for which pairs of integers $(n,k), n>1$ we could have $\phi(n(n+1)(4n-1))=12 \phi(k^2)$?

Question

I have tried to find such pairs of integers $(n,k)$ with $n >1$ for which $\phi(n(n+1)(4n-1))=12 \phi(k^2)$ with $\phi$ is the Euler totionnt function but i didn't get any pairs, In the same time I can't disprove that there is no pairs satisfying that ,it seems that we should prove in the first that $n(n+1)(4n-1)$ is a perfect square such that : $n(n+1)(4n-1)=(6k)^2$ but also no pair I find for $n >1$

Answer 1

Too long for a comment.

First of all, you should note that $\phi(k^2)=k \cdot\phi(k)$ because suppose $k$ has prime factors $p_1,p_2, \dots, p_n$, (regardless of their exponents) $$\phi(k^2)=k^2\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\dots\left(1-\frac{1}{p_n}\right)$$ while $$\phi(k)=k\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\dots\left(1-\frac{1}{p_n}\right)$$ Now, by the euclidean algorithm,$\gcd(4n-1,n)=\gcd(n-1,n)=1$ and of course $\gcd(n,n+1)=1$

while $\gcd(4n-1,n+1)=\gcd(n-4,n+1)=\gcd(n-4,5)$, so we need to check the case when $n-4$ is coprime with $5$ and the other case, suppose it is coprime with $5$, then $$\phi(n(n+1)(4n-1))=\phi(n)\cdot\phi(n+1)\cdot\phi(4n-1)=12k \cdot\phi(k)$$ If it is not, then $$\phi(n) \cdot \phi((n+1)(4n-1))=12k\cdot \phi(k)$$ Then you may suppose different numbers to be prime and use the property $\phi(p)=p-1$, does that help?