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10 ^2+11^2+12^2+13^2+14^2=365*2 設中間數為x,則 (x-2)^2+(x-1)^2+x^2+(x+1)^2+(x+2)^2 =5x^2+10 =5(x^2+2) 令其值為365k,k是整數 則5(x^2+2)=365k=5*73k 只要x^2+2=73k即可 顯然當x=12時,k=2滿足
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