# Number TheoryShow that the number a is not a square of an integer

#### mathmari

##### Well-known member
MHB Site Helper
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 $\mathbb{Z}_8$: $=, [10^2]=, [10^3]==, [10^k]= \text{ for }k \geq 3$ we have:
$$[a]=+ \cdot + \cdot [10^2]+ \cdot [10^3]+...=+ \cdot=+=[3+2]=$$
We suppose that $a$ is a square of an integer, so $a=b^2 \Rightarrow [a]=[b^2]$. So it must be be $[b^2]=$. The possible values of $[b^2]$ are:
$$\left.\begin{matrix} : &  &  &  &  &  & &  & \\ [b^2]:&  & &  &  & &  & &  \end{matrix}\right.$$
Since there is not the value $$, it cannot be true..So $a$ cannot be a square of an integer.

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

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $\mathbb{Z}_8$: $=, [10^2]=, [10^3]==, [10^k]= \text{ for }k \geq 3$ we have:
$$[a]=+ \cdot + \cdot [10^2]+ \cdot [10^3]+...=+ \cdot=+=[3+2]=$$
We suppose that $a$ is a square of an integer, so $a=b^2 \Rightarrow [a]=[b^2]$. So it must be be $[b^2]=$. The possible values of $[b^2]$ are:
$$\left.\begin{matrix} : &  &  &  &  &  & &  & \\ [b^2]:&  & &  &  & &  & &  \end{matrix}\right.$$
Since there is not the value $$, 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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Site Helper
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! 