<s>[INST] Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$? 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.
 [/INST] 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 $$a=c=d,\,\,\,\,\,b=0.$$
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