Let $\varphi(n)$ denote the totient function. I try to work out necessary conditions for the equation $$\varphi(n+46)-\varphi(n)=46$$ for positive integer $n$. The problem arose from the more general equation $$\varphi(n+k)-\varphi(n)=k$$ with even positive integer $k$ and positive integer $n$. The smallest $k$ for which I did not find a composite solution is $k=46$.
I worked out that for $n>4$ , one of $n$ and $n+46$ must be of the form $p^k$ or $2p^k$ with a prime $p$ of the form $4m+3$ and positive integer $k$. This is because not both totient values can be divisible by $4$.
My question is now whether we can rule out the special case $n=2p$ or $n+46=2p$.
So, assume $p\equiv 3\mod 4$ is a prime number. Then , we have $\varphi(2p)=p-1$ and the equation $$\varphi(2p+46)-\varphi(2p)=46$$ transforms into $$2\varphi(p+23)=p+45$$ because $p+23$ is even.
Since $p+23\equiv 2\mod 4$ , we can conclude $$\varphi(\frac{p+23}{2})=\frac{p+45}{2}$$ which is impossible. Now assume $p>24$ of the form $4k+3$ is a prime number and assume $$\varphi(2p)-\varphi(2p-46)=46$$ This transforms to $$2\varphi(p-23)=p-47$$ because $p-23$ is even.
But this time I cannot show that this equation has no solution. There is none for $p\le 10^{10}$. Any ideas ?
Best Answer
Take your last equation $2 \varphi(p-23) = p-47$. Say $p-23 = 2^a b$ where $b$ is odd. Then $2^a \varphi(b) = 2^a b - 24$, so $\varphi(b) = b - 24/2^a$ and $1 \le a \le 3$. Let $r$ be the smallest prime dividing $b$. Then $\varphi(b) \le b(1-1/r)$, so $b/r \le 24/2^a \le 12$. In particular either $b = r$ or $r \le 11$. If $b = r$ then $\varphi(b) = b-1 > b-3 \ge b - 24/2^a$, so this doesn't work. Therefore $r \le 11$ and $b = rx$ for some odd integer $x$ such that $r \le x \le 11$ and such that $x$ has no prime divisors $<r$. There only a handful of possibilities and none of them works. The nearest miss is $b=9$, for which $\varphi(b) = b - 3$, but in this case $23 + 2^3 \cdot b = 95$ is not prime.