# Couldn't solve this one

• Jun 20th 2010, 05:33 AM
darknight
Couldn't solve this one
Positive integers $\displaystyle a, b$ are both relatively prime and less than or equal to 2008. $\displaystyle a^2 + b^2$ is a perfect square.
$\displaystyle b$ has the same digits as $\displaystyle a$ in the reverse order. The number of such ordered pairs $\displaystyle (a, b)$ is _________ .

I started with 2 digits:
Let $\displaystyle a=xy$ and $\displaystyle b=yx$

$\displaystyle (10x+y)^2 + (10y+x)^2$
$\displaystyle 101x^2 + 40xy + 101y^2$
which can't be factorized and isn't a perfect square.

Tried the same with 3 digits and ended up with this:
$\displaystyle 10001x^2 + 200y^2 + 10001z^2 + 400xz + 2020xy + 2020yz$
This also isn't factorizable.

I've simply got no idea of how to proceed from here.
• Jun 20th 2010, 06:25 AM
Wilmer
There simply is NONE. You sure your question is CORRECTLY worded?
• Jun 21st 2010, 01:36 AM
darknight
Quote:

Originally Posted by Wilmer
There simply is NONE. You sure your question is CORRECTLY worded?

Dunno, someone challenged me to solve it. Guess it was his idea of a joke. :(
My Apologies..
• Jun 21st 2010, 02:46 AM
simplependulum
I am not sure but my answer is zero !

It is a famous property ( which is not what i am confused (Happy) ) , $\displaystyle a-b \equiv 0 \bmod{9}$ , the proof is as follows :

Let $\displaystyle a = \sum_{i=0}^n a_i 10^i ~~ a_i \in \{\ 0,1,2,...,9 \}\$

so $\displaystyle b = \sum_{i=0}^n a_i 10^{n-i}$ and $\displaystyle a-b = \sum_{i=0}^n a_i ( 10^i - 10^{n-i} ) \equiv \sum_{i=0}^n a_i ( 1-1) \bmod{9} \equiv 0 \bmod{9}$

We have $\displaystyle a^2 + b^2 = c^2$

Since $\displaystyle a,b$ coprime , they can be expressed as $\displaystyle m^2 - n^2 ~,~ 2mn$ , wlog let $\displaystyle a = m^2 - n^2 ~ b = 2mn$ so we have

$\displaystyle a-b = m^2 - 2mn - n^2 \equiv 0 \bmod{9}$

$\displaystyle (m-n)^2 - 2n^2 \equiv 0 \bmod{9}$

I consider the form $\displaystyle x^2 - 2y^2$ whether it can be the multiple of $\displaystyle 9$

Be caution the quadratic residues are $\displaystyle [R] = \{\ 0,1,4,7 \}$ so $\displaystyle 2[R] = \{\ 0,2,8,14 \}= \{\ 0,2,5,8 \}$ , since the intersection of the sets is just $\displaystyle \{\ 0 \}$ , we conclude $\displaystyle x^2 - 2y^2 \equiv 0 \bmod{9}$ iff $\displaystyle x \equiv y \equiv 0 \bmod{3}$ . Therefore , $\displaystyle m-n \equiv n \equiv 0 \bmod{3} ~ \implies m \equiv n \equiv 0 \bmod{3}$ which is false because , $\displaystyle (a,b)=1 \implies (m,n)=1$ so we can never find out any odered pair $\displaystyle (a,b)$ .

EDIT: I have made it more complicated , in fact we can consider the quadratic residues from here :

$\displaystyle a^2 + b^2 = c^2$

It is easy to show $\displaystyle a \equiv b \bmod{9}$ since we have already proved that $\displaystyle a - b \equiv 0 \bmod{9}$ , so we have :

$\displaystyle 2a^2 \equiv c^2 \bmod{9}$ consider the residues as i mentioned .
• Jun 21st 2010, 06:33 AM
Wilmer
Quote:

Originally Posted by simplependulum
EDIT: I have made it more complicated , in fact we can consider the quadratic residues from here :
$\displaystyle a^2 + b^2 = c^2$

We could simply apply the "pythagorean triplet" rules, couldn't we, SimpleP ?