Question

Hey guy this is a fun question:
Given any nine integers whose prime factors lie in the set {3,7,11}, prove that there must be two whose product is a square....

Answer 1

@rational @AngusV

Answer 2

pigeonhole principle ?

Answer 3

gotchcha

Answer 4

you're in the right path....

Answer 5

Oh so it's 3,7 and 11 without their respective powers ?

Answer 6

Damn that makes things a lot easier.

Answer 7

The box of Dirichlet.

Answer 8

hey @rational not only 1 and 0 there can be many

Answer 9

i read the question wrong haha, let me erase

Answer 10

It's a good answer though. This could pose for a nice 5'th grade question or something.

Answer 11

Ah, got it!

Answer 12

We study the powers and the cases for odd and even.

Any such number will have the form of 3^a * 7*b * 11^c.

The cases for (a,b,c) are :

o o o
o o e
o e o
o e e

e e e
e e o
e o e
e o o

Where o=odd and e=even. Any combination of two numbers that set will of yield an odd number at least at one of the powers of either 3,7 or 11.

However, when we add in a ninth number, regardless of the (odd,even) combination at the powers of 3,7 and 11 - this combination will be found in that list of 8 numbers.

Which means that when we multiply those two numbers whose (odd,even) combination at the powers is the same the resulting combination at the powers will be (e e e) and thus a perfect square.

Answer 13

Nice!

Answer 14

you nailed it!