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- Apr 14, 2013

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Hey! I have to show that the number $a=201340168052123987111222893$ is not a square of an integer, without doing calculations.

Could I solve this in $\mathbb{Z}_8$?

I mean that the number $a$ can be written as followed:

$$a=3+9 \cdot 10 +8 \cdot 10^2 + 2 \cdot 10^3+...$$

Since at $\mathbb{Z}_8$: $[10]=[2], [10^2]=[4], [10^3]=[8]=[0], [10^k]=[0] \text{ for }k \geq 3$ we have:

$$[a]=[3]+[9] \cdot [10]+[8] \cdot [10^2]+[2] \cdot [10^3]+...=[3]+[1] \cdot[2]=[3]+[2]=[3+2]=[5]$$

We suppose that $a$ is a square of an integer, so $a=b^2 \Rightarrow [a]=[b^2]$. So it must be be $[b^2]=[5]$. The possible values of $[b^2]$ are:

$$\left.\begin{matrix}

Could I solve this in $\mathbb{Z}_8$?

I mean that the number $a$ can be written as followed:

$$a=3+9 \cdot 10 +8 \cdot 10^2 + 2 \cdot 10^3+...$$

Since at $\mathbb{Z}_8$: $[10]=[2], [10^2]=[4], [10^3]=[8]=[0], [10^k]=[0] \text{ for }k \geq 3$ we have:

$$[a]=[3]+[9] \cdot [10]+[8] \cdot [10^2]+[2] \cdot [10^3]+...=[3]+[1] \cdot[2]=[3]+[2]=[3+2]=[5]$$

We suppose that $a$ is a square of an integer, so $a=b^2 \Rightarrow [a]=[b^2]$. So it must be be $[b^2]=[5]$. The possible values of $[b^2]$ are:

$$\left.\begin{matrix}

**: & [0] & [1] & [2] & [3] & [4] &[5] & [6] & [7]\\**

[b^2]:& [0] &[1] & [4] & [1] &[0] & [1] &[4] & [1]

\end{matrix}\right.$$

Since there is not the value $[5]$, it cannot be true..So $a$ cannot be a square of an integer.

Is this correct? Or can I not just solve this in $\mathbb{Z}_8$?[b^2]:& [0] &[1] & [4] & [1] &[0] & [1] &[4] & [1]

\end{matrix}\right.$$

Since there is not the value $[5]$, it cannot be true..So $a$ cannot be a square of an integer.

Is this correct? Or can I not just solve this in $\mathbb{Z}_8$?

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