Prove (or provide a counterexample): no pair of primitive Pythagorean triples (a,b,c) and (2a,k,c) exists.

Question

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.

Answer 1

$$c^2=a^2+b^2=4a^2+k^2$$

$$k^2+3a^2=b^2$$

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.