Dear Editor,

I am writing to point out the presence of a major flaw in the solution to problem U118 which you chose to showcase. Mr. Verfasser mentioned in his solution a joint paper by L. Alaoglu and P. Erdös where it is proved that the inequality $\varphi(\sigma(n)) < \varepsilon \cdot n$ holds for every $\varepsilon > 0$ and every natural number $n$ in the complement of a set of density zero. Except for a few typos, everything is just fine with the text up to that point. The problem is that Mr. Verfasser goes on to claim that the previous result implies at once that $$\lim_{n \to \infty} \frac{\varphi(\sigma(n))}{n} = 0.$$ It seems to me that, in reaching such a conclusion, Mr. Verfasser totally disregarded the phrase in the complement of a set of density zero from the thesis of the aforementioned result by Alaoglu and Erdös. Here is a quick, albeit conditional, argument which indicates that the limit $$\lim_{n \to \infty} \frac{\varphi(\sigma(n))}{n}$$ could well be different from zero: if there were infinitely many Mersenne primes, then denoting by $M_{p_{n}}$ the $n$-th such number, we would have that $$\lim_{n \to \infty} \frac{\varphi(\sigma(M_{p_{n}}))}{M_{p_{n}}} = \lim_{n \to \infty} \frac{2^{p_{n}}-2^{p_{n}-1}}{2^{p_{n}}-1} = \lim_{n \to \infty} \frac{1-\frac{1}{2}}{1-\frac{1}{2^{p_{n}}}} = \frac{1}{2},$$ which would in turn yield $$\limsup_{n \to \infty} \frac{\varphi(\sigma(n))}{n} \geq \frac{1}{2}.$$ Clearly enough, this inequality blatantly conflicts with Mr. Verfasser's claim about the limit of $\frac{\varphi(\sigma(n))}{n}$ as $n \to \infty$! Furthermore, and just in case the above conditional argument has not succeeded in convincing you of the inadequacy of Mr. Verfasser's solution to U118, let me add that in R. K. Guy's Unsolved Problems in Number Theory there is a datum that puts the kibosh on the possibility that Mr. Verfasser's crucial claim may be right after all: under entry B42 of the book, Guy writes that A. Mąkowski and A. Schinzel proved in [2] that $$\limsup_{n \to \infty} \frac{\varphi(\sigma(n))}{n} = +\infty.$$ All the best regards,

J. H. S.
(July 20, 2010)


[1] R. K. Guy. Unsolved Problems in Number Theory (2nd edition). Springer Verlag, NY, USA, p. 99.
[2] A. Mąkowski, A. Schinzel. On the functions $\phi(n)$ and $\sigma(n)$. Colloq. Math. 13 (1964-1965), pp. 95-99.
[3] Mathematical Reflections, issue #3, 2009, p. 23.