Radicals and Reciprocals

Observe that for $N = 2^3 \cdot 3^4$ , we have $A_N = 4$ :

• $(N/1)^{1/1} = N = 2^3 \cdot 3^4$
• $(N/2)^{1/2} = (2^2 \cdot 3^4)^{1/2} = 2 \cdot 3^2$
• $(N/3)^{1/3} = (2^3 \cdot 3^3)^{1/3} = 2 \cdot 3$

Thus, $A_N \ge 4$ for this value of $N$ .

Now, we will show $A_N \le 4$ for all $N$ . To do this, suppose for the sake contradiction that it isn't; then, for some $N$ , we have $(N/i)^{1/i}$ to be an integer for all $i = 1, 2, 3, 4$ ; in particular, $(N/2)^{1/2}, (N/4)^{1/4}$ are both integers.

Consider the largest power of 2 that divides $N$ ; suppose this is $2^k$ .

Because $(N/2)^{1/2}$ is an integer, this means $(2^k/2)^{1/2}$ is also an integer. Thus $2^{\frac{k-1}{2}}$ is an integer, so $k$ is odd.

Because $(N/4)^{1/4}$ is an integer, this means $(2^k/4)^{1/4}$ is also an integer. Thus $2^{\frac{k-2}{4}}$ is an integer, so $k \equiv 2 \pmod 4$ . But this means $k$ is even, contradiction.

So no $N$ has $A_N > 4$ . This proves that $A_N \le 4$ for all $N$ , and so $\max A_N = 4$ .

Notice that $A_N \leq 4$ ; if $(\frac {N}{2})^{\frac{1}{2}}$ and $(\frac {N}{4})^{\frac{1}{4}}$ were both integers , then $\frac{N}{2}$ and $\frac{N}{4}$ would both be perfect squares, which is impossible, because $2 = \frac {\frac{N}{2}}{\frac{N}{4}}$ is not the perfect square of a rational.

An example for $A_N = 4$ would be $N= 648$ , as $\frac{N}{1} = 648 , ((\frac{N}{2})^{\frac{1}{2}} = 18$ and $(\frac{N}{3})^{\frac{1}{3}} = 6$ .