Question: $a$ is a natural number and $a \neq 1$. $(x,y,z)$ are the positive integer solutions for $a^x+(a+1)^y=(2a+1)^z$.
Prove or disprove: If $x,y,z$are not all even, $(x,y,z)$ must be $(1,1,1)$
when a=2, it is correct.
Solution:
(1)If $x\ge3$, then, among all solutions $(x,y,z)$, select $(x_0,y_0,z_0)$ which minimizes the sum $x_0+y_0+z_0$. By Modulo 8, $$3^{y_0}\equiv5^{z_0}$$ Note that $3^{y_0}\equiv1\pmod8$ whenever $y_0$ is even, and $3^{y_0}\equiv3\pmod8$ whenever $y_0$ is odd. Also, $5^{z_0}\equiv1\pmod8$ whenever $z_0$ is even, and $5^{z_0}\equiv5\pmod8$ whenever $z_0$ is odd.
Therefore, both $y_0$ and $z_0$ must be even. Let $y_0=2b$ and $z_0=2a$. $$2^{x_0}+3^{2b}=5^{2a}\implies2^{x_0}=(5^a-3^b)(5^a+3^b)$$which shows that $(5^a-3^b)$ and $(5^a+3^b)$ are both powers of $2$. $$\begin{cases}5^a-3^b=2^m\\5^a+3^b=2^n\end{cases}$$This shows that if $(x_0,y_0,z_0)$ is a solution to the original equation, then $(m,b,a)$ is also a solution.
But we already know that $(x_0,y_0,z_0)$ is the smallest solution. This could only mean that $a\le2$.
Case I. $a=1$. $$\begin{cases}5-3^b=2^m\\5+3^b=2^n\end{cases}$$ From this, $2^m+2^n=10$. The only possible solution to this is $(m,n)=(1,3)$ Hence, $3^b=3\implies b=1$. This shows that $(m,b,a)=(1,1,1)\implies(x_0,y_0,z_0)=(4,2,2)$.
Case II. $a=2$. $$\begin{cases}25-3^b=2^m\\25+3^b=2^n\end{cases}$$From this, $2^m+2^n=50$. There are no integer solutions for this equation.
(2)If x=2,note that [1^{z}-(-1)^{y}\equiv 0(mod4)]and [(-1)^{^{z}}\equiv 1(mod3)].So z is even and y is is even.Let $z=2c$ and $y=2d$ .So $4=(5^c-3^d)(5^c+3^d)$ .But $(5^c-3^d)$ and $(5^c+3^d)$ is even ,There are no integer solutions for this equation.
(3)If x=1,assume that $y>1$. You have $5^z \equiv 2 \pmod9$, which implies $z\equiv 5 \pmod6$. Then you have $5^5-3^y \equiv 2 \pmod7$ ,which implies $y\equiv 0 \pmod6$. You have also $-3^y \equiv 2 \pmod5$ ,which implies $y\equiv 1 \pmod4$. A contradiction. So $y$ must be $1$ and hence $z=1$.
But if a>2,is it correct?
This is not a complete answer, just a few observations way too long for a comment.
Let $a>1$ be an integer, and let $x$, $y$ and $z$ be integers such that $$a^x+(a+1)^y=(2a+1)^z.$$
Observation 1: $z>0$.
Proof. Suppose toward a contradiction that $z\leq0$. Then $(2a+1)^z\leq1$, and because $a^x$ and $(a+1)^y$ are positive it follows that $a^x<1$ and $(a+1)^y<1$, meaning that $x,y<0$. Then clearing denominators yields the polynomial equation $$(a+1)^{-y}(2a+1)^{-z}+a^{-x}(2a+1)^{-z}=a^{-x}(a+1)^{-y},$$ where $-x,-y>0$ and $-z\geq0$. We can factor the left hand side to get $$(2a+1)^{-z}(a^{-x}+(a+1)^{-y})=a^{-x}(a+1)^{-y}.$$ The two factors on the right hand side are both coprime to $(2a+1)^{-z}$, so this forces $(2a+1)^{-z}=1$ and hence $z=0$. Then we are left with $$a^{-x}+(a+1)^{-y}=a^{-x}(a+1)^{-y},$$ but the unique solution to $u+v=uv$ in positive integers is is $u=v=2$, which shows that the equation above has no integral solutions, a contradiction. We conclude that $z>0$.$\qquad\square$
Observation 2: $x,y>0$ unless $(a,x,y,z)=(2,2,0,1)$.
Proof. From $z>0$ it follows that $(2a+1)^z>2$. If $x\leq0$ and $y\leq0$ then $$(2a+1)^z=a^x+(a+1)^y\leq1+1=2,$$ a contradiction. So either $x>0$ or $y>0$. If $y>0$ then $x\geq0$ because $$a^x=(2a+1)^z-(a+1)^y,$$ is an integer. Reducing mod $a$ shows that $x>0$.
Similarly, if $x>0$ then $y\geq0$ for the same reason as before. If $y=0$ then $$(2a+1)^z=a^x+(a+1)^y=a^x+1,$$ which shows that $x>1$ as $(2a+1)^z>a+1$, and that $a$ is even. By the binomial theorem we get $$1=(a+1)^y=-a^x+\sum_{i=0}^z\binom{z}{i}(2a)^i,$$ and hence rearranging the terms and dividing by $a$ shows that $$a^{x-1}=\sum_{i=1}^z\binom{z}{i}(2a)^{i-1}.$$ Because $x>1$ the left hand side is divisible by $a$, and clearly all terms on the right hand side with $i>1$ are divisible by $a$, hence so is the term with $i=1$, which is $\binom{z}{1}=z$. In particular $z$ is also even, say $z=2w$, so $$a^x=(2a+1)^z-1=((2a+1)^w-1)((2a+1)^w+1).$$ Of course the gcd of the two factors on the right hand side divides $2$, and because the two factors are congruent mod $2$ and the left hand side is even, their gcd is precisely $2$. Then by unique factorization we have $$(2a+1)^w-1=2^mu^x\qquad\text{ and }\qquad (2a+1)^w+1=2^nv^x,$$ for odd coprime positive integers $u$ and $v$, and positive integers $m$ and $n$ with $m=1$ or $n=1$. It follows that $$1=\frac{((2a+1)^w+1)-((2a+1)^w-1)}{2}=\frac{2^nv^x-2^mu^x}{2}=2^{n-1}v^x-2^{m-1}u^x,$$ which shows that $m=n=1$ and $x=2$. We are left with $$a^2+1=(2a+1)^z,$$ which forces $z<2$ so $z=1$, from which we find $a=2$. This shows that the only solution $(a,x,y,z)$ that does not satisfy $x,y>0$ is $(2,2,0,1)$.$\qquad\square$
A quicker way to 'show' that this is the only solution with $xy=0$ is to apply Mihăilescu's theorem: If either $x=0$ or $y=0$ then $$(2a+1)^z-(a+1)^y=1\qquad\text{ or }\qquad (2a+1)^z-a^x=1,$$ respectively, and because $2a+1>3$ this implies that either $z<2$ or $x,y<2$, from which the proof is quickly finished. But this result seems too advanced to use for an exercise like this.
Observation 3: If $a=2$ then either $x=y=z=1$ or $x$, $y$ and $z$ are all even.
Proof. If $a=2$ and $x\geq3$ then reducing mod $8$ and mod $3$ the equation $$2^x=5^z-3^y,$$ shows first that $y$ and $z$ are both even, and then that also $x$ is even.
If $x=2$ then $4=5^z-3^y$ and reducing mod $8$ shows that $z$ is odd and $y$ is even, say $y=2v$. Then $$5^z=4+3^y=(2+3^vi)(2-3^vi),$$ and the unique factorization $5=(2+i)(2-i)$ in $\Bbb{Z}[i]$ shows that $v=0$, contradiction observation $2$.
If $x=1$ then $2=5^z-3^y$ and reducing mod $8$ shows that $y$ and $z$ are both odd, and reducing mod $7$ shows that $y\equiv z\equiv 1\pmod{6}$. Then $y=1$ as otherwise $5^z\equiv2\pmod{9}$, which implies $z\equiv5\pmod{6}$, a contradiction. It is then clear that also $z=1$.$\quad\square$
Observation 4: $x\equiv z\pmod{2}$.
Proof. Reducing mod $a+1$ shows that $$(-1)^x+0^y\equiv(-1)^z\pmod{a+1},$$ and $a+1>2$, so the result follows.
Observation 5: If $z>1$ then $x>z$ or $y>z$ and $x,y<2z$.
Proof. If $x,y\leq z$ and $z>1$ then by the binomial theorem $$(2a+1)^z> a^z+(a+1)^z\geq a^x+(a+1)^y.$$ If $a>2$ and $x\geq2z$ then $a^2>2a+1$ and so $$a^x\geq(a^2)^z>(2a+1)^z,$$ and similarly $y\geq2z$ leads to a contradiction.