1. ## not integer number

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

2. 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.

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 $\displaystyle n$ 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.

5. 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?

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

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!

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:
$\displaystyle \sqrt{20092010201120122013}=4482411203.93...$

- 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.
$\displaystyle a \equiv 3 \text{ mod } 10 \Rightarrow a \equiv 3 \text{ mod } 5$.

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

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

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