Euler's Totient Dilemma

First, $5544 = 2^3 \cdot 3^2 \cdot 7 \cdot 11$ . Now if $mn$ is divisible by $7^3$ then $\phi(mn)$ is divisible by $7^2$ , which is impossible, so the multiplicity of $7$ in $m$ and $n$ must be exactly $1$ . So we can write $m = 7a$ , $n = 7b$ , where $7 \nmid ab$ , and $a$ and $b$ are coprime.

By the multiplicativity of $\phi$ , we get $\phi(mn) = \phi(49ab) = \phi(49)\phi(a)\phi(b) = 42\phi(a)\phi(b)$ . So $\phi(a)\phi(b) = 132$ . Note that $\phi(m) = 6\phi(a)$ and $\phi(n) = 6 \phi(b)$ .

So we need to choose two factors of $132$ that are closest together, subject to the constraint that the factors are either even or $1$ . Clearly the right choice is $22$ and $6$ . So $\phi(m)$ and $\phi(n)$ are $132$ and $36$ in some order, so the answer is $\fbox{96}$ .

(I guess we should show that these values are attainable--take $m = 161$ and $n = 63$ .)

$5544=2^3 \times 3^2 \times 11 \times 7$

So $7$ appears in $m$ and $n$ only once.

$m=7k$

$n=7j$

$\phi(m)=6\phi(k), \phi(n)=6\phi(j)$

$mn=49kj \Rightarrow \phi(mn)=42\phi(k)\phi(j)=5544$

$\phi(k)\phi(j)=132$

$\phi(k)$ and $\phi(j)$ are both even/or at least one of them is even.

$132=6 \times 22$ is the arrangement which gives us least value.

So answer is $96$