Solve for integers $x$, $y$ $$x^3= y^2-22$$ using only elementary techniques!
I managed to show that there are no solutions using $\sqrt{22}$, but I'd like to explain this to (trained) high school students.
A small note: subtracting $27$ seems to help, but not really.
EDIT: my reasoning in the quadratic field was wrong. Indeed, there is a solution $(3, 7) $ that one can deduce from the "subtract $27$" trick.
You may to substitute $27$ instead of subtracting it. $$x^3= y^2-22\implies y=\pm\sqrt{x^3+22}$$
By inspection, we can see that $\quad\sqrt{3^3+22}=\sqrt{27+22}=\sqrt{49}=\pm7$
suggesting one solution: $\quad (x,y)=(3,\pm7)$
For the $22$-surd alone, if $x=0$, then $y=\sqrt{22}$ which is not an integer. WolframAlpha shows that the only integer solution is $(3,\pm7)$.