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3. $P_{k-1}(t,d)$ divides $A+ P_{k-2}(t,d)d \,I $

Conversely, if $(d,t)$ satisfy (1,2,3), there exists a $B\in M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Precisely, if $P_{k-1}(t,d)\ne0$, it is unique, namely $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big).$$ If $P_{k-1}(t,d)=0$, then $A=mI$, is an integer multiple of the identity, and all the infinitely many $B\in M_2 (\mathbb Z)$ with with characteristic polynomial $z^2-tz+d$ satisfy $B^k=A$.

Proof. Assume $A=B^k$ and $B\in M_2 (\mathbb Z)$ and set $t:=\text{tr}(B)$ and $d:=\det(B)$. Then (1) is $\det(A)=\det(B^k)=\det(B)^k=p^k$. As seen above, the characteristic polynomial of $B$, $p_B(z):=z^2-tz+d $ divides the polynomial $z^k-P_{k-1}(t,d)z+P_{k-2}(t,d)d$, and since by Cayley-Hamilton $B^2-tB+d=0$, we also have $$B^k-P_{k-1}(t,d)B+P_{k-2}(t,d)d\, I=0,$$ so, taking the trace, we have $\text{tr}(A)= P_{k-1}(t,d)t -2P_{k-2}(t,d)d$, which is (2), while $P_{k-1}(t,d)B= A+P_{k-2}(t,d)d\, I$, is (3).

Conversely, assume the above conditions (1,2,3) hold for integers $t,d$. Consider first the case $P_{k-1}(t,d)\ne0$. So one can define $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big),$$ an element of $M_2 (\mathbb Z)$ thanks to (3). The trace and determinant of $B$ are then by (1,2), hidding the variables $(t,d)$ in the $P_j$ $$\text{tr}(B)=\frac{\text{tr}(A)+ 2P_{k-2}d}{P_{k-1} } = \frac{P_k +P_{k-2} d }{P_{k-1} } =\frac{ t P_{k-1} }{P_{k-1} } =t $$ $$\det(B)= \frac{\det\Big(A+ P_{k-2} d \,I \Big)}{P_{k-1} ^2} = \frac{P_{k-2}^2d^2 + \text{tr}(A)P_{k-2} d +\det(A) }{P_{k-1}^2}=$$ $$=\frac{P_{k-2}^2d^2 + \big(P_k-P_{k-2}d\big)P_{k-2}d +d^k }{P_{k-1}^2}= $$ $$=\frac{ P_kP_{k-2}d +d^k }{P_{k-1}^2}= d,$$ because $P_{k-1}^2 -P_kP_{k-2} =d^{k-1}$. Thus the characteristic polynomial of $B$ is $z^2-tz+d$, which implies $B^k=P_{k-1}B-P_{k-2}d\, I=A$. Finally, consider the case $P_{k-1}(t,d)=0$. By (3) $A$ is then a multiple of the identity, $A=m\, I$, for $m:=-P_{k-2}(t,d)d$. If $m=0$, any nilpotent $B$ has the wanted properties. If $m\ne 0$, let $\lambda$ and $\mu$ be the roots of $z^2-tz+d$, so $t=\lambda+\mu$ and $d=\lambda\mu$. Then we have $\lambda\neq\mu$, otherwise $0=P_{k-1}(\lambda+\mu,\lambda\mu)=k\lambda^{k-1}$ and $\lambda=\mu=0=t=d$ and $A=0$. Also (see below) $$0= P_{k-1}(\lambda+\mu,\lambda\mu)= \frac{\lambda^k-\mu^k}{\lambda-\mu} $$ whence $\lambda^k=\mu^k$, and $$m =-\lambda\mu P_{k-2}(\lambda+\mu,\lambda\mu)=-\frac{\lambda^k\mu-\mu^k\lambda}{\lambda-\mu} = \lambda^k=\mu^k.$$ Let $B$ one of the infinitely many matrices in $M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Since $\lambda\ne \mu$, $B$ is diagonalizable, $B=Q^{-1}\text{diag}(\lambda,\mu)Q$, so $$B^k=Q^{-1}\text{diag}(\lambda^k,\mu^k)Q=Q^{-1}mIQ=mI=A,$$ ending the proof.

3. $P_{k-1}(t,d)$ divides $A+ P_{k-2}(t,d)d I $

Conversely, if $(d,t)$ satisfy (1,2,3), there exists a $B\in M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Precisely, if $P_{k-1}(t,d)\ne0$, it is unique, namely $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d I \Big).$$ If $P_{k-1}(t,d)=0$, then $A=mI$, is an integer multiple of the identity, and all the infinitely many $B\in M_2 (\mathbb Z)$ with with characteristic polynomial $z^2-tz+d$ satisfy $B^k=A$.

Proof. Assume $A=B^k$ and $B\in M_2 (\mathbb Z)$ and set $t:=\text{tr}(B)$ and $d:=\det(B)$. Then (1) is $\det(A)=\det(B^k)=\det(B)^k=p^k$. As seen above, the characteristic polynomial of $B$, $p_B(z):=z^2-tz+d $ divides the polynomial $z^k-P_{k-1}(t,d)z+P_{k-2}(t,d)d$, and since by Cayley-Hamilton $B^2-tB+d=0$, we also have $$B^k-P_{k-1}(t,d)B+P_{k-2}(t,d)d I=0,$$ so, taking the trace, we have $\text{tr}(A)= P_{k-1}(t,d)t -2P_{k-2}(t,d)d$, which is (2), while $P_{k-1}(t,d)B= A+P_{k-2}(t,d)d I$, is (3).

Conversely, assume the above conditions (1,2,3) hold for integers $t,d$. Consider first the case $P_{k-1}(t,d)\ne0$. So one can define $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d I \Big),$$ an element of $M_2 (\mathbb Z)$ thanks to (3). The trace and determinant of $B$ are then by (1,2), hidding the variables $(t,d)$ in the $P_j$ $$\text{tr}(B)=\frac{\text{tr}(A)+ 2P_{k-2}d}{P_{k-1} } = \frac{P_k +P_{k-2} d }{P_{k-1} } =\frac{ t P_{k-1} }{P_{k-1} } =t $$ $$\det(B)= \frac{\det\Big(A+ P_{k-2} d I \Big)}{P_{k-1} ^2} = \frac{P_{k-2}^2d^2 + \text{tr}(A)P_{k-2} d +\det(A) }{P_{k-1}^2}=$$ $$=\frac{P_{k-2}^2d^2 + \big(P_k-P_{k-2}d\big)P_{k-2}d +d^k }{P_{k-1}^2}= $$ $$=\frac{ P_kP_{k-2}d +d^k }{P_{k-1}^2}= d,$$ because $P_{k-1}^2 -P_kP_{k-2} =d^{k-1}$. Thus the characteristic polynomial of $B$ is $z^2-tz+d$, which implies $B^k=P_{k-1}B-P_{k-2}d\, I=A$. Finally, consider the case $P_{k-1}(t,d)=0$. By (3) $A$ is then a multiple of the identity, $A=m I$, for $m:=-P_{k-2}(t,d)d$. If $m=0$, any nilpotent $B$ has the wanted properties. If $m\ne 0$, let $\lambda$ and $\mu$ be the roots of $z^2-tz+d$, so $t=\lambda+\mu$ and $d=\lambda\mu$. Then we have $\lambda\neq\mu$, otherwise $0=P_{k-1}(\lambda+\mu,\lambda\mu)=k\lambda^{k-1}$ and $\lambda=\mu=0=t=d$ and $A=0$. Also (see below) $$0= P_{k-1}(\lambda+\mu,\lambda\mu)= \frac{\lambda^k-\mu^k}{\lambda-\mu} $$ whence $\lambda^k=\mu^k$, and $$m =-\lambda\mu P_{k-2}(\lambda+\mu,\lambda\mu)=-\frac{\lambda^k\mu-\mu^k\lambda}{\lambda-\mu} = \lambda^k=\mu^k.$$ Let $B$ one of the infinitely many matrices in $M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Since $\lambda\ne \mu$, $B$ is diagonalizable, $B=Q^{-1}\text{diag}(\lambda,\mu)Q$, so $$B^k=Q^{-1}\text{diag}(\lambda^k,\mu^k)Q=Q^{-1}mIQ=mI=A,$$ ending the proof.

Consider the polynomials $P_k(x,y)\in\mathbb{Z}[x,y]$

$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j.$$

Then

A matrix $A\in M_2 (\mathbb Z)$ is a $k$-th power of an element of $ M_2 (\mathbb Z)$ if and only if there are $t,d$ in $\mathbb Z$ such that

Moreover, if $B\in M_2 (\mathbb Z)$ verifies $B^k=A$ then $d:=\det(B)$ and $t:=\text{tr}(B)$ satisfy 1,2,3. Conversely, if $(d,t)$ satisfy 1,2,3, there exists $B\in M_2 (\mathbb Z)$ such that $B^k=A$, $d:=\det(B)$ and $t:=\text{tr}(B)$.

 Precisely, if $P_{k-1}(t,d)\ne0$, one can take $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big).$$ If $A\ne0$ and $P_{k-1}(t,d)=0$, then $k=2h$, $b=0$ and $d=\pm p^2$ for $h,p$ in $\mathbb{N}$. Then one can take $$B:= \begin{bmatrix} 0 & -p \\ p & 0 \end{bmatrix} \text{ if } d>0,\qquad B:= \begin{bmatrix} 0 & p \\ p & 0 \end{bmatrix} \text{ if } d<0.$$

Edit.Edit, 11.11.2020 The proof.The proof, is essentially routine; yet I post everything for convenience, before I forget all details.

It is convenient to introduce the polynomials
$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j;$$ their relevance in this context being that the polynomial $z^2-xz+y$ divides the polynomial $$z^{k+2}-xP_{k+1}(x,y)z+yP_{k}(x,y)$$ (even as elements of $\mathbb{Z}[x,y,z]$; see below for other properties we need).

Characterization of the $k$-th powers in $ M_2(\mathbb Z)$. Let $k\ge0$. A matrix $A\in M_2 (\mathbb Z)$ is a $k$-th power of an element of $ M_2 (\mathbb Z)$ if and only if there are $t,d$ in $\mathbb Z$ such that

Precisely, if $B\in M_2 (\mathbb Z)$ verifies $B^k=A$ then $d:=\det(B)$ and $t:=\text{tr}(B)$ satisfy (1,2,3).

Conversely, if $(d,t)$ satisfy (1,2,3), there exists a $B\in M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$.  Precisely, if $P_{k-1}(t,d)\ne0$, it is unique, namely $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big).$$ If $P_{k-1}(t,d)=0$, then $A=mI$, is an integer multiple of the identity, and all the infinitely many $B\in M_2 (\mathbb Z)$ with with characteristic polynomial $z^2-tz+d$ satisfy $B^k=A$.

Proof. Assume $A=B^k$ and $B\in M_2 (\mathbb Z)$ and set $t:=\text{tr}(B)$ and $d:=\det(B)$. Then (1) is $\det(A)=\det(B^k)=\det(B)^k=p^k$. As seen above, the characteristic polynomial of $B$, $p_B(z):=z^2-tz+d $ divides the polynomial $z^k-P_{k-1}(t,d)z+P_{k-2}(t,d)d$, and since by Cayley-Hamilton $B^2-tB+d=0$, we also have $$B^k-P_{k-1}(t,d)B+P_{k-2}(t,d)d\, I=0,$$ so, taking the trace, we have $\text{tr}(A)= P_{k-1}(t,d)t -2P_{k-2}(t,d)d$, which is (2), while $P_{k-1}(t,d)B= A+P_{k-2}(t,d)d\, I$, is (3).

Conversely, assume the above conditions (1,2,3) hold for integers $t,d$. Consider first the case $P_{k-1}(t,d)\ne0$. So one can define $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big),$$ an element of $M_2 (\mathbb Z)$ thanks to (3). The trace and determinant of $B$ are then by (1,2), hidding the variables $(t,d)$ in the $P_j$ $$\text{tr}(B)=\frac{\text{tr}(A)+ 2P_{k-2}d}{P_{k-1} } = \frac{P_k +P_{k-2} d }{P_{k-1} } =\frac{ t P_{k-1} }{P_{k-1} } =t $$ $$\det(B)= \frac{\det\Big(A+ P_{k-2} d \,I \Big)}{P_{k-1} ^2} = \frac{P_{k-2}^2d^2 + \text{tr}(A)P_{k-2} d +\det(A) }{P_{k-1}^2}=$$ $$=\frac{P_{k-2}^2d^2 + \big(P_k-P_{k-2}d\big)P_{k-2}d +d^k }{P_{k-1}^2}= $$ $$=\frac{ P_kP_{k-2}d +d^k }{P_{k-1}^2}= d,$$ because $P_{k-1}^2 -P_kP_{k-2} =d^{k-1}$. Thus the characteristic polynomial of $B$ is $z^2-tz+d$, which implies $B^k=P_{k-1}B-P_{k-2}d\, I=A$. Finally, consider the case $P_{k-1}(t,d)=0$. By (3) $A$ is then a multiple of the identity, $A=m\, I$, for $m:=-P_{k-2}(t,d)d$. If $m=0$, any nilpotent $B$ has the wanted properties. If $m\ne 0$, let $\lambda$ and $\mu$ be the roots of $z^2-tz+d$, so $t=\lambda+\mu$ and $d=\lambda\mu$. Then we have $\lambda\neq\mu$, otherwise $0=P_{k-1}(\lambda+\mu,\lambda\mu)=k\lambda^{k-1}$ and $\lambda=\mu=0=t=d$ and $A=0$. Also (see below) $$0= P_{k-1}(\lambda+\mu,\lambda\mu)= \frac{\lambda^k-\mu^k}{\lambda-\mu} $$ whence $\lambda^k=\mu^k$, and $$m =-\lambda\mu P_{k-2}(\lambda+\mu,\lambda\mu)=-\frac{\lambda^k\mu-\mu^k\lambda}{\lambda-\mu} = \lambda^k=\mu^k.$$ Let $B$ one of the infinitely many matrices in $M_2 (\mathbb Z)$ with characteristic polynomial $z^2-tz+d$. Since $\lambda\ne \mu$, $B$ is diagonalizable, $B=Q^{-1}\text{diag}(\lambda,\mu)Q$, so $$B^k=Q^{-1}\text{diag}(\lambda^k,\mu^k)Q=Q^{-1}mIQ=mI=A,$$ ending the proof.

More details. The sequence of polynomials $P_k(x,y)\in\mathbb{Z}[x,y]$ is defined by the two-term recurrence $$\cases{P_{k+2}=xP_{k+1}-yP_k\\ P_0=1 \\ P_{-1}=0.}$$ One easily verifies by induction the expansion $$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j;$$ in fact $P_k$ may also be presented in terms of the Chebyshev polynomials of the first kind as $P_k(x,y^2)=y^kxT_k\big(\frac{x}{2y}\big)\in\mathbb{Z}[x,y^2]$. They verify $$P_k(u+v,uv)=\frac{u^{k+1}-v^{k+1}}{u-v}=\sum_{j=0}^{k} u^jv^{k-j},$$ and, related to that, for all $k\ge0$ one has: $$z^{k+2}-P_{k+1}(x,y)z+yP_{k}(x,y)=\big(z^2-xz+y\big) \sum_{j=0}^kP_{k-j}(x,y)z^j, $$ both easily verified by induction. Finally, since they solve a two-term linear recursion, the Hankel determinant of order $2$ must be a $1$-term linear recurrence, and one finds $$P_{k}(x,y)^2-P_{k+1}(x,y)P_{k-1}(x,y)=y^{k}.$$

Here is the necessary and sufficient condition, in terms of $\det A$ and $\text{tr}(A)$, in order that a $2\times2$ matrix $A$ be the $k$-th power of some matrix with integer coefficients.

Consider the polynomials $P_k(x,y)\in\mathbb{Z}[x,y]$

$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j.$$

Then

A matrix $A\in M_2 (\mathbb Z)$ is a $k$-th power of an element of $ M_2 (\mathbb Z)$ if and only if there are $t,d$ in $\mathbb Z$ such that

1. $\det(A)=d^k$

2. $\text{tr}(A)=P_k(t,d)-P_{k-2}(t,d)d$

3. $P_{k-1}(t,d)$ divides $A+ P_{k-2}(t,d)d \,I $

Moreover, if $B\in M_2 (\mathbb Z)$ verifies $B^k=A$ then $d:=\det(B)$ and $t:=\text{tr}(B)$ satisfy 1,2,3. Conversely, if $(d,t)$ satisfy 1,2,3, there exists $B\in M_2 (\mathbb Z)$ such that $B^k=A$, $d:=\det(B)$ and $t:=\text{tr}(B)$.

Precisely, if $P_{k-1}(t,d)\ne0$, one can take $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big).$$ If $A\ne0$ and $P_{k-1}(t,d)=0$, then $k=2h$ and $d=\pm p^2$ for $h,p$ in $\mathbb{N}$. Then one can take $$B:= \begin{bmatrix} 0 & -p \\ p & 0 \end{bmatrix} \text{ if } d>0,\qquad B:= \begin{bmatrix} 0 & p \\ p & 0 \end{bmatrix} \text{ if } d<0.$$

Here is the necessary and sufficient condition, in terms of $\det A$ and $\text{tr}(A)$, in order that a $2\times2$ matrix $A$ be the $k$-th power of some matrix with integer coefficients.

Consider the polynomials $P_k(x,y)\in\mathbb{Z}[x,y]$

$$P_k(x,y):=\sum_{j\ge0}(-1)^j{ k-j\choose j }x^{k-2j}y^j.$$

Then

A matrix $A\in M_2 (\mathbb Z)$ is a $k$-th power of an element of $ M_2 (\mathbb Z)$ if and only if there are $t,d$ in $\mathbb Z$ such that

1. $\det(A)=d^k$

2. $\text{tr}(A)=P_k(t,d)-P_{k-2}(t,d)d$

3. $P_{k-1}(t,d)$ divides $A+ P_{k-2}(t,d)d \,I $

Moreover, if $B\in M_2 (\mathbb Z)$ verifies $B^k=A$ then $d:=\det(B)$ and $t:=\text{tr}(B)$ satisfy 1,2,3. Conversely, if $(d,t)$ satisfy 1,2,3, there exists $B\in M_2 (\mathbb Z)$ such that $B^k=A$, $d:=\det(B)$ and $t:=\text{tr}(B)$.

Precisely, if $P_{k-1}(t,d)\ne0$, one can take $$B:=\frac1{P_{k-1}(t,d)}\Big(A+ P_{k-2}(t,d)d \,I \Big).$$ If $A\ne0$ and $P_{k-1}(t,d)=0$, then $k=2h$, $b=0$ and $d=\pm p^2$ for $h,p$ in $\mathbb{N}$. Then one can take $$B:= \begin{bmatrix} 0 & -p \\ p & 0 \end{bmatrix} \text{ if } d>0,\qquad B:= \begin{bmatrix} 0 & p \\ p & 0 \end{bmatrix} \text{ if } d<0.$$