# Exist 4 numbers whose product is a perfect square

Let be given $131$ distinct natural numbers, each having prime divisors not exceeding $42$. how to Prove that one can choose four of them whose product is a perfect square.

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What have you tried? –  Jonas Meyer Jul 6 '12 at 3:38
Sounds like the pigeon hole principle will come into play? –  Jyrki Lahtonen Jul 6 '12 at 4:07

2. $\displaystyle{131\choose 2}>2^{13}$.
4. Out comes two pairs of numbers $(a,b), a\neq b,$ and $(c,d), c\neq d,$ such that $abcd$ is a square. If all four are distinct, we are done. If, say $b=d$, then $ac$ is also a square. Repeat without $a$ and $c$ using $$\displaystyle{129\choose2}>2^{13}.$$
You mean $131\choose 4$? –  Jonas Meyer Jul 6 '12 at 4:17