# squareroots of matrices

• Apr 18th 2013, 07:54 AM
siffredi
squareroots of matrices
Hi guys,First of all please pardon my language , I'm from Italy and not fluent in english.
I discoverd a mathematical problem in a book which made me curious: It's about matrices but goes beyond the basic calculation that I'm used to.
We look at the set of 2 x 2 matrices made out of natural numbers only.
Some of these matrices have "squareroots" , i.e. there is some certain 2 x 2 matrix which, multiplicated by itself, gives our original 2 x 2 matrix. The quantity of these square-roots is different from matrix to matrix.
So the question is: What's the quantity of 2 x 2 matrices (filled with natural number, no 0 allowed) which have 2 squareroots AND whose trace is not larger than x.
For example x = 1.000
Any Ideas ?
• Apr 18th 2013, 11:02 AM
Shakarri
Re: squareroots of matrices

$\displaystyle \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \times \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}=\begin{pmatrix} p & q \\ r & s \\ \end{pmatrix}$

You can get 4 simultaneous equations by multiplying out the matrices. And a 5th equation $\displaystyle p+s \leq x$.
Your simultaneous equations will be quadratic so when finding solutions you will get square roots of numbers, you can find more inequalities by ensuring that the square roots are the square root of a positive number.
You might also be able to get equations by knowing that all elements are natural numbers. For example, if you found that $\displaystyle p+r=1+\frac{d}{b+c}$ since p+r must be a natural number d must be a multiple of b+c, so you can say that d=k(b+c)
• Apr 19th 2013, 11:24 AM
siffredi
Re: squareroots of matrices
Quote:

Originally Posted by Shakarri

$\displaystyle \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \times \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}=\begin{pmatrix} p & q \\ r & s \\ \end{pmatrix}$

You can get 4 simultaneous equations by multiplying out the matrices. And a 5th equation $\displaystyle p+s \leq x$.

You mean like:

p = aa + bc
r = ac + dc
q = ab + ab
s = bc + dd

?
and p + s < x because of the trace

Quote:

Your simultaneous equations will be quadratic so when finding solutions you will get square roots of numbers, you can find more inequalities by ensuring that the square roots are the square root of a positive number.
Sry don't get your clue there
Quote:

You might also be able to get equations by knowing that all elements are natural numbers. For example, if you found that $\displaystyle p+r=1+\frac{d}{b+c}$ since p+r must be a natural number d must be a multiple of b+c, so you can say that d=k(b+c)
[/QUOTE]

Ah ok, I understand the schematic here but dont really get how that helps to determine the quantity of the matrices..

Thank you very much for answering, btw I'm not necessarily ambitious in developing a own solution, so if somebody could give me a detailed step by step solution without executing every single calculation , would be great. Just want to see how its done correctly.