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As far as I know, a real square matrix $M$ has a real square root if $M$ is positive semidefinite, i.e., if all eigenvalues are nonnegative. And, in fact, its square root is unique.

I have read some research papers on how to solve the square root of a $3 \times 3$ positive definite matrix using Cayley-Hamilton, the minimal polynomial, and diagonalization.

However, when does a $3 \times 3$ integer matrix $M$ have an integer square root?

Trivially, $M$ must be positive definite to make sure its square root exists and is real. Also, $\det(M)$ must be a a perfect square. Other than that, I am stuck.

Please help me with this. Or just give me a hint or a lead. Thank you.

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    $\begingroup$ Welcome to MathOverflow! MathOverflow is for mathematicians to ask each other questions about their research. See Mathematics to ask general questions in mathematics. $\endgroup$ – Glorfindel Oct 1 '18 at 11:49
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    $\begingroup$ crossposted: math.stackexchange.com/questions/2937733/… --- don't do that, please. $\endgroup$ – Carlo Beenakker Oct 1 '18 at 12:34
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    $\begingroup$ I agree that the question shouldn't be double-posted, but I find @Glorfindel's comment a bit odd. I don't see why this question is off-topic, although it doesn't help that the OP seems to be omitting various adjectives and seems to think that only PSD matrices have square roots $\endgroup$ – Yemon Choi Oct 4 '18 at 0:10
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For an $n\times n$ matrix $X$ with characteristic polynomial $p_X(t)$, the roots of $p_X$ form the multiset of eigenvalues of $X$. If $A=B^2$ then the eigenvalues of $A$, as a multiset, are the squares of the eigenvalues of $B$, and $$p_A(t^2)=p_B(t)p_B(-t)\tag{*}$$ Now, if $A$ and $B$ are integer matrices then their characteristic polynomials are monic polynomials of degree $n$ with integer coefficients, so that the eigenvalues are algebraic integers, and $(*)$ provides a necessary condition in terms of $p_A$.

This condition can be analyzed further by considering the irreducible factorization of $p_B$ over $\mathbb{Z}$. For $n=3$, there are three cases to consider: (a) 3 linear factors, (b) a linear and an irreducible quadratic factors, and (c) an irreducible cubic polynomial.

(a) Eigenvalues of $A$ are perfect square integers.

(b) One eigenvalue of $A$ is a perfect square integer and the other two are conjugate perfect square integers in a quadratic field.

(c) The eigenvalues of $A$ are conjugate perfect square integers in a cubic field.

This condition is sufficient for semisimple matrices, but not in general. For example, an $n\times n$ Jordan block matrix does not admit an integer square root, essentially, because $1$ is odd.

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