Problem 4

Given a set

M

1985

distinct positive integers, none of which has a prime divisor greater than

23

, prove that

M

contains four distinct elements whose product is the fourth power of an integer.

Step 2 of 4: Pair equal vectors to make squares

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In plain words

The first pigeonhole round makes matching parity labels collide in pairs; every collision is exactly a square product.

xy\text{ is a square whenever }x,y\text{ have the same parity vector}

Detailed analysis

Two numbers with the same parity vector have even exponent in every prime when multiplied, so their product is a square. Since $1985>3\cdot2^9+1$ , repeatedly apply the pigeonhole principle to obtain at least $2^9+1$ disjoint pairs $\{x_j,y_j\}$ with $x_jy_j=s_j^2$ . The numerical threshold is sufficient because $2(2^9+1)<3\cdot2^9+1$ .