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#ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#試證明四個連續奇數的乘積加上16必為一個整數的平方 #ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#一個差不多的解法:#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#(2n-3)(2n-1)(2n+1)(2n+3)+16#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#=(4n^2-9)(4n^2-1)+16#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#令y=4n^2-1#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#y^2-8y+16#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#=(y-4)^2#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl#

=(2n+1)(2n+3)(2n+5)(2n+7)+16
=16n^4+128n^3+334n^2+352n+121
=(4n^2+16n+11)^2