It can be shown that an order triple, (a, b, c) is a "Pythagorean triple", a^2+ b^2= c^2, if an only if they can be written in the form a= n^2- m^2, b= 2nm, and c= n^2+ m^2 for some numbers m and n. And, of course, if we want a, b, c integers, m and n must be integers also.

In order that (a, b, c) be a Pythagorean triple, they must be of the form a= n^2- m^2, b= 2nm, and c= n^2+ m^2 for some integers m and n. In order that (b, c, d) also be a Pythagorean triple, they must be of the form b= p^2- q^2, c= 2pq, and d= p^2+ q^2 for some integers p and q. That means we must have b= 2nm= p^2- q^2 and c= n^2+ m^2= 2pq. Adding those equations, m^2+ 2mn+ n^2= (m+n)^2= p^2+ 2pq- q^2 so that p^2+2pq- q^2 is a perfect square. p^2+ 2pq- q^2= p^2+ 2pq+ q^2- 2q^2= (p+q)^2- 2q^2 so p and q must be such that (p+q)^2- 2q^2 is a perfect square.

So the whole question reduces to "can x^2- 2y^2 be a perfect square for any integers x and y?"