Problem from an old Gazeta Matematica

'Gazeta Matematica' is the most well known mathematical journal in Romania, and it has been published continuously ever since 1895. The current owner of the journal is the Romanian Mathematical Society.

I have come across a rather interesting problem on AoPS recently, a problem concerning linear algebra. Here it is:

Let N be an odd number and A and B real N x N matrices such that the equation

\(det(A+xB)=det(B+xA)\)


has only integer roots. Prove that

\(|det(A-B)|=2^{\frac {n-1}{2}}\sqrt {det(A^2-AB)+det(B^2-AB)}\)


This problem first appeared in a number of Gazeta Matematica (we'll call it 'GM' for short) from 1991.

Here's the solution I offered on AoPS:

Consider the polynomial 

\(P(x,y)=det(xA+yB)=det(A)x^n+a_1x^{n-1}y+...+a_{n-1}xy^{n-1}+det(B)y^n\)

Now, if we consider the polynomial
\(F(X)=det(A+XB)-det(XA+B)\)

.


Obviously,

\(det(A+XB)=P(1,X), det(XA+B)=P(X,1)\).

By substituting the two determinants, we find out that 
\(F(X)=(detA-detB)+x(a_1-a_{n-1})+...+x^n(detB-detA)\)   . 


Its roots have product equal to 1, from Viete's formulas. Thus, since all the roots are in , all roots are 1, and
\(F(X)=(detB-detA)(X-1)^n\)
 . 

By writing 
\((detB-detA)(X-1)^n=det(A+XB)-det(XA+B)\).

,

 and replacing x with -1, we get a nicer form of the conclusion.

I found this problem to be a nice linear algebra exercise, as it tested the student's ability to juggle between polynomials and matrices. It wasn't really hard, but it was interesting nonetheless. 

NOTE: I have included the Maths as photos. This process takes a long while, I really hope I figure out a better way of writing the Maths as soon as possible.

Cris.