## Equations: A^2 + B^3 = C^4 and D^3 + E^4 = F^5

Find positive integer solutions to

(1)   $A_1^2 \; \; + \; \; B_1^3 \; = \; C_1^4$
(2)   $A_2^3 \; \; + \; \; B_2^4 \; = \; C_2^5$

For example,

(1)

$3^3 \; = \; 27$                              $(3^3)^2 \; = \; 3^6$
$2 \; 3^2 \; = \; 18$                          $(2 3^2)^3 \; = \; 2^3 \; 3^6$
$3^6 \; + \; 2^3 \; 3^6$
$= 3^6 \; (1 + 2^3)$
$= 3^6 \; 3^2$
$= 3^8$
$= 9^4$

$27^2 \; + \; 18^3 \; = \; 6561 \; = \; 9^4$

$7 \; 3^2 \; = \; 63$                            $(7 \; 3^2)^2 \; = \; 7^2 \; 3^4$
$2^2 \; 3^2 \; = \; 36$                          $(2^2 3^2)^3 \; = \; 2^6 \; 3^6 \; = \; 2^6 \; 3^2 \; 3^4$
$7^2 \; 3^4 \; + \; 2^6 \; 3^2 \; 3^4$
$= 3^4 \; (7^2 + 2^6 3^2)$
$= 3^4 \; 5^4$
$= 15^4$

$63^2 \; + \; 36^3 \; = \; 50625 \; = \; 15^4$

$433$   is a prime number.
$143 = 11\; \times \: 13$

$433^2 \; + \; 143^3 \; = \; 3111696 \; = \; 42^4$

$28^2 \; + \; 8^3\; = \; 1296\; = \; 6^4$

$648^2 \; + \; 108^3\; = \; 1679616\; = \; 36^4$

$110592^2 \; + \; 4608^3\; = \; 110075314176\; = \; 576^4$

——————————————

(2)

$2^8 \; = \; 256$              $(2^8)^3 \; = \; 2^{24}$
$2^6 \; = \; 64$                 $(2^6)^4 \; = \; 2^{24}$
$2^5 \; = \; 32$

$2^{24}\; +\; 2^{24} \; = \; 2^{25}$

$256^3\; +\; 64^4 \; = \; 32^5$

Note that     $7^{15} \; + \; 7^{16} \; = \; 7^{15} \; (1+7) \; = \; 7^{15} \; (2^3)$

$2^4 \; 7^5 \; = \; 268912$              $(2^4 \; 7^5)^3 \; = \; 2^{12} \; 7^{15}$
$2^3 \; 7^4 \; = \; 19208$                 $(2^3 \; 7^4)^4 \; = \; 2^{12} \; 7^{16}$

$2^{12} \; 7^{15} \; + \; 2^{12} \; 7^{16}$
$= \; 2^{12} \; (7^{15} + 7^{16})$
$= \; 2^{12} \; 7^{15} \; (2^3)$
$= \; 2^{15} \; 7^{15}$
$= \; (14^3)^5$
$= \; 2744^5$

$268912^3 \; + \; 19208^4 \; = \; 2744^5$