The relationship of ${\frak m+m=m}$ to AC

Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.)

Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$?
If not, what is the relationship of this statement for all cardinals ${\frak m}$ to AC?

The weaker principle $\aleph_0\le{\frak m\Rightarrow m}+1={\frak m}$ can be derived in ZF, and it is a well-known result of Tarski that $\forall{\frak m}\ge\aleph_0,{\frak m\times m=m}$ implies the axiom of choice, but I am not sure about the intermediate result with ${\frak m+m=m}$.

Edit: A related question: is it consistent with ZF that there exists a cardinal ${\frak m}\ge\aleph_0$ such that ${\frak m=n+p}$ with ${\frak n<m}$ and ${\frak p<m}$? It's hard to say whether ${\frak m+m=m}$ precludes this scenario.

To add slightly on Andres' answer, it is not only that $\frak m+m=m$ does not imply $\sf AC_\omega$, but in the other direction we can have $\sf DC_\kappa$ (for an arbitrary $\kappa$) but $\frak m+m=m$ fails.

We say that $A$ is a $\kappa$-amorphous set if every subset of $A$ has cardinality $<\kappa$, or its complement has such cardinality - but not both.

Note that if $A$ is $\lambda$-amorphous, then for $\kappa>\lambda$, $A$ is $\kappa$-amorphous as well. We say that $A$ is a properly $\kappa$-amorphous set if it is $\kappa$-amorphous, and $\aleph(A)=\kappa$. In other words, if every ordinal smaller than $\kappa$ can be injected into $A$.

For every regular $\kappa$ it is consistent with $\sf ZF+DC_{<\kappa}$ that there exists a [properly] $\kappa$-amorphous set.

Now, if $A$ is a $\kappa$-amorphous set, then $|A|<|A|+|A|$. This is a trivial observation since $A\times2$ can be written as the union of two sets, neither has complement of size $<\kappa$. So there cannot be an injection from $A\times2$ into $A$, but there is an obvious injection in the other direction.

Therefore $\sf ZF+DC_\kappa$ cannot prove $\frak\forall m. m+m=m$.

(On a slightly more confusing note, I am switching between $\sf DC_\kappa$ and $\sf DC_{<\kappa}$, but the former can be thought as $\sf DC_{<\kappa^+}$.)

To the added question, by the way, this is a simple result. If I recall correctly it is by Tarski.

The axiom of choice is equivalent to the assertion that whenever $\frak p,m,n$ are cardinals and $\frak p+m=n$ then $\frak p=n$ or $\frak m=n$.

The proof is simple, let $\frak a$ be a cardinal, and let $\kappa=\aleph(\frak a)$ the Hartogs number of a set of size $\frak a$. Then $\frak a+\kappa=a$ or $\frak a+\kappa=\kappa$. But $\kappa\nleq\frak a$ so the first option is impossible, therefore $\frak a\leq\kappa$, and can be well-ordered. $\square$

In particular if the axiom of choice fails, let $\frak a$ be a non-well ordered cardinal and $\kappa$ its Hartogs number, then $\frak a+\kappa$ is a set which can be decomposed into two strictly smaller cardinals.

It is also consistent with $\sf ZF$ that every cardinal which is not well-orderdable can be written in such way:

Monro, G.P. Decomposable Cardinals. Fund. Math. vol. 80 (1973), no. 2, 101–104.

Suppose that $A$ is infinite and Dedekind finite. Then $\mathfrak m=|A\cup\mathbb N|$ satisfies that $|A|<\mathfrak m$, $\aleph_0<\mathfrak m$, and $\mathfrak m+\mathfrak m>\mathfrak m$.

To see the last inequality, note that if $\mathfrak m+\mathfrak m=\mathfrak m$ then $A\times 2$ embeds into $A\cup\mathbb N$, say via $f$, but only a finite subset of it embeds into $\mathbb N$, so a set strictly larger than $A$ must embed into $A$.

Note that being Dedekind infinite is the same as embedding $\omega$, so if we require $\mathfrak m+\mathfrak m=\mathfrak m$ for all infinite cardinals $\mathfrak m$, or even for all Dedekind-infinite cardinals, then there are no Dedekind-finite sets. But no, Countable Choice is strictly stronger than the lack of infinite Dedekind finite sets. This is due to Pincus, see this MO question.

As for whether $\mathfrak m+\mathfrak m=\mathfrak m$ (the idemmultiple hypothesis) gives us Countable Choice, the answer is again no, as proved by