a) The LHS is either $0$ or $2$ modulo $3$ and the RHS is either $0$ or $1$ modulo $3$. Thus, both sides are divisible by $3$ and therefore $x,z$ are divisible by $3$. Then, the RHS is divisible by $9$, so we must have $3\mid y$ as well. Now, consider the equation modulo $2$. Clearly, $y$ must be even. Now, taking mod $4$, you can see that $2\mid x$. Finally, mod $8$ gives $2\mid z$. Thus, $6$ divides $x,y,z$. $\square$ b) Given any solution $(x,y,z)\equiv (a,b,c)$, we can then find infinite solutions of the form $(ap^6,bp^4,cp^3)$ where $p>1$ is an integer. To find a solution, we can substitute $(x,y,z)\equiv (6a,6b,6c)$ which leads us to the solution $(144,24,12)$. $\blacksquare$

b) If $(a,b,c)$ satisfys the equation, than so does $(64a,16b,8c).$ Because $2a^2+3b^3=4c^4 \implies 2^{12}\cdot(2a^2+3b^3+4c^4)=2^{12}\cdot 4c^4 \implies 2(2^6\cdot a)^2+3(2^4 \cdot b^3)=4(2^3 \cdot c)^4.$

Base solution for b): $(144,24,12).$

(a) Notice $2\mid y.$ Letting $y=2y_1,$ we have $2x^2+3\cdot 2^3y_1^3=4z^4.$ Dividing by $2$ yields $2\mid x.$ Letting $x=2x_1,$ we have $2^2x_1^2+3\cdot 2^2y_1^3=2z^4$ and so $2\mid z.$ Considering modulo $3,$ we have $2x^2\equiv z^4\pmod{3}.$ Since $0,1$ are the only QRs modulo $3,$ we deduce $x\equiv z\equiv 0\pmod{3}.$ Letting $x=3x_2$ and $z=3y_2$ yields $2\cdot 3^2x_2^2+3y^3=4\cdot 3^4z_2^2$ and dividing by $3,$ we have $3\mid y.$ (b) Notice $(x,y,z)=(144k^6,24k^4,12k^3)$ yields $$2x^2+3y^3=2\cdot 12^4\cdot k^{12}+3\cdot 2^3\cdot 12^3\cdot k^{12}=12^4\cdot k^{12}(2+2)=4z^4$$for all $k>0.$ $\square$

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