Solve $1+a^c+b^c=\text{lcm}(a^c,b^c)$ Question: Find all positive integer solutions to $1+a^c+b^c=\text{lcm}(a^c,b^c)$ where LCM denotes the least common multiple.
Attempt: Substitute $a^c=A; b^c=B$. We know that $\text{lcm}(A,B) \cdot \gcd(A,B)=AB$, so if we denote by $d$ the $\gcd$ of A and B, we get that $\text{lcm}(A,B)=\frac{AB}{\gcd(A,B)}=\frac{AB}{D}; \frac{AB}{D}=1+A+B; AB=D(1+A+B)$
$AB=D(1+A+B)$ // $+DAB$
$DAB+AB=D(1+A+B)+DAB$
$(D+1)AB=D(1+A+B+AB)$
$(D+1)AB=D(A+1)(B+1)$
But I am stuck here. I don't even know if any of this is useful. 

Let $a^c=A$ and $b^c=B$. Therefore since $A,B|lcm(A,B)$ we have:$$1+B=lcm(A,B)-A\\1+A=lcm(A,B)-B$$which concludes that $$A|1+B\\B|1+A$$therfore $\gcd(A,B)=1$ and we have $lcm(A,B)=AB$. Embarking on this:$$1+A+B=AB$$ or $$(A-1)(B-1)=2$$which yields to $$(A,B)=(1,2)\\(A,B)=(2,1)$$or equivalently $$a^c=1\\b^c=2$$(the other answer has been ignored hence of symmetry) the only positive integer answer is $a=c=1$ and $b=2$
