) = n-1$$. Also, $$n-1$$ and $$n+1$$ are both even, and thus, $$\varphi(n-1) \le \dfrac{n-1}{2}$$ and $$\varphi(n+1) \le \dfrac{n+1}{2}$$. Do you see why this is true?
Using these results, how large does $$n$$ need to be to guarantee that $$\varphi(n)-\varphi(n-1) > m$$ and $$\varphi(n)-\varphi(n+1) > m$$?
• First of all Thanks you for that Hint, I am understand the hint and I can see that always true because every even number in $\varphi$ great or equal then even number divide by two. For the second line I think that n need to be large as more then m divide by two. I don't sure about that .
• Using $\varphi(n) = n-1$ and $\varphi(n+1) \le \tfrac{n+1}{2}$, can you find a lower bound for $\varphi(n)-\varphi(n+1)$? Then, set that lower bound $> m$ and solve for $n$. Dec 20, 2020 at 22:13
• We want to pick an odd prime $n$ such that both $\varphi(n)-\varphi(n+1) \ge (n-1)-\tfrac{n+1}{2} = \tfrac{n-3}{2} > m$ and $\varphi(n)-\varphi(n-1) \ge (n-1)-\tfrac{n-1}{2} = \tfrac{n-1}{2} > m$ hold. Can you use these inequalities to figure out how big $n$ needs to be? Then the proof goes something like "For any $m \in \mathbb{N}$, pick an odd prime $n > ?$. Then $\varphi(n)-\varphi(n+1) \ge \ldots > m$ and $\varphi(n)-\varphi(n-1) \ge \ldots > m$." Dec 30, 2020 at 21:31