Prove (or provide a counterexample): no pair of primitive Pythagorean triples (a,b,c) and (2a,k,c) exists. A primitive Pythagorean triple is an ordered set of coprime integers (a,b,c) such that $a^2+b^2=c^2$. Show that the system of Diophantine equations
$$a^2+b^2=c^2$$
$$4a^2+k^2=c^2$$
have no solutions.

$$c^2=a^2+b^2=4a^2+k^2$$
$$k^2+3a^2=b^2$$
All solutions of this Diophantine equation are defined parameterization.
$$a=2ps$$
$$k=p^2-3s^2$$
$$b=p^2+3s^2$$
Then the amount of;
$$a^2+b^2=4p^2s^2+p^4+6p^2s^2+9s^4=p^4+10p^2s^2+9s^4$$
A square can not be.
