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

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:

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

• 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.
– Prem
Commented Jul 27, 2023 at 10:20
• Does $15^2+20^2=5^4$ fit requirements? Commented Jul 27, 2023 at 10:26
• @coffeemath, thanks for comment. I have add more restriction. Commented Jul 27, 2023 at 10:43
• 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
– Sgg8
Commented Jul 27, 2023 at 15:09
• I suggest you include restriction $m>1,n>1.$ Commented Jul 27, 2023 at 19:34

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:

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

• @Sggs, Thanks for answer. this is a particular case, I forgot to include in restriction. I had to add $m\neq n$. Commented Jul 27, 2023 at 14:16
• @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:)
– Sgg8
Commented Jul 27, 2023 at 14:53
• 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). Commented Jul 27, 2023 at 17:45
• 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
– Sgg8
Commented Jul 27, 2023 at 18:34
• @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 $a^2+b^3=c$ Commented Jul 28, 2023 at 21:15