Use Pigeonhole and $\left \lceil \frac{1985}{2^9} \right \rceil =4$

mahanmath wrote: Use Pigeonhole and $\left \lceil \frac{1985}{2^9} \right \rceil =4$ Can you explain your solution?

There are only nine primes less than or equal to $26$, that are $2,3,5,7,11,13,17,19$ and $23$. It is desired that the product of four of them is a perfect fourth power. So, let us write the powers as $2^{\text{even}}$ or $2^{\text{odd}}$ and do this for each prime. Thus there are total $2^9$ categories. Now, take $2^9+1$ of them. Out of them there are two, say, $a_1$ and $b_1$ which are of the same category. Now, throw the two out. Repeat the process until we have $513$ pairs $(a_1,b_1),(a_2,b_2),\ldots, (a_{513},b_{513})$. Take $c_i=\sqrt{a_ib_i}$ for $513\ge i\ge 1$. Again note that $c_i$ have the prime divisor set and they correspondingly fall under $2^9=512$ categories. But again by PHP, some $c_i$ and $c_j$ have the same category their geometric mean is an integer. According to my solution $\boxed{1537}$ number would have sufficed.