Is the conjecture A+B=C following correct?

Is the conjecture on A+B=C following correct ?

Conjecture: Let $$A, B, C$$ be three positive integer numbers such that $$A+B=C$$ with $$\gcd(A, B, C) = 1$$. By Fundamental theorem of arithmetic we write:

$$A=a_1^{x_1}a_2^{x_2}...a_n^{x_n}$$,

$$B=b_1^{y_1}b_2^{y_2}...b_m^{y_m}$$,

$$C=c_1^{z_1}c_2^{z_2}...c_k^{z_k}$$

Let $$d = \min\{x_i, y_j, z_h \}$$ where $$1 \le i \le n, 1\ \le j \le m, 1\le h \le k$$ then $$d \le 5$$

PS: I read above one hunded papers, I observed that in any case $$\min\{x_i, y_j, z_h \} \le 3$$

Example 1: Ten solutions of Catalan-Fermat equation

Example 2:

$$2^4.3^5.7^6+5^9.11^8=19^1.23^1.47^1.6679^1.3051977^1$$

• If $\min\{x_i,y_j,z_h\}\geq 6$, then $rad(ABC)\leq \sqrt[6]{ABC}\leq C^{1/2}$, so $C\geq rad(ABC)^2$. The truth of the abc conjecture would imply there are only finitely-many triples $A,B,C$ with $\min\{x_i,y_j,z_h\}\geq 6$. – Julian Rosen Jun 19 '18 at 4:11
• @JulianRosen: In fact Baker conjectured that $C<rad(ABC)^{7/4}$ which would mean that there are no solutions (for coprime $A,B,C$). – GH from MO Jun 19 '18 at 4:12