common perfect square

The square of any number can have unit's digits as $0, 1, 4, 5, 6, 9$ In case of $4n+1$ , n can not take the values 3, 8, 13,... because at these values, the unit's place of $4n+1$ will be 3 which is not permissible in case of squres. Similarly $n$ can not be $4, 9, 14, 19...$ . In case of $6n+1$ the values of n which are not permissible include $1, 6, 11, 16,..$ & $2, 7, 12, 17...$ . So we are left with $n=5k$ where k is positive integer. Putting $n=5,10,15,20..$ We find the minimum value to be $n=\boxed{20}$ .

Let:

4n+1 = a^2

6n + 1 = b^2

Then,

(a+b)(a-b) = -2n

a+b = n a = n-b

and,

a-b = -2

a = b-2

n-b = b-2

n = 2b-2

Then,

6(2b-2) + 1 = b^2

12b - 11 = b^2

(b-11)(b-1) = 0

b = {11, 1}

Substituting.

n = 2b-2

n = 2(11) - 2

n = 20

reject b = 2 because n will be 0..

Therefore, the minimum value of n = 20...