CareerCup

Quick algorithm sketch to solve this:
1. Find all prime factors of x (using the Sieve of Eratosthenes, for example) and store each factor's power in an array prime_power[]
3. Initialize product = 1
2. Iterate over each of the prime factors and do for each prime p: product *= ((prime_power[p]%2 == 0) ? 1 : p)
3. return product

def primes(n):
    primfac = []
    d = 2
    while d * d <= n:
        factor = False
        pow=0
        while (n % d) == 0:
            factor = True  # supposing you want multiple factors repeated
            n /= d
            pow+=1
        if factor:
            primfac.append((d,pow))
        d += 1
    if n > 1:
        primfac.append(n)
    return primfac


def make_perfect_square(n):
    prime_factors=primes(n)
    smallest=1
    for (prime,pow) in prime_factors:
        #pow is odd
        if pow &1 :
            smallest*=prime

    return smallest

In Go. The trial division could be optimized.

package main

import (
	"fmt"
	"math"
	"sort"
)

func main() {
	fmt.Println(findNumber(1))     //1/1 = 1
	fmt.Println(findNumber(2))     //2/2 = 1
	fmt.Println(findNumber(3))     //3/3 = 1
	fmt.Println(findNumber(288))   //288 / 2 = 144 (12*12)
	fmt.Println(findNumber(49392)) //49392 / 7 = 7056 (84*84)
}

func findNumber(x int) int {
	divs := []int{1, x}

	//find all divisors
	d := 2
	for d*d < x {
		if x%d == 0 {
			divs = append(divs, d)
			divs = append(divs, x/d)
		}
		d++
	}

	sort.Ints(divs) //sort factors

	//start with the largest, look for perfect squares
	for i := len(divs) - 1; i >= 0; i-- {
		sq := math.Sqrt(float64(divs[i]))
		if sq == float64(int64(sq)) {
			return x / divs[i] //return the other factor
		}
	}
	return -1 //will never happen, x/x is 1, which is a perfect square
}

The number is 1/x so x/1/x= x^2

the number is 1/x so x/1/x = x^2

public int smallestSquareNumber(int x) {
        // y^2 < x < (y+1)^2
        // find n and y such that y^2 = (x/n) 
        for (int i = x; i > 0; i--) {
            double newSquare = Math.pow(i, 2);
            if (newSquare < x) {
                if (x % newSquare == 0) {
                    return (int) (x / newSquare);
                }
            } else if (newSquare == x) {
                return 1;
            }
        }
        // should never get here.
        return -1;
    }

public int smallestSquareNumber(int x) {
        // y^2 < x < (y+1)^2
        // find n and y such that y^2 = (x/n) 
        for (int i = x; i > 0; i--) {
            double newSquare = Math.pow(i, 2);
            if (newSquare < x) {
                if (x % newSquare == 0) {
                    return (int) (x / newSquare);
                }
            } else if (newSquare == x) {
                return 1;
            }
        }
        // should never get here.
        return -1;
    }

This should do it:

// ZoomBA
def find_square_complement(n){
  if ( n == 0 || n == 1 ) return n
  #(primes,factor) = fold ( [2:n+1] , [set(),1] ) -> {
    cur_primes = $.p.0 ; cur_no = $.o 
    continue ( exists ( cur_primes ) :: {  $.o /? cur_no } )
    $.p.0 += cur_no ; power = 0 ; x = n 
    while ( cur_no /? x ){ power +=1 ; x /= cur_no }
    if ( !(2 /? power ) ){ $.p.1 *= cur_no }
    $.p 
  }
  return ( ( n @ primes ) ? n : factor )
}

public static int perfectSquare(int x)
	{
        int i=1;
		while(i<=x)
		{
		
		  if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
		  {
			  return i;
       		
		  }
		  else
		  {
			 
			  i++;
		  }
			
		}
		
		return x;		
	}

public static int perfectSquare(int x)
	{
                 int i=1;
		while(i<=x)
		{
		
		  if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
		  {
			  return i;
       		
		  }
		  else
		  {
			 
			  i++;
		  }
			
		}
		
		return x;		
	}

public static int perfectSquare(int x)
	{
        int i=1;
		while(i<=x)
		{
		
		  if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
		  {
			  return i;
       		
		  }
		  else
		  {
			 
			  i++;
		  }
			
		}
		
		return x;		
	}

public static int perfectSquare(int x)
{
int i=1;
while(i<=x)
{

if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
{
return i;

}
else
{

i++;
}

}

return x;
}

public static int perfectSquare(int x)
	{
        int i=1;
		while(i<=x)
		{
		
		  if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
		  {
			  return i;
       		
		  }
		  else
		  {
			 
			  i++;
		  }
			
		}
		
		return x;		
	}

public static int perfectSquare(int x)
	{
        int i=1;
		while(i<=x)
		{
		
		  if((x % i==0) && (Math.sqrt(x/i)-(long)(Math.sqrt(x/i))==0))
		  {
			  return i;
       		
		  }
		  else
		  {
			 
			  i++;
		  }
			
		}
		
		return x;		
	}

int find(int n) {
	int rt = (int)sqrt(n);
	if (rt*rt == n) return 1;
       // You can start from bigger to smaller number so that number of iterations become less. For example, 16*16 would have covered 2*2,4*4 and 8*8 when loop reaches 8,4 or 2.. so it will be fast.
	for (int i=rt;i>=2;i--) {
		if (n%(i*i) == 0) 
			n  /= (i*i);
	}
    return n;
}

int find(int n) {
	int rt = (int)sqrt(n);
	if (rt*rt == n) return 1;
	for (int i=rt;i>=2;<blank>) {
		if (n%(i*i) == 0) 
			{ n  /= (i*i); i = (int)sqrt(n);} // If n is divided by i, directly reduce i to current num's sqrt and start downwards from there.
		else 
			{ i--;}
	}
    return n;
}