# An Equation in Determinants

Dan Sitaru has posted the following problem and its solution at the CutTheKnotMath facebook page:

Solve in $M_{4}(\mathbb{Z})$ the equation:

$\det (X^{4}+I_{4})=2013.$

### Solution

Observe that $X^{4}+I_{4}=(X^{2}-X\sqrt{2}+I_{4})(X^{2}+X\sqrt{2}+I_{4})$ from which

$\det (X^{4}+I_{4})=(m-n\sqrt{2})(m+n\sqrt{2})=m^{2}-2n^{2};$ $m,n\in\mathbb{Z}.$

Now, $m^{2}-2n^{2}=2013$ is same as $m^{2}-2013=2n^{2}.$ The latter implies $m\in 2\mathbb{Z}+1,$ and consequently $m^{2}\in 8\mathbb{Z}+1.$ We'll show that this is impossible by considering two cases.

If $n\in 2\mathbb{Z}$ then $2n^{2}\in 8\mathbb{Z}$ and $m^{2}-2n^{2}\in 8\mathbb{Z}+1$ which means that $2013\in 8\mathbb{Z}+1.$ But this is not so because $2013 \mod 8 \equiv 5.$

If $n\in 2\mathbb{Z}+1$ then $n^{2}\in 8\mathbb{Z}$ so that $m^{2}-2n^{2}\in 8\mathbb{Z}+7,$ or $2013\in 8\mathbb{Z}+5,$ but this is as impossible as the previous case.

Thus the equation has no solutions.

### Remark

The problem can be stated as a scalar or a polynomial equation, with only minor typographical changes.

The problem poses the question for the year 2013. The above solution will work for any year whose residue of division by $8$ is neither $1$ nor $7.$ Thus it will not work out (as it was pointed out in the comments) for $2015;$ there is enough time to investigate whether the matrix equation has or does not have a solution for the coming year. The scalar equivalent, obviously, does not have a solution in integers for most of the years.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny69962938