Secondary Special Camp 2011: NT P 1

[Moon]

Problem 1: (a) Prove that $x^2+y^2+z^2=2007^{2011}$ has no integer solution. (2 points)
(b) Find all positive integers $d$ such that $d$ divides both $n^2+1$ and $(n + 1)^2 + 1$ for some integer $n$. (5 points)

[*Mahi*]

Solution:
Part 1:

Let us take them $mod 16$.
Now , $2007^{2011} \equiv 7 (mod16)$ as $2007 \equiv 7 (mod16)$ and $7^2 \equiv 1(mod)16$ and so $7^{2011}\ equiv 7(mod 16)$
Again,for any $a$ $a^2$ can be only $0,1,9 (mod 16)$ and their combination gives us that $x^2+y^2+z^2 \equiv 0/1/2/9/10/11(mod16)$
So , their can't be any integer solutions to the equation.

Part 2:

$d|n^2+1$ and $d|(n+1)^2+1$ or $d|n^2+2n+2$
So,$d|2n+1$
or,$d|(2n+1)^2$
or,$d|4n^2+4n+1$
Again,$d|2n+1$ and $d|n^2+2n+2$
So,$d|n^2+4n+3$
And so,$d|3n^2-2$
And as $d|n^2+1$ , $d|3n^2+3$
So $d|(3n^2+3)-(3n^2-2)$
Or,$d|5$
And so $d=5$

[Masum]

Problem 1: (a) Prove that $x^2+y^2+z^2=2007^{2011}$ has no integer solution. (2 points)
(b) Find all positive integers $d$ such that $d$ divides both $n^2+1$ and $(n + 1)^2 + 1$ for some integer $n$. (5 points)

$\text{Solution to (a)}$ :
$2011\equiv1\pmod 8$ and so $2007^{2011}\equiv7^1\pmod8$. But it is well-known that no tri-square ( I mean sum of three squares) is not of the form $8k+7$. I am leaving the proof since this fact is easy to prove.
$\text{Solution to (b)}$ :
$d|n^2+1,(n+1)^2+1\Longrightarrow d|2n+1|4n^2-1,4n^2+4\Longrightarrow d|5,d=1,5$
But I don't think that this problem can worth $5$ point at all. Also it is a well-known problem from the book $\text{500 Mathematical Challenges.}$

[Masum]

$d|5$
And so $d=5$[/hide]

What about $d=1$? Isn't it a positive integer?

[Moon]

But I don't think that this problem can worth $5$ point at all. Also it is a well-known problem from the book $\text{500 Mathematical Challenges.}$

Masum, you are saying this because you are very experienced in NT. Moreover, it was not a TST, rather a selection test for the extension camp and in this camp we had more than 50 students. Moreover, many student did not have any prior experience of solving hard olympiad problems.

[*Mahi*]

$d|5$
And so $d=5$

What about $d=1$? Isn't it a positive integer?

As $1$ is too much trivial, I left it out .

[tarek like math]

i don't understand this "$djn2+1;(n+1)2+1=)dj2n+1j4n2À1;4n2+4=)dj5;d=1;$" and the method which Masum Billal write.

[Masum]

i don't understand this "$djn2+1;(n+1)2+1=)dj2n+1j4n2À1;4n2+4=)dj5;d=1;$" and the method which Masum Billal write.

And I don't understand also what you have written. Either I am having problems in eyes or my pc is making problems.

[tarek like math]

It's just first line of ur solution (b)

[Labib]

Masum vai,
I guess Tarek doesn't know elementary Divisibility notations.

OR, he didn't get this part...

$d\mid 2n+1\mid 4n^2-1,4n^2+4$

For Tarek, this is how it happens...

$d\mid (2n+1)$
$\Rightarrow d\mid (2n-1)\cdot (2n+1)$
$\Rightarrow d\mid (4n^2-1)$

again,

$d\mid (n^2+1)$
$\Rightarrow d\mid 4\cdot (n^2+1)$
$\Rightarrow d\mid (4n^2+4)$

Masum vai just fit it in one line, that's all!!