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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?

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  • $\begingroup$ Is there a way to go about this without using the divisibility by 4 rule? $\endgroup$ – Al Jebr Nov 1 '13 at 21:28
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Nov 1 '13 at 21:34
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    $\begingroup$ You can compute $\lfloor \sqrt{x}\rfloor = 18257418583505$, and see it's not a square. $\endgroup$ – Daniel Fischer Nov 1 '13 at 21:36
  • $\begingroup$ In addition, we can show that $\underbrace{333\dots33}_{n\text{ }3\textrm{'s}}4$ will never be a square, for any $n$ (except $n=0$). $\endgroup$ – Akiva Weinberger Oct 24 '14 at 3:06
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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.

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  • $\begingroup$ thanks. Is there a way to go about this without using that fact? $\endgroup$ – Al Jebr Nov 1 '13 at 21:16
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    $\begingroup$ 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? $\endgroup$ – Carlos Eugenio Thompson Pinzón Nov 2 '13 at 5:04
  • $\begingroup$ So many upvotes! $\endgroup$ – Sawarnik Jan 6 '14 at 8:22
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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 $34\equiv2\pmod{4}$.

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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. $\blacksquare$

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$ 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.

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