1. ## not integer number

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

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

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

4. Fine, I'll give it

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

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

$\begin{array}{|l|l|}
7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
8 &4 \\
9 &1 \\
\hline
\end{array}$

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 $0, 1, 4, 5, 6$ and $9$. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

$\boxed{\mathrm{QED}}$

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

5. Hello, dhiab!

Prove that number $\sqrt{20092010201120122013}$ 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: . $2009\,2010\,2011\,2012\,2013$

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

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

Any ideas?

6. Originally Posted by Bacterius
Fine, I'll give it

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

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

$\begin{array}{|l|l|}
7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
8 &4 \\
9 &1 \\
\hline
\end{array}$

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 $0, 1, 4, 5, 6$ and $9$. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

$\boxed{\mathrm{QED}}$

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!

7. 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.

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

- Hollywood

9. Originally Posted by 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.
$a \equiv 3 \text{ mod } 10 \Rightarrow a \equiv 3 \text{ mod } 5$.

10. Originally Posted by hollywood
And there's always the brute force method:
$\sqrt{20092010201120122013}=4482411203.93...$

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

11. Code:
   10001000100010001*2009 +
1000200030004 =
20092010201120122013 = 7*2870287171588588859