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A set $L$ consists of $2008$ integers, none of which have a prime divisor larger than $24$. Prove that $L$ has four elements, the product of which is equal to the fourth power of an integer.

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  • $\begingroup$ What have you tried? Do you know how many primes are less than $24$? I would think of each element of $L$ as a vector of the exponents of the primes $\bmod 4$ $\endgroup$ – Ross Millikan Dec 11 '17 at 15:53
  • $\begingroup$ There are 9 primes less than 24 and so any integer in $L$ will have a prime factorization of the form $2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}13^{a_6}17^{a_7}19^{a_8}23^{a_9}$ where $a_i \geq 0$ for each $i $. If each $a_i $ is a multiple of $4$ then we have a fourth power. I haven’t worked out an answer but from here, it is likely not difficult using the suggestion from the previous comment. $\endgroup$ – user328442 Dec 11 '17 at 16:17
  • $\begingroup$ (Except for the obvious change, this is problem 4 from the 1985 IMO.) $\endgroup$ – Misha Lavrov Dec 13 '17 at 15:45
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There are nine primes up to $24$, hence only $2^9=512$ different possibilities for the parities of prime exponents of elements of $L$. In other words, If $L'\subseteq L$ has $513$ elements, there are two elements in $L'$ that have the same exponent parities, which means that the product of these is a square. This way, we can extract pairs with square product from $L$ until less than $513$ elements are left. We end up with $\lceil \frac{2008-512}2\rceil =748$ disjoint pairs $(a_1,b_1), (a_2,b_2),\ldots, (a_{748},b_{748})$ such that $c_k:=\sqrt{a_kb_k}$ is an integer. Within the multiset of these $c_k$, we can again find pairs with square product (because again, we have more than $512$ numbers). Then the product of the original four constituents is a fourth power.

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