Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$?

Question

Question:

Are there $a,b,c,d \in \mathbb N$ such that $$a^2 + b^2 = c^2 \ \ \text{and} \ \ b^2 + c^2 = d^2$$

I'm a bit lost here.

Answer 1

Note that $c^2-b^2=a^2$, $c^2+b^2=d^2$ $\Rightarrow$ $c^4-b^4=(ad)^2$. But here (Wiki) and here (MSE) one can read that equation $$x^4-y^4=z^2$$ has no (pairwise coprime) solutions $x,y,z\in \mathbb{N}$.

Answer 2

Suppose there exist such $a,b,c,d$ with $a\le b\le c\le d,$ then $$c^2-a^2=d^2-c^2=b^2$$ $$2c^2=a^2+d^2$$ Suppose $$a=x-y,\,\,\,\,d=x+y.$$ Then $$c^2=x^2+y^2.$$ Therefore there exist $A,B$ such that $$c=A^2+B^2,\,\,\,x=A^2-B^2,\,\,\,y=2AB.$$ Now we have $$a=A^2-2AB-B^2\le(A-B)^2$$ $$d=A^2+2AB-B^2\le(A+B)^2$$ Also $$c^2-a^2=(A^2+B^2)^2-(A^2-2AB-B^2)^2=4AB(A-B)(A+B)$$ should be a PERFECT SQUARE.
That is all $A,B,A+B,A-B$ are should be perfect squares. Therefore $$A=a_1^2,\,\,\,B=b_1^2$$

This happens only if $B$ is a perfect square and $A=0.$ Hence the only solution is $$a=c=d,\,\,\,\,\,b=0.$$