# not integer number

#### dhiab

Prove that number $$\displaystyle \sqrt{20092010201120122013}$$ is not
integer number .

#### Bacterius

A perfect square never ends in $$\displaystyle 3$$, therefore $$\displaystyle 20092010201120122013$$ is not a perfect square, therefore $$\displaystyle \sqrt{20092010201120122013}$$ is not an integer.

Come on Dhiab, that was too easy, where are the headcracking problems you used to post PS : the proof for my original statement is quite trivial, a quick Google search will have it done.

#### dhiab

Hello : have the demonstration 'A perfect square never ends in '(Talking)(Come on Dhiab, that was too easy)(Evilgrin)
Thanks

#### Bacterius

Fine, I'll give it (Speechless)

Let $$\displaystyle n$$ be any integer number. $$\displaystyle n^2$$ is its square. Let us look at the last digit of $$\displaystyle n^2$$, using arithmetic modulo $$\displaystyle 10$$. The following table enumerates all possible cases :

$$\displaystyle \begin{array}{|l|l|} \hline n \mod 10 &n^2 \mod 10 \\ \hline 0 &0 \\ 1 &1 \\ 2 &4 \\ 3 &9 \\ 4 &6 \\ 5 &5 \\ 6 &6 \end{array}$$
$$\displaystyle \begin{array}{|l|l|} 7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8 &4 \\ 9 &1 \\ \hline \end{array}$$

Sorry, I had to cut the table midway, it was too big for MHF.

This shows that a perfect square may only end by the digits $$\displaystyle 0, 1, 4, 5, 6$$ and $$\displaystyle 9$$. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

$$\displaystyle \boxed{\mathrm{QED}}$$

This implies that if a number terminates by 3, it cannot be a perfect square.

• dhiab

#### Soroban

MHF Hall of Honor
Hello, dhiab!

Prove that number $$\displaystyle \sqrt{20092010201120122013}$$ is not an integer.
I assume that the author forgot (or was unaware of) the Final Digit Rule for integer squares.

If that is true, that number has a special rhythm to it: .$$\displaystyle 2009\,2010\,2011\,2012\,2013$$

We have: .$$\displaystyle N \;=\;10^{16}a + 10^{12}(a+1) + 10^8(a+2) + 10^4(a+3) + (a+4)\;\text{ where }a = 2009$$

And I suspect that we are expected to prove algebraically that $$\displaystyle N$$ is not a square.

Any ideas?

• Bacterius

#### Swlabr

Fine, I'll give it (Speechless)

Let $$\displaystyle n$$ be any integer number. $$\displaystyle n^2$$ is its square. Let us look at the last digit of $$\displaystyle n^2$$, using arithmetic modulo $$\displaystyle 10$$. The following table enumerates all possible cases :

$$\displaystyle \begin{array}{|l|l|} \hline n \mod 10 &n^2 \mod 10 \\ \hline 0 &0 \\ 1 &1 \\ 2 &4 \\ 3 &9 \\ 4 &6 \\ 5 &5 \\ 6 &6 \end{array}$$
$$\displaystyle \begin{array}{|l|l|} 7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8 &4 \\ 9 &1 \\ \hline \end{array}$$

Sorry, I had to cut the table midway, it was too big for MHF.

This shows that a perfect square may only end by the digits $$\displaystyle 0, 1, 4, 5, 6$$ and $$\displaystyle 9$$. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

$$\displaystyle \boxed{\mathrm{QED}}$$

This implies that if a number terminates by 3, it cannot be a perfect square.
You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!

• Bacterius and undefined

#### dhiab

Originally Posted by :Swlabr
"You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
Hello :
you know that the rest of division ,by 10, a integer number is first digit.(Clapping)

#### hollywood

And there's always the brute force method:
$$\displaystyle \sqrt{20092010201120122013}=4482411203.93...$$

- Hollywood

#### Swlabr

Originally Posted by :Swlabr
"You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
Hello :
you know that the rest of division ,by 10, a integer number is first digit.(Clapping)
$$\displaystyle a \equiv 3 \text{ mod } 10 \Rightarrow a \equiv 3 \text{ mod } 5$$.

• Bacterius and dhiab

#### chiph588@

MHF Hall of Honor
And there's always the brute force method:
$$\displaystyle \sqrt{20092010201120122013}=4482411203.93...$$

- Hollywood
Going along these lines you could see that $$\displaystyle 4482411203^2<20092010201120122013$$ and $$\displaystyle 4482411204^2>20092010201120122013$$.

• Bacterius