7. For how many quadruples $(a, b, c, d)$ of non-negative integers are the three expressions $a^{2}-c^{2}, b^{2}-d^{2} \mathrm{~and~}$ $a b+b c+c d+d a$ all equal to 1024?
(A) 0
(B) 1
(C) 4
(D) 9
(E) 11
7. The answer is (B). Let's write the three equations in the form $(a+c)(a-c)=1024,(b+d)(b-d)=1024$ and $(a+c)(b+d)=1024$. Note that $a+c$ and $a-c$ cannot be zero, since $(a+c)(a-c)=1024$. By comparing the first and third equations, we get $b+d=\frac{1024}{a+c}=a-c$, which, when substituted into the second equation, yields $(b-d)(a-c)=1024=(a+c)(a-c)$, from which we have $b-d=a+c$; subtracting (term by term) this equation from $b+d=a-c$ gives $2 d=-2 c$. The numbers $c$ and $d$ are therefore either both zero, or of opposite sign; since by hypothesis $c$ and $d$ are both non-negative, we must then have $c=d=0$. The initial system thus reduces to the equations $a^{2}=b^{2}=a b=1024=32^{2}, c=d=0$, which admit as the only solution in non-negative integers the quadruple $(a, b, c, d)=(32,32,0,0)$.