Use a previous problem to establish that there is no solution in positive integers of simultaneous equations. Problem 4 mentioned in the question is given below:
A system of quadratic simultaneous equations admits no solution in positive integers.
And here is the question:



My questions are:
1- How can I prove the hint?
2- How can I use the hint to solve the question?
 A: Your two equations are
$$x^2 + y^2 = z^2 \tag{1}\label{eq1}$$
$$x^2 + 2y^2 = w^2 \tag{2}\label{eq2}$$
\eqref{eq2} - \eqref{eq1} gives
$$y^2 = w^2 - z^2 \; \iff \; z^2 + y^2 = w^2 \tag{3}\label{eq3}$$
Rearranging \eqref{eq1} gives
$$z^2 - y^2 = x^2 \tag{4}\label{eq4}$$
These are the equations in the hint, so this answers your first question. As you can see, \eqref{eq3} and \eqref{eq4} are the same form as the equations in A system of quadratic simultaneous equations admits no solution in positive integers., where the variable names used are just somewhat different. In both cases, you have the sum of two perfect squares and the difference of two perfect squares each being a perfect squares. Going from these equations to those ones, you have $z$ going to $x$, $y$ staying the same, $w$ going to $z$ and $x$ going to $w$. Making those variable name changes gives you those $2$ other equations of
$$x^2 + y^2 = z^2 \tag{5}\label{eq5}$$
$$x^2 - y^2 = w^2 \tag{6}\label{eq6}$$
As shown in that other question, there are no integral solutions to the simultaneous equations \eqref{eq5} and \eqref{eq6}. Thus, there are no solutions either to \eqref{eq1} and \eqref{eq2} (since, if there were, then the sequence of equations & variable changes given here shows there would be solutions in \eqref{eq5} and \eqref{eq6}). This answers your second question regarding how to use the hint to solve the problem.
A: 1) To prove the hint just substitute x²+y² by z² in the second equation. You will get z²+y²=w².
2) Adding the two equations in the hint would give 2*z²=w² which is not possible.
