Question

已知a,b,c为正整数，且1<a<b<c，b²=ac，a+b+c=111,求b

Answer 1

a+b+c=111,∴c=111-(a+b),
∴b=√(ac)<(a+c)/2,b^2=ac=a[111-(a+b)],
∴3b<111,b<37,b^2+ab+a^2-111a=0,①
△=a^2-4(a^2-111a)=444a-3a^2=a(444-3a)是平方数,
b=(-a+√△)/2<=36,
∴√△<=72+a,
平方得444a-3a^2<=5184+144a+a^2,
4a^2-300a+5184>=0,
a^2-75a+1296>=0,
∴a>=48(舍),或a<=27.
a=2n时△=2n(444-6n)=12n(74-n),n<=13,3|n(74-n),∴n≠1,4,7,10,13.
n...2...3...5...6...8..9..11..12
△..x...x...x.x.x...x...x.x
其中x表示非平方数.
a=2n-1时△=(2n-1)(447-6n)=-12n^2+900n-447=3[5476-(2n-75)^2],是平方数,
3|5476-（2n-75)^2,∴3|（2n-75)^2-1=(2n-74)(2n-76)=4(n-37)(n-38),
∴n≠3,6,9,12,
n.2...4..5..7..8..10..11..13...14
△..x.x...x...x..x.x.x.x.99^2
仅当n=14,a=27时△=99^2,是平方数,这时b=36（由①),c=48.