TITLE: Finding the maximum number with a certain Euler's totient value
QUESTION [5 upvotes]: Euler's totient function has a lower bound for large values, but is there any way to pick out maximums for specific values of the function?
That is, how would I find the maximum number n such that phi(n) = 1000, for example?

REPLY [3 votes]: Combining Theorem 329 in Hardy and Wright with the upper bound on $\sigma(n)$ in Robin (1984), we get
$$  \phi(n) > \frac{6 n \log \log n}{\pi^2 \left( e^\gamma (\log \log n)^2 + 0.6482 \right)},$$ where $\gamma = 0.5772156649...$ is the Euler-Masceroni constant.
As a result, $\phi(n) > 1000$ unless $ n \leq 6872.$ 
I'm just saying. 
1111 = 11 * 101  phi =  1000 = 2^3 * 5^3
1255 = 5 * 251  phi =  1000 = 2^3 * 5^3
1375 = 5^3 * 11  phi =  1000 = 2^3 * 5^3
1875 = 3 * 5^4  phi =  1000 = 2^3 * 5^3
2008 = 2^3 * 251  phi =  1000 = 2^3 * 5^3
2222 = 2 * 11 * 101  phi =  1000 = 2^3 * 5^3
2500 = 2^2 * 5^4  phi =  1000 = 2^3 * 5^3
2510 = 2 * 5 * 251  phi =  1000 = 2^3 * 5^3
2750 = 2 * 5^3 * 11  phi =  1000 = 2^3 * 5^3
3012 = 2^2 * 3 * 251  phi =  1000 = 2^3 * 5^3
3750 = 2 * 3 * 5^4  phi =  1000 = 2^3 * 5^3
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 

One thing I noticed after the fact is that 41 does not appear. This is as it should be. For 2 and powers of 2, we have $\phi(2^{k+1} = 2^k.$ However, for any odd prime $p$ such as 5, it is not possible to have any $\phi(n) = p$ or  $\phi(n) = p^2$ or, in general,  $\phi(n) = p^k.$ In particular it is impossible to have $\phi(n) = 25.$ As a result, if $\phi(n) = 1000,$ it is impossible to have $41 | n.$ Oh, any prime dividing this $n$ must have $(p-1) | 1000,$ and if the exponent is going to be larger than 1 we also need $p | 1000.$ So, with 41, the exponent is 1, and we then need to find some relatively prime $m$ with $\phi(m) = 25,$ impossible.