I have solved a problem in number theory which I am providing the details below how I did. Request to all of you, please check it and advise me if I made any mistake.
Let $m_1, m_2$ be two odd positive integers such that $$\varphi(m_1)=2^\alpha=\varphi(m_2),~~\text{for some}~~\alpha\in \{1,2,\cdots, 31\}$$ where $\varphi$ is Euler's phi function that counts the total number of relatively prime integers less than $n$ given. Then $m_1=m_2$.
Here is what I have done.
If possible let $m_1\neq m_2$. In that case, the canonical form of $m_2$ must contain a different prime factor than in the canonical form of $m_1$.
So without loss of generality we assume that $p_1, p_1'$ be two distinct primes such that $p_1|m_1, p_1\not\mid m_2, p_1'\not\mid m_1, p_1'|m_2$ and let the canonical form of $m_1, m_2$ be given by \begin{align*} &m_1=p_1^{a_1}\prod_{i=2}^{r}p_i^{a_i}\\ &m_2=p_1'^{a_1'}\prod_{i=2}^{s}p_i^{a_i} \end{align*} where each $a_1, a_1', a_2, \cdots, a_r, \cdots, a_s\in \mathbb{N}$ with $r\leqslant s$.
Now it is easy to show that each $a_1, a_1', a_i$ etc will be equal to 1. In other words, $m_1, m_2$ must be square free. I managed to prove that part. Please ignore it.
So we can write then \begin{align*} &m_1=p_1p_2p_3\cdots p_r \\ &m_2=p_1'p_2p_3\cdots p_s \end{align*} where $r\leqslant s$. Since $2^\alpha=\varphi(m_1)=\varphi(m_2)$, we must have \begin{align*} 2^\alpha=(p_1-1)\prod_{i=2}^{r}(p_i-1)=(p_1'-1)\prod_{i=2}^{s}(p_i-1) \end{align*} If $r<s$ then we get $p_1-1=(p_1'-1)\prod\limits_{i=r+1}^{s}(p_i-1)$ which is contradiction as RHS can be necessarily divided by $2^{s-r+1}$ but LHS is not.
If $r=s$ then we get $p_1-1=p_1'-1$ i.e. $p_1=p_1'$, contradiction again.
hence the proof.
Please tell me if I made any mistake. In case, if any mistake you find, please suggest me how to correct that.
Thank you in advance