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n^4+2n^3+5n^2+12n+5为完全平方,n为正整数,求n

题目详情
n^4+2n^3+5n^2+12n+5为完全平方,n为正整数,求n
▼优质解答
答案和解析
n^4+2n^3+5n^2+12n+5-(n^2+n+1)^2=10n+4≥14>0
(n^2+n+4)^2-(n^4+2n^3+5n^2+12n+5)=9n^2-9n+11≥11>0
所以n^4+2n^3+5n^2+12n+5只能等于(n^2+n+2)^2或(n^2+n+3)^2
若n^4+2n^3+5n^2+12n+5=(n^2+n+2)^2 则5n²-13n-1=0 此方程无整数解
若n^4+2n^3+5n^2+12n+5=(n^2+n+3)^2 则7n²-11n+4=0
(7n-4)(n-1)=0 此方程整数解n=1
n=1
验算 n=1时 n^4+2n^3+5n^2+12n+5=25=5²