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Does equation $a^m+b^n =c^{m+n}$ have integral solutions?

Equation $a^m+b^n=c^t $, for a and b as a power of 2 have solutions, for example $4^7+4^7=8^5$ , or $4^{13}+4^{13}=8^9$ . Also may have solution like $7^1+1^n=2^3$. But for:

$\begin{cases} a, b \neq 2^k\\a, b \neq 1\\(a, b, c)=1\\(m, n)=1\\m\neq n\end {cases}$

equation $a^m+b^n=c^t$ has no solution. Equation $a^m+b^n=c^{m+n}$ has solutions of the form $(2^k-1)^1+1^2=2^k$, only with $k=3$ without theses restrictions. With these restrictions I could not find any by brute force. If it is true is there a way or article showing this analytically?

Update: We can consider particular cases to modify the equation to linear form so that the solution become easier. For example suppose:

$a=ma_1+1 \Rightarrow a^m=km+1$

$b=nb_1-1; n \text{ is odd} ; \Rightarrow b^n=l n -1$

$c=(m+n)s+1 \Rightarrow c^{m+n}=t(m+n)+1$

then we get:

$km+ln=tm+tn+1$

We must optimize the solution of this equation subjected to our initial equation, that is following system of Diophantine equation mus be consistent:

$\begin{cases}(k-t)m+(l-t)n=1\\a^m+b^n=c^{m+n}\end{cases}$

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    $\begingroup$ When $GCD(m,n)$ is larger than $2$ , it will Contradict FLT. Hence we will have $GCD(m,n)=1$ , $GCD(m,n)=2$ left over. $\endgroup$
    – Prem
    Jul 27 at 10:20
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    $\begingroup$ Does $15^2+20^2=5^4$ fit requirements? $\endgroup$
    – coffeemath
    Jul 27 at 10:26
  • $\begingroup$ @coffeemath, thanks for comment. I have add more restriction. $\endgroup$
    – sirous
    Jul 27 at 10:43
  • $\begingroup$ The second family of solutions I provided suggests a general method of generating large families of solutions, yet I feel like the general solution is much much harder to find and prove $\endgroup$
    – Sgg8
    Jul 27 at 15:09
  • $\begingroup$ I suggest you include restriction $m>1,n>1.$ $\endgroup$
    – coffeemath
    Jul 27 at 19:34

1 Answer 1

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For an arbitrary Pythagorean tripple $(a, b, c)$ with $a^2 + b^2 = c^2$, we have $(ca)^2 + (bc)^2 = c^4$, so $(ca, cb, c)$ forms a solution with $m = n = 2$.

Another solution:

If $a^2+b^3=c$, then multiplying by $c^{24}$ we get:

$$(ac^{12})^2+(bc^8)^3=(c^5)^5$$

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  • $\begingroup$ @Sggs, Thanks for answer. this is a particular case, I forgot to include in restriction. I had to add $m\neq n$. $\endgroup$
    – sirous
    Jul 27 at 14:16
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    $\begingroup$ @sirous I think you should create a new question asking for a general solution to this, rather than adding restrictions indefinitely. The equation looks very interesting to me, however, obtaining a general solution is likely a difficult task, possibly incredibly difficult. Nevertheless, I'm interested too:) $\endgroup$
    – Sgg8
    Jul 27 at 14:53
  • $\begingroup$ Your findings are great, I could not see them if I had put more restriction. But steel waiting for an analytic solution to show there are/ or are not more solutions than what I mentioned in question.(+1). $\endgroup$
    – sirous
    Jul 27 at 17:45
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    $\begingroup$ That's why I suggested that you create a new question. Your question, as it is, was about merely finding a solution, not all of them $\endgroup$
    – Sgg8
    Jul 27 at 18:34
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    $\begingroup$ @Sgg8 What is really important in these kind of problems, is that the bases $a,b,c$ be coprime. Projectively, the point given by the equality $(ac^{12})^2+(bc^8)^3=(c^5)^5$ is the same that the one given by $a^2+b^3=c$ $\endgroup$
    – Piquito
    Jul 28 at 21:15

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