A problem by Chan Ming Xian

$(ab+cd)^{2}+(ad-bc)^{2}$ = $a^{2}b^{2}+2abcd+c^{2}d^{2}+a^{2}d^{2}-2abcd+b^{2}c^{2}$ = $a^{2}(b^{2}+d^{2})+c^{2}(b^{2}+d^{2})$ = $(a^{2}+c^{2})(b^{2}+d^{2})$ We know that the prime factorization of 2009 is 2009 = $7^{2} \times 41$ So , $(a^{2}+c^{2})(b^{2}+d^{2})=7^{2} \times 41$ Say $(a^{2}+c^{2})\leq(b^{2}+d^{2})$ If $(a^{2}+c^{2})=1, (b^{2}+d^{2})=2009$ If $(a^{2}+c^{2})=7 , (b^{2}+d^{2})=287$ If $(a^{2}+c^{2})=41, (b^{2}+d^{2})=49$ . Therefore, $a^{2}+b^{2}+c^{2}+d^{2}=2010 ,294, 90$ . $2010+294+90=2394$ The last three digits of 2394 , which is the answer is $\boxed{394}$