Yes
No

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Multiplying by 2, we have $2(a^2+b^2+c^2+d^2)-2(ab+bc+cd+da)=0.$ Group the terms to get $(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2cd+d^2)+(d^2-2da+a^2)=0.$ Note that all the terms are squares. Thus the equation can be written as $(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0 \implies a-b=b-c=c-d=d-a=0.$ This implies that $a=b=c=d$ , contradicting the fact that $a,b,c,d$ are

distinct.