Odd binomial sum equality has only trivial solution? Suppose $$\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2} = \sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}.$$
Does $m=n=1$?
Clearly $m \leq n$, and for every $n$ there is at most one $m$.
 A: Here we find a closed expression for the series and    show that a solution with $m$ even is not possible.

We use the convention $\binom{n}{k}=0$ if $0\leq n < k$ and start with the left-hand side. We obtain
  \begin{align*}
\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2}&= \sum_{k=0}^n \binom{n}{2k+1} 2^k\tag{1}\\
&=\frac{1}{\sqrt{2}}\sum_{k=0}^n\binom{n}{2k+1}\left(\sqrt{2}\right)^{2k+1}\tag{2}\\
\end{align*}

Comment:


*

*In (1) we replace $k$ with $2k+1$ in the summands without changing anything since we add only zeros.

*In (2) we do a rearrangement which is helpful for further steps.

We consider
\begin{align*}
  f(x)=\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n
  \end{align*}
  and get the odd part of $f(x)$ via
  \begin{align*}
  \frac{1}{2}\left(f(x)-f(-x)\right)=\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}
  \end{align*}
We obtain from (2)
  \begin{align*}
\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2}&=\frac{1}{2\sqrt{2}}\left(f(\sqrt{2})-f(-\sqrt{2})\right)\\
&=\frac{1}{2\sqrt{2}}\left(\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n\right)\tag{3}
\end{align*}
Similarly we can transform the right-hand side:
\begin{align*}
\sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}&=3^{(m-1)/2}\sum_{k\ {\rm odd}}^m {m \choose k} \left(\frac{2}{3}\right)^{(k-1)/2}\\
&=3^{m/2}\frac{1}{2\sqrt{2}}\left(f\left(\sqrt{\frac{2}{3}}\right)
-f\left(-\sqrt{\frac{2}{3}}\right)\right)\\
&=3^{m/2}\frac{1}{2\sqrt{2}}\left(\left(1+\sqrt{\frac{2}{3}}\right)^m-\left(1-\sqrt{\frac{2}{3}}\right)^m\right)\\
&=\frac{1}{2\sqrt{2}}\left(\sqrt{3}+\sqrt{2}\right)^m-\left(\sqrt{3}-\sqrt{2}\right)^m\tag{4}
\end{align*}

We conclude from (3) and (4) OPs equation is equivalent with

\begin{align*}
  \left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n
  =\left(\sqrt{3}+\sqrt{2}\right)^m-\left(\sqrt{3}-\sqrt{2}\right)^m\tag{5}
  \end{align*}

In case $m=n=1$ we obtain
\begin{align*}
  \left(1+\sqrt{2}\right)-\left(1-\sqrt{2}\right)&=2\sqrt{2}\\
\\
  \left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{3}-\sqrt{2}\right)&=2\sqrt{2}
  \end{align*}
and equality holds.

The  left-hand side of (5) has the representation
  \begin{align*}
\left(1+\sqrt{2}\right)^n&-\left(1-\sqrt{2}\right)^n\\
&=\sum_{k=0}^n\binom{n}{k}\left(\sqrt{2}\right)^k-\sum_{k=0}^n\binom{n}{k}(-1)^k\left(\sqrt{2}\right)^k\tag{6}\\
&=A\sqrt{2}\qquad\qquad  \text{ with }A\in\mathbb{N}
\end{align*}
Since odd $k$ only provide contributions to the expression in (6) we observe that the resulting value is an integer multiple of $\sqrt{2}$. 
The right-hand side has the representation
  \begin{align*}
\left(\sqrt{3}+\sqrt{2}\right)^n&-\left(\sqrt{3}-\sqrt{2}\right)^n\\
&=\sum_{k=0}^n\binom{n}{k}\left(\sqrt{2}\right)^k\left(\sqrt{3}\right)^{n-k}
-\sum_{k=0}^n\binom{n}{k}(-1)^k\left(\sqrt{2}\right)^k\left(\sqrt{3}\right)^{n-k}\tag{7}\\
&=
\begin{cases}
B\sqrt{2}\qquad\qquad  &\text{ with }B\in\mathbb{N} \text { if } n \text { is odd}\\
C\sqrt{6}\qquad\qquad  &\text{ with }C\in\mathbb{N} \text { if } n \text { is even}\\
\end{cases}
\end{align*}
Again odd $k$ only provide contributions to the expression in (7). If $n$ is odd, $n-k$ is even and the resulting value is an integer multiple of $\sqrt{2}$. Otherwise, if $n$ is even we get additionally a factor $\sqrt{3}$ resulting in an integer multiple of $\sqrt{6}$.

Conclusion: OPs equation is not valid if $m$ is even.
The case $m$ odd needs further investigations.
A: Ricardo,
Your equation only has finite number of solutions, but in order to check whether $m=n=1$ is the unique solution, you would need some numerical verification and the existing bounds might not be sufficiently effective.
Relevant references:


*

*M. Laurent. Équations exponentielles polynômes et suites récurrentes linéaires. Astérisque 147–148 (1987), 121–139.

*H. P. Schlickewei and W. Schmidt. Linear equations in members of recurrence sequences. Ann. Scuola Norm. Sup. Pisa Cl. Sci. 20 (1993), 219–246.

*H. P. Schlickewei and W. Schmidt. The intersection of recurrence sequences. Acta Arith. 72 (1995), 1–44.


The equivalence of your problem to finding intersections of two linear recursive sequences $u_n=v_p$, where $p=2m+1$ and
$u_{n+1}=2u_{n}+u_{n-1}$, with $u_0=0$, $u_1=1$;
$v_{p+1}=10v_{p}-v_{p-1}$, with $v_0=1$, $v_1=11$,
can be seen by either directly using the original representation in sum or using the closed-form solutions in Markus Scheuer’s answer.
This set-up has been studied in much greater generality in the reference above.  Since $u$ and $v$ are not related, there are only finite solutions.  There is some bound of the size of the solutions in 3 but I am not sure how effective they are.  You might want to go through more recent work, e.g.,


*M. Bennett and P. Ákos. Intersections of recurrence sequences. Proceedings of the American Mathematical Society 143, no. 6 (2015), 2347-2353.


to see if any improvement has been developed.
