The equation $3^n+4^m=5^k$ in positive integer numbers Please help me to prove that the equation $3^n + 4^m = 5^k$
where $n$, $m$, $k$ are positive integer numbers has only the solution $n=m=k=2$.
I know how to prove it for $n=m=k$.
If $3^x + 4^x = 5^x$ then $(3/4)^x + 1 = (5/4)^x$,
and this equation has at most one solution
since the function $(3/4)^x + 1$ decreases and the function $(5/4)^x$ increases
for all real $x$. So the solution $x = 2$ is unique.
Thank you very much in advance!
 A: Considering the equation modulo 3 and 4 separately, we see that $2 | n \implies n = 2a$ and $2 | k \implies k = 2c$ for some positive integers $a, c$. Rewriting, we get
$$\begin{align}
3^{2a} + 4^m &= 5^{2c}\\
2^{2m} &= (5^c)^2 - (3^a)^2\\
2^{2m} &= (5^c + 3^a)(5^c - 3^a)\end{align}$$
We thus allow $5^c + 3^a = 2^y$ and $5^c - 3^a = 2^x$ for non-negative integers $x, y$ such that $x + y = 2m$. Summing both up, we get:
$$2\cdot5^c = 2^x + 2^y$$
If $x = 0$, then $2\cdot5^c = 1 + 2^y\implies y = 0 \implies m = 0$. But the question states that $m > 0$, so the case where $x = 0$ is impossible $\implies x > 0 \implies y > 0$. 
This allows us to divide throughout by $2$ to get:
$$5^c = 2^{x - 1} + 2^{y - 1}$$
Taking both sides modulo $2$, we find that the above is impossible for $x-1 > 0$. Hence, $x - 1 = 0$, or $x = 1$. From this, we get:
$$\begin{align}5^c = 1 + 2^{y - 1}\\
5^c - 2^{y - 1} = 1\end{align}$$
This smells of Catalan's Conjecture (now promoted to a theorem), which implies (in this context) that either $c$ or $y - 1$ must be equal to $1$.
If $y - 1 = 1$, then $5^c - 2^1 = 1 \implies 5^c = 3$, which has no solution.
If $c = 1$, then $k = 2c = 2$ and $5 - 2^{y - 1} = 1 \implies 2^{y - 1} = 4 \implies y = 3$. But we deduced that $x = 1$, so by substituting back into the relation $x + y = 2m$, we get $m = 2$, which after substituting back into the original equation gives us directly $n = 2$.
We therefore conclude that there is only one solution, that is, $(n, m, k)$ = $(2, 2, 2)$.
Perhaps the above approach could somehow be used to solve the mess I made over here at this thread.
A: Reducing the equation mod $3$ shows that $k$ must be even, and reducing mod $4$ shows the same for $n$, so let's rewrite the whole thing as
$$3^{2n}+2^{2m}=5^{2k}$$
and show that $(n,m,k)=(1,2,1)$ is the only solution in positive integers.
Now
$$2^{2m}=5^{2k}-3^{2n}=(5^k-3^n)(5^k+3^n)$$
implies $5^k-3^n=2^a$ and $5^k+3^n=2^b$ with $a\lt b=2m-a$.  Subtracting these gives
$$2\cdot3^n=2^a(2^{b-a}-1)$$
which implies $a=1$ and $3^n=2^{b-a}-1$.  It remains to show that the only positive power of $3$ of the form $2^u-1$ is $3^1=2^2-1$.  
If $3$ divides $2^u-1$, then $u$ must be even, so let $u=2v$.  In that case, $2^u-1=(2^v-1)(2^v+1)$.  In order for $3$ to divide $2^v+1$, $v$ must be odd.  But if $v$ is odd then $2^v-1$ is not divisible by $3$, so the only way it can be a power of $3$ is if it's equal to $1$, which implies $v=1$, which is exactly what we need.
Added later:  One might alternatively rewrite the (modified) equation as
$$3^{2n}=5^{2k}-2^{2m}=(5^k-2^m)(5^k+2^m)$$
which implies $5^k-2^m=3^a$ and $5^k+2^m=3^b$ with $a\lt b=2n-a$.  In this case subtracting gives
$$2\cdot2^m=3^a(3^{b-1}-1)$$
which implies $a=0$, so that $5^k-2^m=1$, or $5^k=2^m+1$.  This leads to $3^{2n}=2^{m+1}+1$, or $2^{m+1}=(3^n-1)(3^n+1)$.  As before, we must have $3^n-1=2^a$ and $3^n+1=2^b$ with $a\lt b$, but this time subtracting gives $2=2^a(2^{b-1}-1)$, which forces $a=1$, hence $n=1$.
Note, the only role of $5$ in either approach is to be congruent to $2$ mod $3$ and $1$ mod $4$ (so that we can replace $n$ and $k$ with $2n$ and $2k$), so the same proof works with anything congruent to $5$ mod $12$, e.g., $17$, $29$, $41$, etc.  The only difference is that in these cases the final conclusion is that there are no solutions:  Once you've got $n=1$ and $m=2$ on the left hand side, you can only have $5^2$ on the right.
