If p is a prime factor of a perfect square,
then the maximum k for which pk is a factor,
then k MUST be an even number.

We prime factor 2352 with a factor tree:

          2352
          /  \
         2  1176 
            /  \
           2   588
               / \
              2  294
                 / \
                2  147
                  /  \
                 3   49
                    /  \
                   7    7

So 2352 = 24∙3∙72

The primes 2 and 7 already have even powers,
so we must multiply by 3 so that the 3 will
also have an even power.

So we must multiply by 3:

2352∙3 = 24∙32∙72

The answer is 3.  Then the perfect square will be
7056, which is 24∙32∙72 
and (22∙3∙7)2 = (4∙3∙7)2 = 842

Edwin