# Number Theory (1)

• Nov 21st 2009, 09:56 PM
NonCommAlg
Number Theory (1)
This interesting question is taken from an Iranian ((Cool)) competetion (high school level): (Have fun!)

Suppose $\displaystyle x,y,z$ are positive integers with $\displaystyle xy=z^2+1.$ Prove that there exist integers $\displaystyle a,b,c,d$ such that $\displaystyle x=a^2+b^2, \ y=c^2+d^2,$ and $\displaystyle z=ac+bd.$
• Nov 21st 2009, 11:45 PM
simplependulum
you mean

$\displaystyle (a^2 + b^2 )(c^2 + d^2 ) = (ac + bd )^2 + 1 \implies$

$\displaystyle (ad - bc )^2 = 1$

$\displaystyle ad - bc = 1$ or $\displaystyle bc - ad = 1$

I remember that If $\displaystyle (A,B) = 1$ , then there exists integers

$\displaystyle A,B$ such that $\displaystyle Ax + By = 1$ , $\displaystyle (x,y) \in \mathbb{N}$

Is it correct if $\displaystyle (a,b) = (a,c) = (b,d) = (c,d) = 1$

there also exists integers $\displaystyle a,b,c,d$ satisfying

$\displaystyle ad - bc = 1$ or $\displaystyle bc - ad = 1$ ?

For exmaple , set $\displaystyle a = 5 , c = 3$

$\displaystyle 5d - 3b = 1$ , i obtain $\displaystyle b = 3 + 5t , d = 2 + 3t$

let $\displaystyle t = 0$

therefore

$\displaystyle x = a^2 + b^2 = 25 + 9 = 34$

$\displaystyle y = c^2 + d^2 = 9 + 4 = 13$

$\displaystyle z = ac + bd = 15 + 6 = 21$

$\displaystyle 34 \times 13 = 442 = z^2 + 1$

It is also true when $\displaystyle t = 1$

$\displaystyle a = 5 , b= 8 , c=3 , d=5$

$\displaystyle (5^2 + 8^2 )( 3^2 + 5^2 ) = \left ( (5)(3) + (8)(5) \right)^2 + 1$
• Nov 22nd 2009, 03:07 PM
Bruno J.
Oops :)
• Nov 22nd 2009, 05:31 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.

...and $\displaystyle z=ac+bd$ follows by simple computation.