Lemma: If $a\in \mathbb{R}, a\neq 0, a\neq 1, a\neq -1; m, n, p, q \in \mathbb{N}; a^m+a^n= a^p+a^q (*); a^{3m}+a^{3n}= a^{3p}+a^{3q}(**)\implies m+n= p+q$ and also $a^{xm}+a^{xn}= a^{xp}+a^{xq}$ $\forall x\in \mathbb{N}$ (1) Proof: Click to reveal hidden textAt the beginning, we have, from $(**) a^{3m}+a^{3n}= a^{3p}+a^{3q}$. So, let's rewrite (**) as $(a^m)^3+ (a^n)^3= (a^p)^3+(a^q)^3\implies (a^m+a^n)(a^{2m}-a^{m+n}+a^{2n})= (a^p+a^q)(a^{2p}-a^{p+q}+a^{2q})$. From $(*)\implies a^m+a^n= a^p+a^q$, and as $a\neq 0, a\neq 1, a\neq -1$, (*) can't be zero, so we will have: $\cancel{(a^m+a^n)}(a^{2m}-a^{m+n}+a^{2n})=\cancel{(a^p+a^q)}(a^{2p}-a^{p+q}+a^{2q})$ $\implies a^{2m}-a^{m+n}+a^{2n}= a^{2p}-a^{p+q}+a^{2q}\implies (a^m+a^n)^2-3a^{m+n}= (a^p+a^q)^2-3a^{p+q}$. Again, we can cut in both sides $(a^m+a^n)$ and $(a^p+a^q)\implies \cancel{-3a}^{m+n}= \cancel{-3a}^{p+q}\implies a^{m+n}= a^{p+q}$ (2). As $a\neq 0, a\neq -1, a\ne 1)\implies$ doesn't have a "period" for his powers, it means it's strictly increasing or decreasing. So, if $w>k\implies a^w>a^k$ (if $a$ is positive or $a$ is negative and $w$ is even) or $a^w<a^k$( if $a$ is negative and $w$ is odd). Now, as we guaranteed this, from (2), we have $m+n= p+q$ (we prove the first part of the lemma). For the second part, we can do induction in $x$. For initial cases, consider (*) and (**). Suppose that the property (1) is valid for $1, 2, ..., k, \in \mathbb{N}$. Let's prove for $(k+1)$. According to the supposition, we know that $a^{km}+a^{kn}= a^{kp}+a^{kq}$. By (*) $\implies (a^m+a^n) (a^{km}+a^{kn})= (a^{kp}+a^{kq}) (a^p+a^q)$ $\implies (a^{km+m}+a^{km+n})+ (a^{kn+n}+a^{kn+m})= (a^{kp+p}+a^{kp+q})+ (a^{kq+q}+a^{kq+p})$ $\implies a^{(k+1)m}+a^{km+n}+a^{(k+1)n}+a^{kn+m}= a^{(k+1)p}+a^{kp+q}+a^{(k+1)q}+a^{kq+p}$ $\implies a^{(k+1)m}+a^{(k+1)n}+a^{m+n}(a^{(k-1)m}+a^{(k-1)n})= a^{(k+1)p}+a^{(k+1)q}+a^{p+q}(a^{(k-1)p}+a^{(k-1)q})$(3) But we can see that, by (2) $a^{m+n}= a^{p+q}$ and by the supposition we know that is valid $(a^{(k-1)m}+a^{(k-1)n})=(a^{(k-1)p}+a^{(k-1)q}) \implies a^{m+n}(a^{(k-1)m}+a^{(k-1)n})=a^{p+q}(a^{(k-1)p}+a^{(k-1)q})$. So, by (3), $a^{(k+1)m}+a^{(k+1)n}+\cancel{a^{m+n}(a^{(k-1)m}+a^{(k-1)n})}= a^{(k+1)p}+a^{(k+1)q}+\cancel{a^{p+q}(a^{(k-1)p}+a^{(k-1)q})}$ $\implies a^{(k+1)m}+a^{(k+1)n}= a^{(k+1)p}+a^{(k+1)q}$, as we wanted. And the result follows by induction. Therefore, for every $x\in \mathbb{N}, a^{xm}+a^{xn}= a^{xp}+a^{xq}$. By the lemma, we can put $x= p+q $, for LHS and $x= m+n $, for RHS. So, we now have $a^{(p+q)m}+a^{(p+q)n}= a^{(m+n)p}+a^{(m+n)q}$ $\implies a^{pm+qm}+a^{pn+qn}= a^{pm+pn}+a^{qm+qn}$ $\implies a^{pm+qm}-a^{pm+pn}+a^{pn+qn}-a^{qm+qn}=0$ $\implies (a^{qm}-a^{pn})(a^{pm}-a^{qn})=0$ As the product is 0, we know that $(a^{qm}-a^{pn})$ or $(a^{pm}-a^{qn})$ is zero. Let's do one case (the other one is similar) Case 1: $(a^{qm}-a^{pn})=0$ As we see before in the proof of the Lemma, we must have $qm= pn$. We also know that $m+n= p+q\implies m-q= p-n$( if $m= q\implies p=n\implies m\cdot n= q\cdot p= p\cdot q$ and the problem would finish) $\implies (m-q)^2= (p-n)^2$ $\implies m^2-\cancel{2mq}+q^2= p^2-\cancel{2pn}+n^2$ $\implies m^2-n^2= p^2-q^2$ $\implies (m-n)(m+n)= (p+q)(p-q)$ Let's separate now into two subcases: Case 1.1 $m=n\implies p=q$. By (*), we would have: $\cancel{2}a^m= \cancel{2}a^p\implies a^m= a^p \implies m=p \implies m=n=p=q\implies m\cdot n= m^2= p\cdot q$ Here, the problem is finish. Case 1.2 If $m\neq n$ and $p\neq q$ $\implies (m-n)\neq 0, (p-q)\neq 0$. As $(m+n)$ and $(p+q)$ are both $>0$ and $m+n= p+q$, we will have $(m-n)\cancel{(m+n)}= \cancel{(p+q)}(p-q)\implies m-n= p-q$. We now have the following system: So $\begin{cases} m+n= p+q (4) \\ m-n= p-q (5)\end{cases}$ Summing (4) and (5), $2m= 2p\implies m= p$. In (*), let's check: $\cancel{a^m}+a^n= \cancel{a^m}+a^q\implies a^n= a^q \Leftrightarrow n=q\therefore m\cdot n= p\cdot q$, as we wanted.