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n+2為完全平方數設為a^2,同樣n-1設為b^2,於是a^2-b^2=(a+b)*(a-b)=n+2-(n-1)=3,由於a,b均為自然數,所以有a+b=3,a-b=1,解得a=2,b=1,於是n=2,代入可知4n-n^2-3=4*2-2*2-3=1>0,成立
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