Are these squares perfect?

All positive integers can be written in the form of $2k$ and $2k + 1$ , when $k$ is a positive integer.

Then $(2k)^{2} = 4k^2 \equiv 0 (mod 4)$

Then $(2k + 1)^{2} = 4(k^{2} + k) + 1 \equiv 1 (mod 4)$

Thus, let $a$ be a positive integer,

$\boxed{a^{2} \equiv 0, 1 (mod 4) }$

$364829362819214 \equiv 2 (mod 4)$

$2618361628328171 \equiv 3 (mod 4)$

$17283926283618266 \equiv 2 (mod 4)$

$17282626151715219 \equiv 3 (mod 4)$

$172736373827271731 \equiv 3 (mod 4)$

Thus all number cannot be a perfect square.

Get the digital root of the number by adding all its digits, if the number exceeds 9, add the digits again until you reach the single unit number. A perfect square has digital roots of 1, 4, 7 or 9.

1) Any square number which has 6 as last digit will have odd number as its penultimate digit.

2) Any square number that doesnt have 6 as its last digit has an Even number as its penultimate digit.

A longer way wd be to calculate the digital root since squares have only 1 4 7 9 as their digital roots.

every perfect square is in form of 4p or 4p+1......

Therefore these numbers must be divisible by 4 or leave remainder 1 when divided.......

Since no one follows therefore 0 is the answer

Every perfect square is of the form 4k+1....and....4k+0.....observe that the numbers given are not according to the above form....hence,NO number is a perfect square....