Is the number $333{,}333{,}333{,}333{,}333{,}333{,}333{,}333{,}334$ a perfect square? I know that if the number is a perfect square then it will be congruent to $0$ or $1$ (mod $4$).  Now since the number is even, I know that it is either $0$ or $2$ (mod $4$).  How would I go about answering this?
 A: A number is divisible by $4$ if and only if the number made of its last two digits is divisible by $4$; this is immediate from the fact that $100$ is divisible by $4$.
The last two digits are $34 = 2 \cdot 17$, so our number is divisible by $2$ only.
A: More generally. "Find all positive integers $k,n$ such that $\frac{(6n-2)^k-1}{3}+1$ is a square."
Answer. Clearly such a square, if it exists, has to be odd. Suppose then that there exists a positive integer $x$ such that
$$\frac{1}{3}((6n-2)^k-1)+1=(2x+1)^2 \implies 12x(x+1)=(6n-2)^k-1.$$
As far as one between $x$ and $x+1$ has to be even, then $\upsilon_2((6n-2)^k-1) \ge 3$, which is clearly impossible since it's a odd number. $\blacksquare$
A: The square of an even number is $0\pmod{4}$. The square of an odd number is $1\pmod{4}$. Thus, a perfect square is either $0$ or $1\pmod{4}$.
This number is $2$ mod $4$ since it is $n\times 100+34$ and $34\equiv2\pmod{4}$.
A: 
$ clisp -q
[1]> (isqrt (read-from-string (remove #\, "333,333,333,333,333,333,333,333,334")))
18257418583505

The last digit of the integer square root approximation is 5. So the square of that approximation must end in 5 and thus it is not exact, meaning that the original number isn't a square.
