# 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?

• Is there a way to go about this without using the divisibility by 4 rule? Commented Nov 1, 2013 at 21:28
• Sure, you can see that it's $\equiv 5 \pmod{7}$, so it's not a square. But looking at it modulo $4$ is the easiest and fastest way to see it, one would need a very good reason to not use it. Commented Nov 1, 2013 at 21:34
• You can compute $\lfloor \sqrt{x}\rfloor = 18257418583505$, and see it's not a square. Commented Nov 1, 2013 at 21:36
• In addition, we can show that $\underbrace{333\dots33}_{n\text{ }3\textrm{'s}}4$ will never be a square, for any $n$ (except $n=0$). Commented Oct 24, 2014 at 3:06

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.

• thanks. Is there a way to go about this without using that fact? Commented Nov 1, 2013 at 21:16
• Divide by two: you get: 166,666,666,666,666,666,666,666,667, which is odd, then 333,333,333,333,333,333,333,333,334 is not divisible by 4. But why dividing the whole number if you can just check the last to digits and get the same result? Commented Nov 2, 2013 at 5:04
• So many upvotes! Commented Jan 6, 2014 at 8:22

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 $34\equiv2\pmod{4}$.

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. $\blacksquare$

\$ 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.