Find all $2{\times}2$ matrices $M$ such that, for some $k > 1$, both $M$ and $M^k$ have entries in arithmetic progression. Let $S$ be the set of $2{\times}2$ matrices 
$$
\pmatrix{
a&b\\
c&d\\
}
$$
such that $a,b,c,d$ are in arithmetic progression.

Find all $M\in S$ such that $M^k\in S$, for some integer $k > 1$.
 A: Let $S$ be the set of matrices $A\in M_2(\mathbb{R})$ such that the sequence
$$a_{11},\;a_{12},\;a_{21},\;a_{22}$$
is an arithmetic sequence.

Clearly, $S$ is a subspace of $M_2(\mathbb{R})$.

Let $T=\{A\in S\mid A^k\in S,\;\text{for some integer}\;k > 1\}$.

Our goal is to find all elements of $T$.

Since
$\pmatrix{
1&1\\
1&1\\
}^2
=
\pmatrix{
2&2\\
2&2\\
}
$,
we have
$
\pmatrix{
1&1\\
1&1\\
}\in T$.

Since
$\pmatrix{
-3&-1\\
1&3\\
}^3
=
\pmatrix{
-24&-8\\
8&24\\
}
$,
we have
$\pmatrix{
-3&-1\\
1&3\\
}\in T$.

It's clear that for all $c\in\mathbb{R}$, if $A\in T$, then $cA\in T$.

Hence for all nonzero $c\in\mathbb{R}$, if $cA\in T$, then $A\in T$.

Claim:$\;A\in T$ if and only if $A=cB$, for some $c\in\mathbb{R}$, where 
$B=\pmatrix{
1&1\\
1&1\\
}\;$
or
$\;B=\pmatrix{
-3&-1\\
1&3\\
}
$.

Proof:

Suppose $A\in T$ is such that not all entries of $A$ are equal.

Then for some nonzero $c\in\mathbb{R}$, we can write $A=cB$, where
$B=\pmatrix{
b-3&b-1\\
b+1&b+3\\
}$, for some $b\in\mathbb{R}$.

To prove the stated claim, it suffices to show $b=0$.

Since $B\in T$, we have $B^k\in S$, for some integer $k > 1$.

Noting that $\det(B)=-8$, it follows that $B$ is nonsingular.

Let $I$ denote the $2{\times}2$ identity matrix.

Let $f\in\mathbb{R}[x]$ be the minimal monic polynomial for $B$ over $\mathbb{R}$.

Since $B$ is not a scalar multiple of $I$, we get $\deg(f)=2$.

It follows that $f$ is equal to the characteristic polynomial of $B$, which yields 
$$
f(x)
=
\det(xI-B)
=
\det\pmatrix{
x-(b-3)&-(b-1)\\
-(b+1)&x-(b+3)\\
}
=
x^2-2bx-8
$$
By the Division Algorithm, we can write
$$x^k = q(x)f(x) + r(x)$$
where $q,r\in\mathbb{R}[x]$, and $\deg(r) < 2$.

Thus, we can write $r(x)=ux+v$, for some $u,v\in\mathbb{R}$.

Then $B^k=q(B)f(B)+r(B)=r(B)$ (since $f(B)=0$), hence $B^k=uB+vI$.

Since $B\in S$, we get $uB\in S$, hence, since $B^k\in S$, we get $B^k-uB\in S$, so $vI$ in $S$, which implies $v=0$.

Thus, $B^k=uB$, hence, since $B$ is non-singular, $B^{k-1}=uI$, and $u\ne 0$.

Let $n=k-1$, and let $g(x)=x^n-u$.

Since $B^n=uI$, and the minimal polynomial for $B$ is $f$, which has degree $2$, it follows that $n\ge 2$ and $f{|}g$ in $\mathbb{R}[x]$.

Since the discriminant of $f$ is $4b^2+32$, it follows that $f$ has two distinct real roots.

But the roots of $f$ are also roots of $g$, hence, since $u\ne 0,\;$for $g$ to have two real roots, $n$ must be even and $u > 0$.

Since $n$ is even and $u > 0$, $g$ has exactly two real roots, each the negation of the other, hence their sum is zero.

Since $f,g$ both have exactly two distinct real roots, it follows that the real roots of $f$ are the same as the real roots of $g$.

Since $f=x^2-2bx-8$, the sum of the roots of $f$ is $2b$, hence we must have $b=0$.

This completes the proof.
