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For which integers $n$ is the relation $\varphi(n)^3\le n^2$ true where $\varphi$ is the totient fuction of Euler.

I used the property of $\varphi$ function ($\varphi(n)=n\prod_{p|n}(1-\frac1 p)$ ) to obtain $n\prod_{p|n}(p-1)^3\le \prod_{p|n}p^3$ Now, how do we proceed further. Thanks beforehand.

This is problem U391 from the Problem column of Mathematical Reflections - Issue 6 2016.

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    How did you get your second inequality? I think you should have $$n^3\prod_{p\mid n}\left(1-\frac1p\right)^3\le n^2\iff n\prod_{p\mid n}\left(1-\frac1p\right)^3\le 1\iff n\prod_{p\mid n}\left(p-1\right)^3\le\prod_{p\mid n}p^3$$ – 2017-01-05
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    @DonAntonio yes, but isnt $\prod_{p|n}p=n$? – 2017-01-05
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    @vidyarthi only if $n$ is squarefree – 2017-01-05
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    @AlexMacedo thanks, modified the post – 2017-01-05
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    Now that you have the correct form, perhaps it will be clearer that to characterise the natural numbers that fulfill that is not, apparently, that terribly messy. Of course, it is true for any prime and some products of primes, but for products of large powers of primes it is false. For example, it is false for $\;36=2^2\cdot3^2\;$ , but true for $\;12=2^2\cdot3\;,\;\;24=2^3\cdot3\;$ ... – 2017-01-05
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    @DonAntonio so is it true for numbers with atmost one non-squarefree prime factor? – 2017-01-05
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    @DonAntonio. It is not true for primes in general. In fact, $(p - 1)^3 \leq p^2$ is true only for $p = 2$ or $3$. – 2017-01-05
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    I ran a quick computation on sage and the only $n \leq 410000$ satisfying $\varphi(n) ^3 \leq n^2$ are: $2$, $3$, $2^2$, $2 \cdot 3$, $2^3$, $2\cdot 5$, $2^2\cdot 3$, $2 \cdot 3^2$, $2^3 \cdot 3$, $2 \cdot 3 \cdot 5$, $2 \cdot 3 \cdot 7$, – 2017-01-05
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    If I had to guess, maybe we could prove that first such $n$ cannot have more than $3$ prime factors. Then for 3 prime factors we can try to show that $n$ has to be square free and we then bound the size of the largest prime in its factorization. Then repeat the same kind of argument for 2 prime factors. – 2017-01-05
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    @AlexMacedo Of course, thanks. I missed the $\;n\;$ on the left of the inequality when writing in this uncomfortable comment space and remained with $\;\prod\limits_{p\mid n}(p-1)^3\le p^3\;$ ...The inequality is true for very few natural numbers – 2017-01-05
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    What does "(Mathematical Reflections)" in the title of the question means? If that's related to the source of the question, probably adding [more details about the source](http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960) in the question might be useful. – 2017-01-05
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    @MartinSleziak It is the name of a magazine. – 2017-01-05
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    @vidyarthi Yes, and it is also a name of a book. Still it would be useful to explain where the problem comes from and what do *Mathematical Reflections* have to do with this question. (Or remove that title from the question if it has nothing to do with it.) – 2017-01-05
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    @MartinSleziak I thought it moral to name the source of problem. But, if you think it confuses, then, next time onwards i would not cite the source. Is it fine? – 2017-01-05
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    @vidyarthi Exactly the opposite. I am saying that you should *add* the source. Simply saying *Mathematical Reflections* has many possible explanations. It might mean that it is from the book Hilton, Holton, Pedersen: *[Mathematical Reflection](https://books.google.sk/books?id=hcXcBwAAQBAJ)*. It might mean that it is from the magazine you've mentioned in the previous comment. I would expect a more exact identification of the source (link, journal volume and year, page numbers, something like that). – 2017-01-05
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    Anyway, these comments have digressed from the topic of the post - I would suggest that we could [continue this discussion in chat](http://chat.stackexchange.com/transcript/19138/2017/1/5) if needed. – 2017-01-05
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    The deadline for this problem (January 15, 2017) has not yet passed. I will be requesting the moderators to lock this post until then. – 2017-01-13

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We know that $\varphi(n)$ is multiplicative (if $n,m$ are coprime then $\varphi(nm)=\varphi(n)\varphi(m)$), therefore so is $f(n)=\varphi(n)^3/n^2$. We're looking for numbers with $f(n)\leq 1$, and so if we write $n$ as a product of distinct prime powers we can just multiply the $f$ values together. They key is that there are very few prime powers with $f(p^k)\leq 1$.

If $k>1$ then $f(p^k)=p^{k-1}f(p)$, and it is easy to check that $p=2$ and $p=3$ are the only primes with $f(p)\leq 1$. It follows that $2$, $4$, $8$ and $3$ are the only prime powers with this property. Then you can enumerate a few prime powers which give the smallest $f(p^k)>1$ -- the $f$ values get too large to be relevant very quickly -- and complete the classification.

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    The function $f$ should be $f(n)=\frac{\varphi(n)^3}{n^2}$ – 2017-01-05
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    Again, $f(p^k)=p^{k-3}f(p)$,isnt it? – 2017-01-05
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    I don't think so, e.g. $f(16)=8^3/16^2=2$ and $f(2)=1/4$. – 2017-01-05