On $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ Euler's totient function $\varphi$ is multiplicative, and it plays important roles in number theory.
QUESTION: Is it true that for each integer $m>6$ we have $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ for some positive integer $n$?
For every $m=7,\ldots,10^4$, I have found the smallest positive integer $n$ such that  $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ (cf. http://oeis.org/A248007). For example, for $m=10$ the least positive integer $n$  with $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ is $14$; in fact,
$$\varphi(10)\varphi(14)=4\times 6\equiv0\pmod{10+14}.$$
I formulated the above question in 2014 and conjectured that the answer should be positive. I have ever mentioned this question to some number theorists, but there is no substantial progress on the question.
I don't think that the question is very difficult. Any ideas towards its solution?
 A: If there are counterexamples, they are likely to be square free and very large.  Here's why.
Suppose $ t^2 $ divides $ m$ and $ t$ is larger than $O( \log p ),$ with $ p$ the largest prime divisor of $m$ (it is enough that $ t$ is larger than $ m/\phi(m) )$. Then $n= t\phi(m) - m$ is a positive multiple of $t^2$, so $t$ divides $ \phi(n)$, and $n,m$ then solves the desired congruence.  One can fine tune this to smaller $t$ for certain $m$, but the point now is to look at solving the problem for a given and mostly squarefree $m$.  
Edit 2018.06.27 GRP
A couple more thoughts.
Let $t$ be a common factor of $m$ and $\phi(m)$.  If $n=\phi(t)\phi(m) - m$ is positive, this is another solution which works.
Otherwise, $m$ and $\phi(m)$ are mostly coprime, and we seek an integer $t$ such that $n$ above is positive and so that $n$ is a multiple of $t$. (More generally, for $d$ a large divisor of $\phi(m)$ we could instead look at $n=\phi(t)d -m$ being a multiple of $t$, but I won't do that now.) By running through $t$ (and thus $\phi(t)$) being 3-smooth numbers, we may encounter an $n$ having enough prime factors of the form 6k+1 distinct from factors of $m$ and factors $\phi(m)$ that will yield a solution. As remarked by Stanley above, it is not clear how to guarantee enough of these factors of $n$ to appear.
END Edit 2018.06.27 GRP
Gerhard "Sure Euler Could Solve It" Paseman, 2018.06.24.
