Prove A contains at least one subset of 4# (bi , distinct integers) / b1*b2*b3*b4=m^4 
Let A be a set of 2015 distinct positive integers such that no element
  of A is divisible by any prime larger than 25.    Prove that A
  contains at least one subset of four distinct integers whose product
  is the fourth power of an integer.

I try to solve this exercise, but I have no luck. I divided the logical steps to understand more what I need to prove.

(1) A=2015 positive integers. No element of A is divisible by any
  prime larger than 25.

Therefore, I said:
B={Set of  2015  distinct integers}
P=primes greater that 25={29,31,37,41,47,53,59,61,67,71,…,2011,2015}
M=multiples of P
Then A=B-P-M

(2) There is at least one subset B included in A, B[b1,b2,b3,b4] such
  that b1*b2*b3*b4=m^4 where b1, b2, b3, and b4 are distinct, and m
  integer.

I believe, I need to find A, with the restriction given, and after doing it I can took at least 1 set {b1,b2,b3,b4} such as (2) is true and then I will be done?? Is it right? Please I need help to understand such exercise. Thanks!
 A: The number $2015$ is very weak.We will prove the stronger problem:
Problem: Let $A$ be a set of $1537$ distinct positive integers such that no element of $A$ is divisible by any prime larger than $25$. Prove that $A$ contains at least one subset of four distinct integers whose product is the fourth power of an integer.
Solution: Every element of $A$ is of the form $2^{\alpha_1}\cdot 3^{\alpha_2}\cdot...\cdot 19^{\alpha_8}\cdot 23^{\alpha_9}$.Now every $\alpha_i$ leaves residue $0$ or $1$ when divided by $2$,so by Pigeon Hole Principle from any $2^9+1=513$ elements of $A$ there will be two of the form $2^{\alpha_1}\cdot 3^{\alpha_2}\cdot...\cdot 19^{\alpha_8}\cdot 23^{\alpha_9}$(we will call this number $a_1$) and $2^{\beta_1}\cdot 3^{\beta_2}\cdot...\cdot 19^{\beta_8}\cdot 23^{\beta_9}$(we will call this number $b_1$) such that $\alpha_i\equiv\beta_i,i=\overline{1,9}$.From here it follows that $a_1b_1$ is the square of an integer.
Now take those two numbers away from $A$.We choose other $513$ elements of $A$.Again,by Pigeon Hole Principle there will be two numbers $a_2,b_2$ such that $a_2b_2$ is the square of an integer.Take those two numbers away from $A$.
We will procced this way until $A$ is left with only $513$ elements.From those final $513$ elements,By Pigeon Hole Principle,there will be two such that their product is the square of an integer.
Therefore we will have $\frac{1537-513}{2}+1=513$ pairs $(a_i,b_i)$ of elements of $A$ such that $c_i=a_ib_i,i=\overline{1,513}$ is the square of an integer.
Observe that $\sqrt{c_i}\in\mathbb{N}$ and $\sqrt{c_i}$ is of the form $2^{\alpha_1}\cdot 3^{\alpha_2}\cdot...\cdot 19^{\alpha_8}\cdot 23^{\alpha_9}$,so again by Pigeon Hole Principle there will be two numbers $1\le i\le j\le 513$ such that $\sqrt{c_i}\cdot\sqrt{c_j}=a^2$ for some integer $a$.From here it follows that $a^4=c_ic_j=a_ib_ia_jb_j$.Now just note that $a_i,b_i,a_j,b_j\in A$.This ends our proof.
