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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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    $\begingroup$ yes. If $3|n$ and $n$ is a perfect square, then $9|n$. $\endgroup$
    – N. S.
    Dec 28, 2012 at 15:16
  • $\begingroup$ In fact, $10200300040000100004000300201= 3 · 23 · 1613867 · 91600135179863204287$ has no square factors. $\endgroup$
    – lhf
    Feb 17, 2017 at 20:35

4 Answers 4

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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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  • $\begingroup$ How does the fact that it is divisible by 9 mean that it is not a perfect square? $\endgroup$
    – Airdish
    Feb 22, 2016 at 18:50
  • $\begingroup$ @S.Mo That's the contradiction. This number is divisible by 3 but NOT divisible by 9. $\endgroup$
    – N. S.
    Feb 22, 2016 at 19:43
  • $\begingroup$ Ah, I see, didn't notice that. Thanks. $\endgroup$
    – Airdish
    Feb 23, 2016 at 1:20
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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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