# prove that $10200300040000100004000300201$ is not perfect square

How can I prove that $10200300040000100004000300201$ is not a perfect square ? This number is divisible with $3$ only one time. Is it a good reason and it is enough ?

thanks :)

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yes. If $3|n$ and $n$ is a perfect square, then $9|n$. – N. S. Dec 28 '12 at 15:16

Yes. Assume by contradiction that your number is a perfect square $n^2$.

Since $3|n \cdot n$ and $3$ is prime, it follows that $3|n$. Then $n=3k$ and hence $n^2=9k^2$. Thus, your number is divisible by $9$, contradiction.

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How does the fact that it is divisible by 9 mean that it is not a perfect square? – Airdish Feb 22 at 18:50
@S.Mo That's the contradiction. This number is divisible by 3 but NOT divisible by 9. – N. S. Feb 22 at 19:43
Ah, I see, didn't notice that. Thanks. – Airdish Feb 23 at 1:20

Assume the distinct prime factors of $n$ are $p_1, p_2, \cdots, p_k$, as:

$$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$

For $n$ to be a perfect square, a necessary and sufficient condition is $\alpha_1, \alpha_2, \cdots, \alpha_k$ even, such that $\sqrt{p_i^{\alpha_i}}$ be an integer $\implies$ $\sqrt{n}$ be an integer. Otherwise, $\sqrt{n}$ is irrational and hence $n$ is not a perfect square.

If $3$ divides $n$ only one time, then we must have:

$$n = 3^1 p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$

$1$ is not even, therefore $n$ is not a perfect square.

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HINT: $3|n \wedge (n=k^2,k\in\mathbb{N})$ follows that $9|n$.

HINT 2: Use the divisibility test of $9$.

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The digital root of a square is $1, 4, 7$, or $9$. Yours is $3$...

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