Can these two equations both be squares in $\Bbb N^+$?

Question

Are there any examples where $$a^2+2ab+2b^2=p^2,2a^2+2ab+b^2=q^2\qquad{a,b,p,q\in\Bbb N^+}\tag1$$ If not, can $(1)$ be disproven?

Testing by brute force brings up nothing for small integers!

Answer 1

Rewrite the equations: $$(a+b)^2+b^2=p^2$$ $$(a+b)^2+a^2=q^2$$ Two Pythagorean triples. If a + b is odd, one of a and b must be odd and the triangle with two odd legs cannot be Pythagorean. If a and b are even and part of a solution they can be divided by two to yield another solution. Thus, we consider the case when a and b are odd.

Write $a=m^2-n^2$ and $a+b=2mn$, $m>n\geq1$, so the (a, a + b) triangle is Pythagorean: $$b=-m^2+2mn+n^2$$ This must now be a perfect square for the (b, a + b) triangle to be Pythagorean: $$(a + b)^2+b^2$$ $$=m^4-4m^3n+6m^2n^2+4mn^3+n^4$$ However, this polynomial is irreducible over the integers and thus has no square root with the conditions on m and n imposed above. The original equations have no solution.

(We can safely assume that both triples are primitive, since any factor k introduced into the first triple will necessarily propagate into the second triple and can be divided out afterwards. Thanks to Joffan for pointing this out.)