If $n^x$ + $n^y$ + $n^z$ = $n^t$. Solve for positive integers $(x, y, z, t, n)$.
I have realised that there should be infinite solutions when I examined the case $x = y = z$, but I don't know how to proceed further and prove it for all cases. I have also assumed the following : $t>x\geq y\geq z$ Then after simplification, I get the following result: $n^ {(x-z)} + n^ {(y-z)} + 1 = n^ {(t-z)}$
After this, I just have no idea on how to continue further. I am quite new to these types of questions and know very rudimentary number theory. I would appreciate a complete solution using elementary number theory.
Let $x,y,z,t$ be 4 distinct positive integers. There are two cases: either $t$ is the minimum or one of $x,y,z$ is minimum where we can take $x$ as the minimum by symmetry. The former case leads to $$n^{x-t}+n^{y-t}+n^{z-t}=1$$which is impossible since $x-t,y-t,z-t\in\Bbb N$. The latter case also leads to $1+n^{y-x}+n^{z-x}=n^{t-x}$ which yields $n|1$ or $n=1$ and this is impossible. Therefore all $x,y,z,t$ can't be distinct. Let 2 of them be equal.
at this case $$2n^x+n^z=n^t$$where $x,z,t$ are distinct. By the same argument either $t$ is minimum or $x$ or $z$. The only possible case is where $x$ is minimum (check it!) therefore $$2+n^{z-x}=n^{t-x}$$this means that $n|2$ so either $n=1$ which is impossible or $n=2$. Here by substitution we get$$2+2^{z-x}=2^{t-x}$$the only valid answer is $z-x=1$ and $t-x=2$ therefore one general answer is $(x,x,x+1,x+2,2)$
This case doesn't hold since $n^y+n^z=0$
The other cases to investigate are $x=y=z$ or $x=y=t$ or $x=y=z=t$. The two latter cases don't hold (why?) and for $x=y=z$ we have$$3n^x=n^t$$ or $$n^{t-x}=3$$where the only answer is $n=3$ and $t-x=1$ therefore we have two general answers to this question$$(x,x,x+1,x+2,2)\\(x,x,x,x+1,3)$$for which the primitive ones are $$(1,1,2,3,2)\\(1,1,1,2,3)$$