# $\varphi(m+n)\mid n$ for some positive integer $n$

Let $$\varphi$$ be Euler's totient function. If $$p$$ is a prime, then $$\varphi(1+n)=n$$ for $$n=p-1$$.

Question. Is it true that for each integer $$m>1$$ there is a positive integer $$n\le m^2-m$$ such that $$\varphi(m+n)\mid n$$ ?

I conjecture that the question has a positive answer. See http://oeis.org/A248568 for related data. For example, $$\varphi(10+40)=20$$ divides $$40$$.