## 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. $143 = 11\; \times \: 13$ $433^2 \; + \; 143^3 \; = \; 3111696 \; = \; 42^4$ $28^2 \; + \; 8^3\; = \; 1296\; = \; 6^4$ $648^2 \; + \; 108^3\; = \; 1679616\; = \; 36^4$ $110592^2 \; + \; 4608^3\; = \; 110075314176\; = \; 576^4$

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(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 $7^{15} \; + \; 7^{16} \; = \; 7^{15} \; (1+7) \; = \; 7^{15} \; (2^3)$ $2^4 \; 7^5 \; = \; 268912$ $(2^4 \; 7^5)^3 \; = \; 2^{12} \; 7^{15}$ $2^3 \; 7^4 \; = \; 19208$ $(2^3 \; 7^4)^4 \; = \; 2^{12} \; 7^{16}$ $2^{12} \; 7^{15} \; + \; 2^{12} \; 7^{16}$ $= \; 2^{12} \; (7^{15} + 7^{16})$ $= \; 2^{12} \; 7^{15} \; (2^3)$ $= \; 2^{15} \; 7^{15}$ $= \; (14^3)^5$ $= \; 2744^5$ $268912^3 \; + \; 19208^4 \; = \; 2744^5$ 1. Paul says: