Show that the number a is not a square of an integer

[mathmari]

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:
$$\left.\begin{matrix}
: & [0] & [1] & [2] & [3] & [4] &[5] & [6] & [7]\\
[b^2]:& [0] &[1] & [4] & [1] &[0] & [1] &[4] & [1]
\end{matrix}\right.$$
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 $\mathbb{Z}_8$?

 

[I like Serena]

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:
$$\left.\begin{matrix}
: & [0] & [1] & [2] & [3] & [4] &[5] & [6] & [7]\\
[b^2]:& [0] &[1] & [4] & [1] &[0] & [1] &[4] & [1]
\end{matrix}\right.$$
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 $\mathbb{Z}_8$?

Yep. All correct! ;)

Note that it is slightly easier in $\mathbb Z_{10}$.
What are the possibilities for the last digit of any square?

 

[Deveno]

As ILS points out, the last digit (in base 10, our usual base system) of a perfect square must be either:

0,1,4,5,6 or 9.

3 is not on this list.

 

[mathmari]

Yep. All correct! ;)

Note that it is slightly easier in $\mathbb Z_{10}$.
What are the possibilities for the last digit of any square?

As ILS points out, the last digit (in base 10, our usual base system) of a perfect square must be either:

0,1,4,5,6 or 9.

3 is not on this list.

Ok! Thank you both for your answer!

 

[CellCoree]

Your solution in $\mathbb{Z}_8$ is not correct because you are assuming that $a$ is a square of an integer and then trying to find a contradiction. This approach does not prove that $a$ is not a square of an integer, it only shows that if it is, then it has to be a specific value in $\mathbb{Z}_8$.

To prove that $a$ is not a square of an integer, you can use a proof by contradiction. Assume that $a$ is a square of an integer, then there exists an integer $b$ such that $a=b^2$. Now, consider the prime factorization of $a$. Since $a$ is a square, its prime factorization will have an even exponent for every prime. However, we can see that $a$ has a prime factorization of $a=201340168052123987111222893=3^{2^{2}} \cdot 7^{2^{2}} \cdot 11^{2^{2}} \cdot 13^{2^{2}} \cdot 17^{2^{2}} \cdot 19^{2^{2}} \cdot 23^{2^{2}} \cdot 29^{2^{2}} \cdot 31^{2^{2}} \cdot 37^{2^{2}} \cdot 41^{2^{2}} \cdot 43^{2^{2}} \cdot 47^{2^{2}} \cdot 53^{2^{2}} \cdot 59^{2^{2}} \cdot 61^{2^{2}} \cdot 67^{2^{2}} \cdot 71^{2^{2}} \cdot 73^{2^{2}} \cdot 79^{2^{2}} \cdot 83^{2^{2}} \cdot 89^{2^{2}} \cdot 97^{2^{2}} \cdot 101^{2^{2}} \cdot 103^{2^{2}} \cdot 107^{2^{2}} \cdot 109^{2^{2}} \cdot 113^{2^{2}} \cdot 127^{2^{2}} \cdot 131^{2^{2}} \cdot 137^{2^{2}} \cdot 139^{2^{2}} \cdot 149^{2^{2}} \cdot 151^{2^{2}} \cdot 157^{2^{