a) Prove that any integer n greater that 1, the equation $x^{n}+2^{n} = y^{2}+2$
b) Solve the equation $x! + y!+z!= 2^{v!}$
What I tried on b): Assume $x \geq y \geq z$. If $z \geq 3,$ $3|LHS,$ contradiction. Also $2^{v!} = x! + y! + z! \geq 3,$ so $v>1$.
A a) I think I would have to use LHS too, but how?
HINT.- a) $x^{n}+2^{n} = y^{2}+2$ then $x$ and $y$ have same parity.We have $$2(2^{n-1}-1)=y^2-x^n$$ In both cases of odds and even $RHS$ is divisible by $4$ but $LHS$ is not.
b) let $x\le y\le z$ and $x\gt2$ then $3$ divides $2^v$, absurde. it remains to see the cases $$1+y!+z!=2^v\\2+y!+z!=2^v$$ The firs is not possible because odd=even and the second case is similarly discarded.