Find a minimum positive integer, which is multiplied by 2 is a complete square number, multiplied by 3 is a complete cubic number

Find a minimum positive integer, which is multiplied by 2 is a complete square number, multiplied by 3 is a complete cubic number

2*2*2*3*3=72

Three consecutive positive integers, the middle one is a complete square number, the product of such three consecutive positive integers is called "wonderful number", ask all less than 2008
Three consecutive positive integers, the middle one is a complete square number, the product of such three consecutive positive integers is called "wonderful number", what is the greatest common factor of all wonderful numbers less than 2008?

Let the number in the middle be x ^ 2 (x is an integer greater than 1)
The wonderful number can be expressed as (x ^ 2-1) · x ^ 2 · (x ^ 2 + 1) (x ≥ 2)
Obviously, the smallest wonderful number is 60 (x = 2,3 × 4 × 5 = 60), so the greatest common factor of all wonderful numbers must be less than or equal to 60. Now we prove that the greatest common factor is 60
Since 60 = 2 × 2 × 3 × 5, it can be proved that the wonderful number is a multiple of 3,4,5, then the proposition is proved
First of all, we get any wonderful number (x ^ 2-1) · x ^ 2 · (x ^ 2 + 1) = (x-1) (x + 1) · x · x · (x ^ 2 + 1)
X-1, x, x + 1 are three continuous positive integers. One of them must be a multiple of 3, so their product must also be a multiple of 3
Then if x is even, then the square of X must be a multiple of 4
If x is odd, then both X + 1 and X-1 are even, and their product is also a multiple of 4
Finally, it is proved that the product is a multiple of 5
If x is a multiple of 5, the product is a multiple of 5
If the remainder of X divided by 5 is 1, then X-1 is a multiple of 5, and the product must be a multiple of 5
If the remainder of X divided by 5 is 4, then x + 1 is a multiple of 5, and the product must be a multiple of 5
If the remainder of X divided by 5 is 2, then x ^ 2 + 1 is a multiple of 5 [because (5K + 2) ^ 2 + 1 = 25K ^ 2 + 20K + 5]
If the remainder of X divided by 5 is 3, then x ^ 2 + 1 is also a multiple of 5 [because (5K + 3) ^ 2 + 1 = 25K ^ 2 + 30K + 10]
So anyway, the product must be a multiple of 5
To sum up, the greatest common factor is 60

How to prove that 2008 is the sum of K (k is a positive integer) complete squares to find the minimum value of K?

2008=44*44+8*8+2*2+2*2
First find the largest square less than or equal to 2008, which is 44
The remaining 72, and then find the largest square less than or equal to 72, is 8
The remaining 8 are expressed as two 2 * 2,
How to prove that I feel very difficult, there is no idea